Proof verification of $lim_{n to infty}frac{q^n}{n} = 0$ for $|q| < 1$ using $epsilon$ definition
Prove
$$
lim_{n to infty}frac{q^n}{n} = 0
$$ for $|q| < 1$ using $epsilon$ definition.
Using the definition of a limit:
$$
lim_{nto infty}frac{q^n}{n} = 0 stackrel{text{def}}{iff} { forallepsilon>0 ,exists Ninmathbb N, forall n > N : left|frac{q^n}{n} - 0right| < epsilon }
$$
Consider the following:
$$
left|frac{q^n}{n}right| < epsilon iff frac{|q|^n}{n} < epsilon
$$
Redefine $|q|^n$:
$$
|q|^n = frac{1}{(1+t)^n} lefrac{1}{(1+nt)}
$$
Thus:
$$
frac{|q|^n}{n} < frac{1}{n(1+nt)} < frac{1}{n^2t} < frac{1}{n} < epsilon
$$
So from this we may find $N$ such that:
$$
frac{1}{n} < frac{1}{N} < epsilon
$$
Thus the limit is $0$.
Is it a correct proof?
calculus limits proof-verification limits-without-lhopital epsilon-delta
asked Nov 21 '18 at 16:28
roman
1,93921221
add a comment |
Prove
$$
lim_{n to infty}frac{q^n}{n} = 0
$$ for $|q| < 1$ using $epsilon$ definition.
Using the definition of a limit:
$$
lim_{nto infty}frac{q^n}{n} = 0 stackrel{text{def}}{iff} { forallepsilon>0 ,exists Ninmathbb N, forall n > N : left|frac{q^n}{n} - 0right| < epsilon }
$$
Consider the following:
$$
left|frac{q^n}{n}right| < epsilon iff frac{|q|^n}{n} < epsilon
$$
Redefine $|q|^n$:
$$
|q|^n = frac{1}{(1+t)^n} lefrac{1}{(1+nt)}
$$
Thus:
$$
frac{|q|^n}{n} < frac{1}{n(1+nt)} < frac{1}{n^2t} < frac{1}{n} < epsilon
$$
So from this we may find $N$ such that:
$$
frac{1}{n} < frac{1}{N} < epsilon
$$
Thus the limit is $0$.
Is it a correct proof?
calculus limits proof-verification limits-without-lhopital epsilon-delta
asked Nov 21 '18 at 16:28
roman
1,93921221
add a comment |
Prove
$$
lim_{n to infty}frac{q^n}{n} = 0
$$ for $|q| < 1$ using $epsilon$ definition.
Using the definition of a limit:
$$
lim_{nto infty}frac{q^n}{n} = 0 stackrel{text{def}}{iff} { forallepsilon>0 ,exists Ninmathbb N, forall n > N : left|frac{q^n}{n} - 0right| < epsilon }
$$
Consider the following:
$$
left|frac{q^n}{n}right| < epsilon iff frac{|q|^n}{n} < epsilon
$$
Redefine $|q|^n$:
$$
|q|^n = frac{1}{(1+t)^n} lefrac{1}{(1+nt)}
$$
Thus:
$$
frac{|q|^n}{n} < frac{1}{n(1+nt)} < frac{1}{n^2t} < frac{1}{n} < epsilon
$$
So from this we may find $N$ such that:
$$
frac{1}{n} < frac{1}{N} < epsilon
$$
Thus the limit is $0$.
Is it a correct proof?
calculus limits proof-verification limits-without-lhopital epsilon-delta
asked Nov 21 '18 at 16:28
roman
1,93921221
Prove
$$
lim_{n to infty}frac{q^n}{n} = 0
$$ for $|q| < 1$ using $epsilon$ definition.
Using the definition of a limit:
$$
lim_{nto infty}frac{q^n}{n} = 0 stackrel{text{def}}{iff} { forallepsilon>0 ,exists Ninmathbb N, forall n > N : left|frac{q^n}{n} - 0right| < epsilon }
$$
Consider the following:
$$
left|frac{q^n}{n}right| < epsilon iff frac{|q|^n}{n} < epsilon
$$
Redefine $|q|^n$:
$$
|q|^n = frac{1}{(1+t)^n} lefrac{1}{(1+nt)}
$$
Thus:
$$
frac{|q|^n}{n} < frac{1}{n(1+nt)} < frac{1}{n^2t} < frac{1}{n} < epsilon
$$
So from this we may find $N$ such that:
$$
frac{1}{n} < frac{1}{N} < epsilon
$$
Thus the limit is $0$.
Is it a correct proof?
calculus limits proof-verification limits-without-lhopital epsilon-delta
calculus limits proof-verification limits-without-lhopital epsilon-delta
asked Nov 21 '18 at 16:28
roman
1,93921221
asked Nov 21 '18 at 16:28
roman
1,93921221
edited Nov 21 '18 at 16:51
asked Nov 21 '18 at 16:28
roman
1,93921221
asked Nov 21 '18 at 16:28
roman
1,93921221
asked Nov 21 '18 at 16:28
roman
1,93921221
1,93921221
add a comment |
add a comment |
3 Answers
3
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votes
Your proof is almost correct, but there is a small problem concerning the inequality $dfrac1{n^2t}<dfrac1n$. This is equivalent to $nt>1$. Why would that be true? All you know about $t$ is that $t>0$. So, you should deal with the inequality $dfrac1{nt}<varepsilon$. That is, choose $N$ such that $dfrac1{Nt}<varepsilon$.
answered Nov 21 '18 at 16:33
José Carlos Santos
151k22123224
That's a nice catch, thank you!
– roman
Nov 21 '18 at 16:50
add a comment |
Yes, your proof is fine. You should mention, that the inequality $frac{1}{n^2t }<frac{1}{n}$ does not hold for all $n $, but for almost all $n $.
answered Nov 21 '18 at 16:39
Fred
44.2k1845
add a comment |
You have::
$dfrac{|q|^n}{n} < dfrac{1}{n^2t}< dfrac{1}{nt},$ $t>0$.
Let $epsilon$ be given.
Archimedean principle:
There is a $N$, positive interger, s.t.
$N >dfrac{1}{t epsilon}$.
For $n ge N$:
$dfrac{|q|^n}{n} <dfrac{1}{nt} le dfrac{1}{Nt} <epsilon$.
answered Nov 21 '18 at 16:56
Peter Szilas
10.7k2720
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
Your proof is almost correct, but there is a small problem concerning the inequality $dfrac1{n^2t}<dfrac1n$. This is equivalent to $nt>1$. Why would that be true? All you know about $t$ is that $t>0$. So, you should deal with the inequality $dfrac1{nt}<varepsilon$. That is, choose $N$ such that $dfrac1{Nt}<varepsilon$.
answered Nov 21 '18 at 16:33
José Carlos Santos
151k22123224
That's a nice catch, thank you!
– roman
Nov 21 '18 at 16:50
add a comment |
Your proof is almost correct, but there is a small problem concerning the inequality $dfrac1{n^2t}<dfrac1n$. This is equivalent to $nt>1$. Why would that be true? All you know about $t$ is that $t>0$. So, you should deal with the inequality $dfrac1{nt}<varepsilon$. That is, choose $N$ such that $dfrac1{Nt}<varepsilon$.
answered Nov 21 '18 at 16:33
José Carlos Santos
151k22123224
That's a nice catch, thank you!
– roman
Nov 21 '18 at 16:50
add a comment |
Your proof is almost correct, but there is a small problem concerning the inequality $dfrac1{n^2t}<dfrac1n$. This is equivalent to $nt>1$. Why would that be true? All you know about $t$ is that $t>0$. So, you should deal with the inequality $dfrac1{nt}<varepsilon$. That is, choose $N$ such that $dfrac1{Nt}<varepsilon$.
answered Nov 21 '18 at 16:33
José Carlos Santos
151k22123224
Your proof is almost correct, but there is a small problem concerning the inequality $dfrac1{n^2t}<dfrac1n$. This is equivalent to $nt>1$. Why would that be true? All you know about $t$ is that $t>0$. So, you should deal with the inequality $dfrac1{nt}<varepsilon$. That is, choose $N$ such that $dfrac1{Nt}<varepsilon$.
answered Nov 21 '18 at 16:33
José Carlos Santos
151k22123224
answered Nov 21 '18 at 16:33
José Carlos Santos
151k22123224
answered Nov 21 '18 at 16:33
José Carlos Santos
151k22123224
answered Nov 21 '18 at 16:33
José Carlos Santos
151k22123224
151k22123224
That's a nice catch, thank you!
– roman
Nov 21 '18 at 16:50
add a comment |
That's a nice catch, thank you!
– roman
Nov 21 '18 at 16:50
That's a nice catch, thank you!
– roman
Nov 21 '18 at 16:50
That's a nice catch, thank you!
– roman
Nov 21 '18 at 16:50
add a comment |
Yes, your proof is fine. You should mention, that the inequality $frac{1}{n^2t }<frac{1}{n}$ does not hold for all $n $, but for almost all $n $.
answered Nov 21 '18 at 16:39
Fred
44.2k1845
add a comment |
Yes, your proof is fine. You should mention, that the inequality $frac{1}{n^2t }<frac{1}{n}$ does not hold for all $n $, but for almost all $n $.
answered Nov 21 '18 at 16:39
Fred
44.2k1845
add a comment |
Yes, your proof is fine. You should mention, that the inequality $frac{1}{n^2t }<frac{1}{n}$ does not hold for all $n $, but for almost all $n $.
answered Nov 21 '18 at 16:39
Fred
44.2k1845
Yes, your proof is fine. You should mention, that the inequality $frac{1}{n^2t }<frac{1}{n}$ does not hold for all $n $, but for almost all $n $.
answered Nov 21 '18 at 16:39
Fred
44.2k1845
answered Nov 21 '18 at 16:39
Fred
44.2k1845
answered Nov 21 '18 at 16:39
Fred
44.2k1845
answered Nov 21 '18 at 16:39
Fred
44.2k1845
44.2k1845
add a comment |
add a comment |
You have::
$dfrac{|q|^n}{n} < dfrac{1}{n^2t}< dfrac{1}{nt},$ $t>0$.
Let $epsilon$ be given.
Archimedean principle:
There is a $N$, positive interger, s.t.
$N >dfrac{1}{t epsilon}$.
For $n ge N$:
$dfrac{|q|^n}{n} <dfrac{1}{nt} le dfrac{1}{Nt} <epsilon$.
answered Nov 21 '18 at 16:56
Peter Szilas
10.7k2720
add a comment |
You have::
$dfrac{|q|^n}{n} < dfrac{1}{n^2t}< dfrac{1}{nt},$ $t>0$.
Let $epsilon$ be given.
Archimedean principle:
There is a $N$, positive interger, s.t.
$N >dfrac{1}{t epsilon}$.
For $n ge N$:
$dfrac{|q|^n}{n} <dfrac{1}{nt} le dfrac{1}{Nt} <epsilon$.
answered Nov 21 '18 at 16:56
Peter Szilas
10.7k2720
add a comment |
You have::
$dfrac{|q|^n}{n} < dfrac{1}{n^2t}< dfrac{1}{nt},$ $t>0$.
Let $epsilon$ be given.
Archimedean principle:
There is a $N$, positive interger, s.t.
$N >dfrac{1}{t epsilon}$.
For $n ge N$:
$dfrac{|q|^n}{n} <dfrac{1}{nt} le dfrac{1}{Nt} <epsilon$.
answered Nov 21 '18 at 16:56
Peter Szilas
10.7k2720
You have::
$dfrac{|q|^n}{n} < dfrac{1}{n^2t}< dfrac{1}{nt},$ $t>0$.
Let $epsilon$ be given.
Archimedean principle:
There is a $N$, positive interger, s.t.
$N >dfrac{1}{t epsilon}$.
For $n ge N$:
$dfrac{|q|^n}{n} <dfrac{1}{nt} le dfrac{1}{Nt} <epsilon$.
answered Nov 21 '18 at 16:56
Peter Szilas
10.7k2720
answered Nov 21 '18 at 16:56
Peter Szilas
10.7k2720
answered Nov 21 '18 at 16:56
Peter Szilas
10.7k2720
answered Nov 21 '18 at 16:56
Peter Szilas
10.7k2720
10.7k2720
add a comment |
add a comment |
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