Finding if $sum_{n=1}^inftyfrac{n^n+1/n}{(n+1/n)^n}$ is convergent or divergent .












1












$begingroup$


Is this series convergent or divergent ?
(The sum will be infinite or it will converge to a certain number?)



$$sum_{n=1}^inftyfrac{n^n+1/n}{(n+1/n)^n}$$






calculus sequences-and-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    Welcome to MathStackExchange! Please take a look at the tour page and consider editing your question to comply with the homework questions guidelines. In particular, it is good practice to give some context on what course you are taking and show your work in attempting to solve this question.
    $endgroup$
    – Mefitico
    Nov 28 '18 at 14:43
















1












$begingroup$


Is this series convergent or divergent ?
(The sum will be infinite or it will converge to a certain number?)



$$sum_{n=1}^inftyfrac{n^n+1/n}{(n+1/n)^n}$$






calculus sequences-and-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    Welcome to MathStackExchange! Please take a look at the tour page and consider editing your question to comply with the homework questions guidelines. In particular, it is good practice to give some context on what course you are taking and show your work in attempting to solve this question.
    $endgroup$
    – Mefitico
    Nov 28 '18 at 14:43














1












1








1





$begingroup$


Is this series convergent or divergent ?
(The sum will be infinite or it will converge to a certain number?)



$$sum_{n=1}^inftyfrac{n^n+1/n}{(n+1/n)^n}$$






calculus sequences-and-series







share|cite|improve this question











$endgroup$




Is this series convergent or divergent ?
(The sum will be infinite or it will converge to a certain number?)



$$sum_{n=1}^inftyfrac{n^n+1/n}{(n+1/n)^n}$$






calculus sequences-and-series




calculus sequences-and-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 28 '18 at 15:45









Asaf Karagila♦

303k32429761




303k32429761










asked Nov 28 '18 at 14:36









NegarNegar

397




397












  • $begingroup$
    Welcome to MathStackExchange! Please take a look at the tour page and consider editing your question to comply with the homework questions guidelines. In particular, it is good practice to give some context on what course you are taking and show your work in attempting to solve this question.
    $endgroup$
    – Mefitico
    Nov 28 '18 at 14:43


















  • $begingroup$
    Welcome to MathStackExchange! Please take a look at the tour page and consider editing your question to comply with the homework questions guidelines. In particular, it is good practice to give some context on what course you are taking and show your work in attempting to solve this question.
    $endgroup$
    – Mefitico
    Nov 28 '18 at 14:43
















$begingroup$
Welcome to MathStackExchange! Please take a look at the tour page and consider editing your question to comply with the homework questions guidelines. In particular, it is good practice to give some context on what course you are taking and show your work in attempting to solve this question.
$endgroup$
– Mefitico
Nov 28 '18 at 14:43




$begingroup$
Welcome to MathStackExchange! Please take a look at the tour page and consider editing your question to comply with the homework questions guidelines. In particular, it is good practice to give some context on what course you are taking and show your work in attempting to solve this question.
$endgroup$
– Mefitico
Nov 28 '18 at 14:43










2 Answers
2






active

oldest

votes


















4












$begingroup$

You can certainly ignore the term $dfrac1n$ at the numerator and



$$sumfrac{n^n}{(n+frac1n)^n}=sumfrac1{(1+frac1{n^2})^n}$$ diverges as the general term tends to $1$.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    At first, I lost the n^n term at the numerator.
    $endgroup$
    – gimusi
    Nov 28 '18 at 14:52










  • $begingroup$
    Shouldn’t we use one of the series tests to answer this question?
    $endgroup$
    – Negar
    Nov 28 '18 at 15:12










  • $begingroup$
    @Negarrezaei: it is enough to note that as $(1+1/n^2)^n=sqrt[n]{(1+1/n^2)^{n^2}}$, the general term goes to $1$.
    $endgroup$
    – Yves Daoust
    Nov 28 '18 at 15:15












  • $begingroup$
    @YvesDaoust even befor simplifying , the sum of the very first 3 characters of the series would be 2.45. How can the sum of the whole characters of the series from 1 to infinitive would be convergent to 1 ?
    $endgroup$
    – Negar
    Nov 28 '18 at 15:39










  • $begingroup$
    @Negar: I am talking about the general term, not the partial sums. And I said that the sum diverges !!!
    $endgroup$
    – Yves Daoust
    Nov 28 '18 at 16:00





















-1












$begingroup$

HINT



We have that



$$(n+1/n)^n=e^{nlog (n+1/n)}=e^{nlog n+nlog(1+1/n^2)}sim n^n$$



$$frac{n^n+1/n}{(n+1/n)^n}sim1$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    We have sum in the question.I don’t get how the sum of the whole characters of the series is converge to 1 when even the sum of the very fisrt 3 characters would be 2,45 ?
    $endgroup$
    – Negar
    Nov 28 '18 at 15:57










  • $begingroup$
    @Negar We are claiming that for n large the terms of the series is equal to $1$ and then the series diverges.
    $endgroup$
    – gimusi
    Nov 28 '18 at 15:59










  • $begingroup$
    I got that thanks.
    $endgroup$
    – Negar
    Nov 28 '18 at 16:02











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

You can certainly ignore the term $dfrac1n$ at the numerator and



$$sumfrac{n^n}{(n+frac1n)^n}=sumfrac1{(1+frac1{n^2})^n}$$ diverges as the general term tends to $1$.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    At first, I lost the n^n term at the numerator.
    $endgroup$
    – gimusi
    Nov 28 '18 at 14:52










  • $begingroup$
    Shouldn’t we use one of the series tests to answer this question?
    $endgroup$
    – Negar
    Nov 28 '18 at 15:12










  • $begingroup$
    @Negarrezaei: it is enough to note that as $(1+1/n^2)^n=sqrt[n]{(1+1/n^2)^{n^2}}$, the general term goes to $1$.
    $endgroup$
    – Yves Daoust
    Nov 28 '18 at 15:15












  • $begingroup$
    @YvesDaoust even befor simplifying , the sum of the very first 3 characters of the series would be 2.45. How can the sum of the whole characters of the series from 1 to infinitive would be convergent to 1 ?
    $endgroup$
    – Negar
    Nov 28 '18 at 15:39










  • $begingroup$
    @Negar: I am talking about the general term, not the partial sums. And I said that the sum diverges !!!
    $endgroup$
    – Yves Daoust
    Nov 28 '18 at 16:00


















4












$begingroup$

You can certainly ignore the term $dfrac1n$ at the numerator and



$$sumfrac{n^n}{(n+frac1n)^n}=sumfrac1{(1+frac1{n^2})^n}$$ diverges as the general term tends to $1$.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    At first, I lost the n^n term at the numerator.
    $endgroup$
    – gimusi
    Nov 28 '18 at 14:52










  • $begingroup$
    Shouldn’t we use one of the series tests to answer this question?
    $endgroup$
    – Negar
    Nov 28 '18 at 15:12










  • $begingroup$
    @Negarrezaei: it is enough to note that as $(1+1/n^2)^n=sqrt[n]{(1+1/n^2)^{n^2}}$, the general term goes to $1$.
    $endgroup$
    – Yves Daoust
    Nov 28 '18 at 15:15












  • $begingroup$
    @YvesDaoust even befor simplifying , the sum of the very first 3 characters of the series would be 2.45. How can the sum of the whole characters of the series from 1 to infinitive would be convergent to 1 ?
    $endgroup$
    – Negar
    Nov 28 '18 at 15:39










  • $begingroup$
    @Negar: I am talking about the general term, not the partial sums. And I said that the sum diverges !!!
    $endgroup$
    – Yves Daoust
    Nov 28 '18 at 16:00
















4












4








4





$begingroup$

You can certainly ignore the term $dfrac1n$ at the numerator and



$$sumfrac{n^n}{(n+frac1n)^n}=sumfrac1{(1+frac1{n^2})^n}$$ diverges as the general term tends to $1$.






share|cite|improve this answer









$endgroup$



You can certainly ignore the term $dfrac1n$ at the numerator and



$$sumfrac{n^n}{(n+frac1n)^n}=sumfrac1{(1+frac1{n^2})^n}$$ diverges as the general term tends to $1$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 28 '18 at 14:42









Yves DaoustYves Daoust

126k672226




126k672226








  • 1




    $begingroup$
    At first, I lost the n^n term at the numerator.
    $endgroup$
    – gimusi
    Nov 28 '18 at 14:52










  • $begingroup$
    Shouldn’t we use one of the series tests to answer this question?
    $endgroup$
    – Negar
    Nov 28 '18 at 15:12










  • $begingroup$
    @Negarrezaei: it is enough to note that as $(1+1/n^2)^n=sqrt[n]{(1+1/n^2)^{n^2}}$, the general term goes to $1$.
    $endgroup$
    – Yves Daoust
    Nov 28 '18 at 15:15












  • $begingroup$
    @YvesDaoust even befor simplifying , the sum of the very first 3 characters of the series would be 2.45. How can the sum of the whole characters of the series from 1 to infinitive would be convergent to 1 ?
    $endgroup$
    – Negar
    Nov 28 '18 at 15:39










  • $begingroup$
    @Negar: I am talking about the general term, not the partial sums. And I said that the sum diverges !!!
    $endgroup$
    – Yves Daoust
    Nov 28 '18 at 16:00
















  • 1




    $begingroup$
    At first, I lost the n^n term at the numerator.
    $endgroup$
    – gimusi
    Nov 28 '18 at 14:52










  • $begingroup$
    Shouldn’t we use one of the series tests to answer this question?
    $endgroup$
    – Negar
    Nov 28 '18 at 15:12










  • $begingroup$
    @Negarrezaei: it is enough to note that as $(1+1/n^2)^n=sqrt[n]{(1+1/n^2)^{n^2}}$, the general term goes to $1$.
    $endgroup$
    – Yves Daoust
    Nov 28 '18 at 15:15












  • $begingroup$
    @YvesDaoust even befor simplifying , the sum of the very first 3 characters of the series would be 2.45. How can the sum of the whole characters of the series from 1 to infinitive would be convergent to 1 ?
    $endgroup$
    – Negar
    Nov 28 '18 at 15:39










  • $begingroup$
    @Negar: I am talking about the general term, not the partial sums. And I said that the sum diverges !!!
    $endgroup$
    – Yves Daoust
    Nov 28 '18 at 16:00










1




1




$begingroup$
At first, I lost the n^n term at the numerator.
$endgroup$
– gimusi
Nov 28 '18 at 14:52




$begingroup$
At first, I lost the n^n term at the numerator.
$endgroup$
– gimusi
Nov 28 '18 at 14:52












$begingroup$
Shouldn’t we use one of the series tests to answer this question?
$endgroup$
– Negar
Nov 28 '18 at 15:12




$begingroup$
Shouldn’t we use one of the series tests to answer this question?
$endgroup$
– Negar
Nov 28 '18 at 15:12












$begingroup$
@Negarrezaei: it is enough to note that as $(1+1/n^2)^n=sqrt[n]{(1+1/n^2)^{n^2}}$, the general term goes to $1$.
$endgroup$
– Yves Daoust
Nov 28 '18 at 15:15






$begingroup$
@Negarrezaei: it is enough to note that as $(1+1/n^2)^n=sqrt[n]{(1+1/n^2)^{n^2}}$, the general term goes to $1$.
$endgroup$
– Yves Daoust
Nov 28 '18 at 15:15














$begingroup$
@YvesDaoust even befor simplifying , the sum of the very first 3 characters of the series would be 2.45. How can the sum of the whole characters of the series from 1 to infinitive would be convergent to 1 ?
$endgroup$
– Negar
Nov 28 '18 at 15:39




$begingroup$
@YvesDaoust even befor simplifying , the sum of the very first 3 characters of the series would be 2.45. How can the sum of the whole characters of the series from 1 to infinitive would be convergent to 1 ?
$endgroup$
– Negar
Nov 28 '18 at 15:39












$begingroup$
@Negar: I am talking about the general term, not the partial sums. And I said that the sum diverges !!!
$endgroup$
– Yves Daoust
Nov 28 '18 at 16:00






$begingroup$
@Negar: I am talking about the general term, not the partial sums. And I said that the sum diverges !!!
$endgroup$
– Yves Daoust
Nov 28 '18 at 16:00













-1












$begingroup$

HINT



We have that



$$(n+1/n)^n=e^{nlog (n+1/n)}=e^{nlog n+nlog(1+1/n^2)}sim n^n$$



$$frac{n^n+1/n}{(n+1/n)^n}sim1$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    We have sum in the question.I don’t get how the sum of the whole characters of the series is converge to 1 when even the sum of the very fisrt 3 characters would be 2,45 ?
    $endgroup$
    – Negar
    Nov 28 '18 at 15:57










  • $begingroup$
    @Negar We are claiming that for n large the terms of the series is equal to $1$ and then the series diverges.
    $endgroup$
    – gimusi
    Nov 28 '18 at 15:59










  • $begingroup$
    I got that thanks.
    $endgroup$
    – Negar
    Nov 28 '18 at 16:02
















-1












$begingroup$

HINT



We have that



$$(n+1/n)^n=e^{nlog (n+1/n)}=e^{nlog n+nlog(1+1/n^2)}sim n^n$$



$$frac{n^n+1/n}{(n+1/n)^n}sim1$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    We have sum in the question.I don’t get how the sum of the whole characters of the series is converge to 1 when even the sum of the very fisrt 3 characters would be 2,45 ?
    $endgroup$
    – Negar
    Nov 28 '18 at 15:57










  • $begingroup$
    @Negar We are claiming that for n large the terms of the series is equal to $1$ and then the series diverges.
    $endgroup$
    – gimusi
    Nov 28 '18 at 15:59










  • $begingroup$
    I got that thanks.
    $endgroup$
    – Negar
    Nov 28 '18 at 16:02














-1












-1








-1





$begingroup$

HINT



We have that



$$(n+1/n)^n=e^{nlog (n+1/n)}=e^{nlog n+nlog(1+1/n^2)}sim n^n$$



$$frac{n^n+1/n}{(n+1/n)^n}sim1$$






share|cite|improve this answer











$endgroup$



HINT



We have that



$$(n+1/n)^n=e^{nlog (n+1/n)}=e^{nlog n+nlog(1+1/n^2)}sim n^n$$



$$frac{n^n+1/n}{(n+1/n)^n}sim1$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 28 '18 at 14:48

























answered Nov 28 '18 at 14:42









gimusigimusi

92.8k84494




92.8k84494












  • $begingroup$
    We have sum in the question.I don’t get how the sum of the whole characters of the series is converge to 1 when even the sum of the very fisrt 3 characters would be 2,45 ?
    $endgroup$
    – Negar
    Nov 28 '18 at 15:57










  • $begingroup$
    @Negar We are claiming that for n large the terms of the series is equal to $1$ and then the series diverges.
    $endgroup$
    – gimusi
    Nov 28 '18 at 15:59










  • $begingroup$
    I got that thanks.
    $endgroup$
    – Negar
    Nov 28 '18 at 16:02


















  • $begingroup$
    We have sum in the question.I don’t get how the sum of the whole characters of the series is converge to 1 when even the sum of the very fisrt 3 characters would be 2,45 ?
    $endgroup$
    – Negar
    Nov 28 '18 at 15:57










  • $begingroup$
    @Negar We are claiming that for n large the terms of the series is equal to $1$ and then the series diverges.
    $endgroup$
    – gimusi
    Nov 28 '18 at 15:59










  • $begingroup$
    I got that thanks.
    $endgroup$
    – Negar
    Nov 28 '18 at 16:02
















$begingroup$
We have sum in the question.I don’t get how the sum of the whole characters of the series is converge to 1 when even the sum of the very fisrt 3 characters would be 2,45 ?
$endgroup$
– Negar
Nov 28 '18 at 15:57




$begingroup$
We have sum in the question.I don’t get how the sum of the whole characters of the series is converge to 1 when even the sum of the very fisrt 3 characters would be 2,45 ?
$endgroup$
– Negar
Nov 28 '18 at 15:57












$begingroup$
@Negar We are claiming that for n large the terms of the series is equal to $1$ and then the series diverges.
$endgroup$
– gimusi
Nov 28 '18 at 15:59




$begingroup$
@Negar We are claiming that for n large the terms of the series is equal to $1$ and then the series diverges.
$endgroup$
– gimusi
Nov 28 '18 at 15:59












$begingroup$
I got that thanks.
$endgroup$
– Negar
Nov 28 '18 at 16:02




$begingroup$
I got that thanks.
$endgroup$
– Negar
Nov 28 '18 at 16:02


















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