Prove that the limit of the sequence $a_n:=frac{6n+1}{2n+5}$ is 3, using the Epsilon definition.












3














I want to show that for sequence $(a_n)$, defined as $a_n=frac{6n+5}{2n+5}$, the following statement is true:



$$lim_{n rightarrow infty} {frac{6n+1}{2n+5}}=3$$



I started with stating that there exists an arbritary but fixed $ε>0$ and assumed that for an $n$ and $N_ε$, $n > N_ε$. I then used the Epsilon definition to prove that the limit is 3.



$$|a_n-3|<ε=|frac{6n+1}{2n+5}-3|<ε iff (frac{6n+1}{2n+5}-3)=frac{-7}{n}<ε iff frac{-7}{ε}=n$$



Here I set $frac{-7}{ε}=:N_ε$ and since we know that $n>N_ε$ that means that $$frac{1}{n}<frac{1}{ε}=frac{1}{frac{-7}{ε}}=frac{ε}{-7} ≠ ε$$



I don't understand where I went wrong in this proof, I've tried the same thing with other examples and get stuck in the same place.










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    3














    I want to show that for sequence $(a_n)$, defined as $a_n=frac{6n+5}{2n+5}$, the following statement is true:



    $$lim_{n rightarrow infty} {frac{6n+1}{2n+5}}=3$$



    I started with stating that there exists an arbritary but fixed $ε>0$ and assumed that for an $n$ and $N_ε$, $n > N_ε$. I then used the Epsilon definition to prove that the limit is 3.



    $$|a_n-3|<ε=|frac{6n+1}{2n+5}-3|<ε iff (frac{6n+1}{2n+5}-3)=frac{-7}{n}<ε iff frac{-7}{ε}=n$$



    Here I set $frac{-7}{ε}=:N_ε$ and since we know that $n>N_ε$ that means that $$frac{1}{n}<frac{1}{ε}=frac{1}{frac{-7}{ε}}=frac{ε}{-7} ≠ ε$$



    I don't understand where I went wrong in this proof, I've tried the same thing with other examples and get stuck in the same place.










    share|cite|improve this question



























      3












      3








      3







      I want to show that for sequence $(a_n)$, defined as $a_n=frac{6n+5}{2n+5}$, the following statement is true:



      $$lim_{n rightarrow infty} {frac{6n+1}{2n+5}}=3$$



      I started with stating that there exists an arbritary but fixed $ε>0$ and assumed that for an $n$ and $N_ε$, $n > N_ε$. I then used the Epsilon definition to prove that the limit is 3.



      $$|a_n-3|<ε=|frac{6n+1}{2n+5}-3|<ε iff (frac{6n+1}{2n+5}-3)=frac{-7}{n}<ε iff frac{-7}{ε}=n$$



      Here I set $frac{-7}{ε}=:N_ε$ and since we know that $n>N_ε$ that means that $$frac{1}{n}<frac{1}{ε}=frac{1}{frac{-7}{ε}}=frac{ε}{-7} ≠ ε$$



      I don't understand where I went wrong in this proof, I've tried the same thing with other examples and get stuck in the same place.










      share|cite|improve this question















      I want to show that for sequence $(a_n)$, defined as $a_n=frac{6n+5}{2n+5}$, the following statement is true:



      $$lim_{n rightarrow infty} {frac{6n+1}{2n+5}}=3$$



      I started with stating that there exists an arbritary but fixed $ε>0$ and assumed that for an $n$ and $N_ε$, $n > N_ε$. I then used the Epsilon definition to prove that the limit is 3.



      $$|a_n-3|<ε=|frac{6n+1}{2n+5}-3|<ε iff (frac{6n+1}{2n+5}-3)=frac{-7}{n}<ε iff frac{-7}{ε}=n$$



      Here I set $frac{-7}{ε}=:N_ε$ and since we know that $n>N_ε$ that means that $$frac{1}{n}<frac{1}{ε}=frac{1}{frac{-7}{ε}}=frac{ε}{-7} ≠ ε$$



      I don't understand where I went wrong in this proof, I've tried the same thing with other examples and get stuck in the same place.







      real-analysis sequences-and-series limits proof-writing epsilon-delta






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      share|cite|improve this question













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      edited Nov 22 '18 at 13:07









      José Carlos Santos

      152k22123225




      152k22123225










      asked Nov 22 '18 at 12:28









      xxxtentacion

      372112




      372112






















          3 Answers
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          active

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          3














          You made one algebra error, and a number of other errors in the way your proof is organized and presented.



          The algebra error, as José Carlos Santos notes, is that ${6n+1over2n+5}-3={-14over2n+5}$, not $-7over n$. So that of course affects what comes after. However, let's suppose $a_n-3={-7over n}$ is correct, and go from there.



          The first, minor, problem is either a typo or a misuse of the equal sign: where you write



          $$|a_n-3|ltepsilon=|{6n+1over2n+5}-3|ltepsilon$$



          you really mean is $iff$ instead of $=$, i.e.,



          $$|a_n-3|ltepsiloniff|{6n+1over2n+5}-3|ltepsilon$$



          Also, at the other end of the display, where you write ${-7overepsilon}=n$, it should be ${-7overepsilon}lt n$.



          The second, more serious, problem is that $|a_n-3|ltepsilon$ is not equivalent to $(a_n-3)ltepsilon$; the absolute value sign is important. So the first set of displayed expressions should end in



          $$|a_n-3|=left|-7over nright|ltepsiloniff {7overepsilon}lt n$$



          and thus we should set $N_epsilon={7overepsilon}$.



          The next problem is merely, I think, a typo: where you say that $ngt N_epsilon$ means that ${1over n}lt{1overepsilon}$, you pretty clearly meant to write ${1over n}lt{1over N_epsilon}$, since you go on to substitute the expression you derived for $N_epsilon$ in the continuation.



          The final, most significant, problem is one of logical relevance. You note, correctly, that $epsilonover-7$ (which, as noted above, shouldn't have the minus sign) is not equal to $epsilon$. But that doesn't matter. What you really need here is a set of implications along the lines of



          $$ngt N_epsilonimplies{1over n}lt{1over N_epsilon}={epsilonover7}implies|a_n-3|={7over n}lt7cdot{epsilonover7}=epsilon$$



          Of these, I'd say, the final problem is the most important: It looks like you were trying to derive the inequality ${1over n}ltepsilon$ instead of $|a_n-3|ltepsilon$. In other words, it looks like you lost track of what it is you are supposed to prove. It might help, for future proofs of this type, to keep in mind that, come hell or high water, you need to wrap things up with a logical sentence of the form $ngt N_epsilonimplies|a_n-L|ltepsilon$.






          share|cite|improve this answer































            3














            For some reason, your absolute value vanished. It turns ou that$$leftlvertfrac{6n+1}{2n+5}-3rightrvert=frac{14}{5+2n}.$$And$$frac{14}{5+2n}<varepsiloniff n>frac{14-5varepsilon}varepsilon.$$






            share|cite|improve this answer





























              2














              Be careful:




              When we're given an arbitrary $varepsilon>0$ we have to find $N_varepsilon$ such that, for $n>N_varepsilon$,
              $$
              left|frac{6n+1}{2n+5}-3right|<varepsilon
              $$

              We don't need to do “$iff$” as you seem to want. Now consider that
              $$
              left|frac{6n+1}{2n+5}-3right|=
              left|frac{6n+1-3(2n+5)}{2n+5}right|=
              left|frac{-14}{2n+5}right|=
              frac{14}{2n+5}<frac{14}{2n}=frac{7}{n}
              $$

              We know have $7/n<varepsilon$ provided $n>7/varepsilon$, so we can take $N_varepsilon$ the largest integer less than or equal to $7/varepsilon$. For $n>N_varepsilon$ we thus have
              $$
              varepsilon>frac{7}{n}geleft|frac{6n+1}{2n+5}-3right|
              $$

              which ends our proof.




              Be careful when you remove the absolute value sign and also how you look for upper bounds.






              share|cite|improve this answer





















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                3 Answers
                3






                active

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                3 Answers
                3






                active

                oldest

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                active

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                active

                oldest

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                3














                You made one algebra error, and a number of other errors in the way your proof is organized and presented.



                The algebra error, as José Carlos Santos notes, is that ${6n+1over2n+5}-3={-14over2n+5}$, not $-7over n$. So that of course affects what comes after. However, let's suppose $a_n-3={-7over n}$ is correct, and go from there.



                The first, minor, problem is either a typo or a misuse of the equal sign: where you write



                $$|a_n-3|ltepsilon=|{6n+1over2n+5}-3|ltepsilon$$



                you really mean is $iff$ instead of $=$, i.e.,



                $$|a_n-3|ltepsiloniff|{6n+1over2n+5}-3|ltepsilon$$



                Also, at the other end of the display, where you write ${-7overepsilon}=n$, it should be ${-7overepsilon}lt n$.



                The second, more serious, problem is that $|a_n-3|ltepsilon$ is not equivalent to $(a_n-3)ltepsilon$; the absolute value sign is important. So the first set of displayed expressions should end in



                $$|a_n-3|=left|-7over nright|ltepsiloniff {7overepsilon}lt n$$



                and thus we should set $N_epsilon={7overepsilon}$.



                The next problem is merely, I think, a typo: where you say that $ngt N_epsilon$ means that ${1over n}lt{1overepsilon}$, you pretty clearly meant to write ${1over n}lt{1over N_epsilon}$, since you go on to substitute the expression you derived for $N_epsilon$ in the continuation.



                The final, most significant, problem is one of logical relevance. You note, correctly, that $epsilonover-7$ (which, as noted above, shouldn't have the minus sign) is not equal to $epsilon$. But that doesn't matter. What you really need here is a set of implications along the lines of



                $$ngt N_epsilonimplies{1over n}lt{1over N_epsilon}={epsilonover7}implies|a_n-3|={7over n}lt7cdot{epsilonover7}=epsilon$$



                Of these, I'd say, the final problem is the most important: It looks like you were trying to derive the inequality ${1over n}ltepsilon$ instead of $|a_n-3|ltepsilon$. In other words, it looks like you lost track of what it is you are supposed to prove. It might help, for future proofs of this type, to keep in mind that, come hell or high water, you need to wrap things up with a logical sentence of the form $ngt N_epsilonimplies|a_n-L|ltepsilon$.






                share|cite|improve this answer




























                  3














                  You made one algebra error, and a number of other errors in the way your proof is organized and presented.



                  The algebra error, as José Carlos Santos notes, is that ${6n+1over2n+5}-3={-14over2n+5}$, not $-7over n$. So that of course affects what comes after. However, let's suppose $a_n-3={-7over n}$ is correct, and go from there.



                  The first, minor, problem is either a typo or a misuse of the equal sign: where you write



                  $$|a_n-3|ltepsilon=|{6n+1over2n+5}-3|ltepsilon$$



                  you really mean is $iff$ instead of $=$, i.e.,



                  $$|a_n-3|ltepsiloniff|{6n+1over2n+5}-3|ltepsilon$$



                  Also, at the other end of the display, where you write ${-7overepsilon}=n$, it should be ${-7overepsilon}lt n$.



                  The second, more serious, problem is that $|a_n-3|ltepsilon$ is not equivalent to $(a_n-3)ltepsilon$; the absolute value sign is important. So the first set of displayed expressions should end in



                  $$|a_n-3|=left|-7over nright|ltepsiloniff {7overepsilon}lt n$$



                  and thus we should set $N_epsilon={7overepsilon}$.



                  The next problem is merely, I think, a typo: where you say that $ngt N_epsilon$ means that ${1over n}lt{1overepsilon}$, you pretty clearly meant to write ${1over n}lt{1over N_epsilon}$, since you go on to substitute the expression you derived for $N_epsilon$ in the continuation.



                  The final, most significant, problem is one of logical relevance. You note, correctly, that $epsilonover-7$ (which, as noted above, shouldn't have the minus sign) is not equal to $epsilon$. But that doesn't matter. What you really need here is a set of implications along the lines of



                  $$ngt N_epsilonimplies{1over n}lt{1over N_epsilon}={epsilonover7}implies|a_n-3|={7over n}lt7cdot{epsilonover7}=epsilon$$



                  Of these, I'd say, the final problem is the most important: It looks like you were trying to derive the inequality ${1over n}ltepsilon$ instead of $|a_n-3|ltepsilon$. In other words, it looks like you lost track of what it is you are supposed to prove. It might help, for future proofs of this type, to keep in mind that, come hell or high water, you need to wrap things up with a logical sentence of the form $ngt N_epsilonimplies|a_n-L|ltepsilon$.






                  share|cite|improve this answer


























                    3












                    3








                    3






                    You made one algebra error, and a number of other errors in the way your proof is organized and presented.



                    The algebra error, as José Carlos Santos notes, is that ${6n+1over2n+5}-3={-14over2n+5}$, not $-7over n$. So that of course affects what comes after. However, let's suppose $a_n-3={-7over n}$ is correct, and go from there.



                    The first, minor, problem is either a typo or a misuse of the equal sign: where you write



                    $$|a_n-3|ltepsilon=|{6n+1over2n+5}-3|ltepsilon$$



                    you really mean is $iff$ instead of $=$, i.e.,



                    $$|a_n-3|ltepsiloniff|{6n+1over2n+5}-3|ltepsilon$$



                    Also, at the other end of the display, where you write ${-7overepsilon}=n$, it should be ${-7overepsilon}lt n$.



                    The second, more serious, problem is that $|a_n-3|ltepsilon$ is not equivalent to $(a_n-3)ltepsilon$; the absolute value sign is important. So the first set of displayed expressions should end in



                    $$|a_n-3|=left|-7over nright|ltepsiloniff {7overepsilon}lt n$$



                    and thus we should set $N_epsilon={7overepsilon}$.



                    The next problem is merely, I think, a typo: where you say that $ngt N_epsilon$ means that ${1over n}lt{1overepsilon}$, you pretty clearly meant to write ${1over n}lt{1over N_epsilon}$, since you go on to substitute the expression you derived for $N_epsilon$ in the continuation.



                    The final, most significant, problem is one of logical relevance. You note, correctly, that $epsilonover-7$ (which, as noted above, shouldn't have the minus sign) is not equal to $epsilon$. But that doesn't matter. What you really need here is a set of implications along the lines of



                    $$ngt N_epsilonimplies{1over n}lt{1over N_epsilon}={epsilonover7}implies|a_n-3|={7over n}lt7cdot{epsilonover7}=epsilon$$



                    Of these, I'd say, the final problem is the most important: It looks like you were trying to derive the inequality ${1over n}ltepsilon$ instead of $|a_n-3|ltepsilon$. In other words, it looks like you lost track of what it is you are supposed to prove. It might help, for future proofs of this type, to keep in mind that, come hell or high water, you need to wrap things up with a logical sentence of the form $ngt N_epsilonimplies|a_n-L|ltepsilon$.






                    share|cite|improve this answer














                    You made one algebra error, and a number of other errors in the way your proof is organized and presented.



                    The algebra error, as José Carlos Santos notes, is that ${6n+1over2n+5}-3={-14over2n+5}$, not $-7over n$. So that of course affects what comes after. However, let's suppose $a_n-3={-7over n}$ is correct, and go from there.



                    The first, minor, problem is either a typo or a misuse of the equal sign: where you write



                    $$|a_n-3|ltepsilon=|{6n+1over2n+5}-3|ltepsilon$$



                    you really mean is $iff$ instead of $=$, i.e.,



                    $$|a_n-3|ltepsiloniff|{6n+1over2n+5}-3|ltepsilon$$



                    Also, at the other end of the display, where you write ${-7overepsilon}=n$, it should be ${-7overepsilon}lt n$.



                    The second, more serious, problem is that $|a_n-3|ltepsilon$ is not equivalent to $(a_n-3)ltepsilon$; the absolute value sign is important. So the first set of displayed expressions should end in



                    $$|a_n-3|=left|-7over nright|ltepsiloniff {7overepsilon}lt n$$



                    and thus we should set $N_epsilon={7overepsilon}$.



                    The next problem is merely, I think, a typo: where you say that $ngt N_epsilon$ means that ${1over n}lt{1overepsilon}$, you pretty clearly meant to write ${1over n}lt{1over N_epsilon}$, since you go on to substitute the expression you derived for $N_epsilon$ in the continuation.



                    The final, most significant, problem is one of logical relevance. You note, correctly, that $epsilonover-7$ (which, as noted above, shouldn't have the minus sign) is not equal to $epsilon$. But that doesn't matter. What you really need here is a set of implications along the lines of



                    $$ngt N_epsilonimplies{1over n}lt{1over N_epsilon}={epsilonover7}implies|a_n-3|={7over n}lt7cdot{epsilonover7}=epsilon$$



                    Of these, I'd say, the final problem is the most important: It looks like you were trying to derive the inequality ${1over n}ltepsilon$ instead of $|a_n-3|ltepsilon$. In other words, it looks like you lost track of what it is you are supposed to prove. It might help, for future proofs of this type, to keep in mind that, come hell or high water, you need to wrap things up with a logical sentence of the form $ngt N_epsilonimplies|a_n-L|ltepsilon$.







                    share|cite|improve this answer














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                    share|cite|improve this answer








                    edited Nov 22 '18 at 15:30

























                    answered Nov 22 '18 at 13:46









                    Barry Cipra

                    59.3k653125




                    59.3k653125























                        3














                        For some reason, your absolute value vanished. It turns ou that$$leftlvertfrac{6n+1}{2n+5}-3rightrvert=frac{14}{5+2n}.$$And$$frac{14}{5+2n}<varepsiloniff n>frac{14-5varepsilon}varepsilon.$$






                        share|cite|improve this answer


























                          3














                          For some reason, your absolute value vanished. It turns ou that$$leftlvertfrac{6n+1}{2n+5}-3rightrvert=frac{14}{5+2n}.$$And$$frac{14}{5+2n}<varepsiloniff n>frac{14-5varepsilon}varepsilon.$$






                          share|cite|improve this answer
























                            3












                            3








                            3






                            For some reason, your absolute value vanished. It turns ou that$$leftlvertfrac{6n+1}{2n+5}-3rightrvert=frac{14}{5+2n}.$$And$$frac{14}{5+2n}<varepsiloniff n>frac{14-5varepsilon}varepsilon.$$






                            share|cite|improve this answer












                            For some reason, your absolute value vanished. It turns ou that$$leftlvertfrac{6n+1}{2n+5}-3rightrvert=frac{14}{5+2n}.$$And$$frac{14}{5+2n}<varepsiloniff n>frac{14-5varepsilon}varepsilon.$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Nov 22 '18 at 12:35









                            José Carlos Santos

                            152k22123225




                            152k22123225























                                2














                                Be careful:




                                When we're given an arbitrary $varepsilon>0$ we have to find $N_varepsilon$ such that, for $n>N_varepsilon$,
                                $$
                                left|frac{6n+1}{2n+5}-3right|<varepsilon
                                $$

                                We don't need to do “$iff$” as you seem to want. Now consider that
                                $$
                                left|frac{6n+1}{2n+5}-3right|=
                                left|frac{6n+1-3(2n+5)}{2n+5}right|=
                                left|frac{-14}{2n+5}right|=
                                frac{14}{2n+5}<frac{14}{2n}=frac{7}{n}
                                $$

                                We know have $7/n<varepsilon$ provided $n>7/varepsilon$, so we can take $N_varepsilon$ the largest integer less than or equal to $7/varepsilon$. For $n>N_varepsilon$ we thus have
                                $$
                                varepsilon>frac{7}{n}geleft|frac{6n+1}{2n+5}-3right|
                                $$

                                which ends our proof.




                                Be careful when you remove the absolute value sign and also how you look for upper bounds.






                                share|cite|improve this answer


























                                  2














                                  Be careful:




                                  When we're given an arbitrary $varepsilon>0$ we have to find $N_varepsilon$ such that, for $n>N_varepsilon$,
                                  $$
                                  left|frac{6n+1}{2n+5}-3right|<varepsilon
                                  $$

                                  We don't need to do “$iff$” as you seem to want. Now consider that
                                  $$
                                  left|frac{6n+1}{2n+5}-3right|=
                                  left|frac{6n+1-3(2n+5)}{2n+5}right|=
                                  left|frac{-14}{2n+5}right|=
                                  frac{14}{2n+5}<frac{14}{2n}=frac{7}{n}
                                  $$

                                  We know have $7/n<varepsilon$ provided $n>7/varepsilon$, so we can take $N_varepsilon$ the largest integer less than or equal to $7/varepsilon$. For $n>N_varepsilon$ we thus have
                                  $$
                                  varepsilon>frac{7}{n}geleft|frac{6n+1}{2n+5}-3right|
                                  $$

                                  which ends our proof.




                                  Be careful when you remove the absolute value sign and also how you look for upper bounds.






                                  share|cite|improve this answer
























                                    2












                                    2








                                    2






                                    Be careful:




                                    When we're given an arbitrary $varepsilon>0$ we have to find $N_varepsilon$ such that, for $n>N_varepsilon$,
                                    $$
                                    left|frac{6n+1}{2n+5}-3right|<varepsilon
                                    $$

                                    We don't need to do “$iff$” as you seem to want. Now consider that
                                    $$
                                    left|frac{6n+1}{2n+5}-3right|=
                                    left|frac{6n+1-3(2n+5)}{2n+5}right|=
                                    left|frac{-14}{2n+5}right|=
                                    frac{14}{2n+5}<frac{14}{2n}=frac{7}{n}
                                    $$

                                    We know have $7/n<varepsilon$ provided $n>7/varepsilon$, so we can take $N_varepsilon$ the largest integer less than or equal to $7/varepsilon$. For $n>N_varepsilon$ we thus have
                                    $$
                                    varepsilon>frac{7}{n}geleft|frac{6n+1}{2n+5}-3right|
                                    $$

                                    which ends our proof.




                                    Be careful when you remove the absolute value sign and also how you look for upper bounds.






                                    share|cite|improve this answer












                                    Be careful:




                                    When we're given an arbitrary $varepsilon>0$ we have to find $N_varepsilon$ such that, for $n>N_varepsilon$,
                                    $$
                                    left|frac{6n+1}{2n+5}-3right|<varepsilon
                                    $$

                                    We don't need to do “$iff$” as you seem to want. Now consider that
                                    $$
                                    left|frac{6n+1}{2n+5}-3right|=
                                    left|frac{6n+1-3(2n+5)}{2n+5}right|=
                                    left|frac{-14}{2n+5}right|=
                                    frac{14}{2n+5}<frac{14}{2n}=frac{7}{n}
                                    $$

                                    We know have $7/n<varepsilon$ provided $n>7/varepsilon$, so we can take $N_varepsilon$ the largest integer less than or equal to $7/varepsilon$. For $n>N_varepsilon$ we thus have
                                    $$
                                    varepsilon>frac{7}{n}geleft|frac{6n+1}{2n+5}-3right|
                                    $$

                                    which ends our proof.




                                    Be careful when you remove the absolute value sign and also how you look for upper bounds.







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Nov 22 '18 at 14:43









                                    egreg

                                    179k1484201




                                    179k1484201






























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