Root test for series
up vote
3
down vote
favorite
Find if the series converges or diverges:
$$
a_n=sum_{n=1}^{infty}left(frac{1}{n}-frac{1}{n^2}right)^n
$$
Simplifying the series expression we get
$$
left(frac{n-1}{n^2}right)^n=frac{left(1+frac{-1}{n}right)^n}{(n)^{2n}},
$$
conducting Root test, taking $n$-th root of the simplified expression as $n to infty$, $e^{-1}$.
Is this methods correct? OR as the author has done by taking the $n^{th}$ root of the original expression of $a_n$, we get
$$
lim_{ntoinfty}left(frac{1}{n}-frac{1}{n^2}right)
=0 Rightarrow a_n
$$ converges?
calculus sequences-and-series limits convergence
edited Nov 17 at 9:55
idea
2,20121024
asked Apr 7 '12 at 6:11
Vikram
3,61243044
|
show 2 more comments
up vote
3
down vote
favorite
Find if the series converges or diverges:
$$
a_n=sum_{n=1}^{infty}left(frac{1}{n}-frac{1}{n^2}right)^n
$$
Simplifying the series expression we get
$$
left(frac{n-1}{n^2}right)^n=frac{left(1+frac{-1}{n}right)^n}{(n)^{2n}},
$$
conducting Root test, taking $n$-th root of the simplified expression as $n to infty$, $e^{-1}$.
Is this methods correct? OR as the author has done by taking the $n^{th}$ root of the original expression of $a_n$, we get
$$
lim_{ntoinfty}left(frac{1}{n}-frac{1}{n^2}right)
=0 Rightarrow a_n
$$ converges?
calculus sequences-and-series limits convergence
edited Nov 17 at 9:55
idea
2,20121024
asked Apr 7 '12 at 6:11
Vikram
3,61243044
1
If you got $e^{-1}$ as $ntoinfty$ you did something wrong.
– anon
Apr 7 '12 at 6:21
2
Not that it matters, but it should be $n^n$ in the denominator, not $n^{2n}$.
– André Nicolas
Apr 7 '12 at 7:17
1
Your $e^{-1}$ is the limit of the $n$th root of the numerator only, you should be computing the $n$th root of the fraction. The denominator should be $n^n$ (and not your $n^{2n}$) and its $n$th root is $n$, hence the $n$th root of the fraction is $sim1/(en)$, which goes to zero.
– Did
Apr 7 '12 at 8:23
@Andre, can you pls throw some light on why the denominator has to be n^n, any article.
– Vikram
Apr 7 '12 at 11:37
1
No article needed here, you simply made the mistake of replacing $(n-1)/n^2$ by $(1-1/n)/n^2$ instead of $(1-1/n)/n$.
– Did
Apr 7 '12 at 11:42
|
show 2 more comments
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Find if the series converges or diverges:
$$
a_n=sum_{n=1}^{infty}left(frac{1}{n}-frac{1}{n^2}right)^n
$$
Simplifying the series expression we get
$$
left(frac{n-1}{n^2}right)^n=frac{left(1+frac{-1}{n}right)^n}{(n)^{2n}},
$$
conducting Root test, taking $n$-th root of the simplified expression as $n to infty$, $e^{-1}$.
Is this methods correct? OR as the author has done by taking the $n^{th}$ root of the original expression of $a_n$, we get
$$
lim_{ntoinfty}left(frac{1}{n}-frac{1}{n^2}right)
=0 Rightarrow a_n
$$ converges?
calculus sequences-and-series limits convergence
edited Nov 17 at 9:55
idea
2,20121024
asked Apr 7 '12 at 6:11
Vikram
3,61243044
Find if the series converges or diverges:
$$
a_n=sum_{n=1}^{infty}left(frac{1}{n}-frac{1}{n^2}right)^n
$$
Simplifying the series expression we get
$$
left(frac{n-1}{n^2}right)^n=frac{left(1+frac{-1}{n}right)^n}{(n)^{2n}},
$$
conducting Root test, taking $n$-th root of the simplified expression as $n to infty$, $e^{-1}$.
Is this methods correct? OR as the author has done by taking the $n^{th}$ root of the original expression of $a_n$, we get
$$
lim_{ntoinfty}left(frac{1}{n}-frac{1}{n^2}right)
=0 Rightarrow a_n
$$ converges?
calculus sequences-and-series limits convergence
calculus sequences-and-series limits convergence
edited Nov 17 at 9:55
idea
2,20121024
asked Apr 7 '12 at 6:11
Vikram
3,61243044
edited Nov 17 at 9:55
idea
2,20121024
asked Apr 7 '12 at 6:11
Vikram
3,61243044
edited Nov 17 at 9:55
idea
2,20121024
edited Nov 17 at 9:55
idea
2,20121024
edited Nov 17 at 9:55
idea
2,20121024
2,20121024
asked Apr 7 '12 at 6:11
Vikram
3,61243044
asked Apr 7 '12 at 6:11
Vikram
3,61243044
asked Apr 7 '12 at 6:11
Vikram
3,61243044
3,61243044
1
If you got $e^{-1}$ as $ntoinfty$ you did something wrong.
– anon
Apr 7 '12 at 6:21
2
Not that it matters, but it should be $n^n$ in the denominator, not $n^{2n}$.
– André Nicolas
Apr 7 '12 at 7:17
1
Your $e^{-1}$ is the limit of the $n$th root of the numerator only, you should be computing the $n$th root of the fraction. The denominator should be $n^n$ (and not your $n^{2n}$) and its $n$th root is $n$, hence the $n$th root of the fraction is $sim1/(en)$, which goes to zero.
– Did
Apr 7 '12 at 8:23
@Andre, can you pls throw some light on why the denominator has to be n^n, any article.
– Vikram
Apr 7 '12 at 11:37
1
No article needed here, you simply made the mistake of replacing $(n-1)/n^2$ by $(1-1/n)/n^2$ instead of $(1-1/n)/n$.
– Did
Apr 7 '12 at 11:42
|
show 2 more comments
1
If you got $e^{-1}$ as $ntoinfty$ you did something wrong.
– anon
Apr 7 '12 at 6:21
2
Not that it matters, but it should be $n^n$ in the denominator, not $n^{2n}$.
– André Nicolas
Apr 7 '12 at 7:17
1
Your $e^{-1}$ is the limit of the $n$th root of the numerator only, you should be computing the $n$th root of the fraction. The denominator should be $n^n$ (and not your $n^{2n}$) and its $n$th root is $n$, hence the $n$th root of the fraction is $sim1/(en)$, which goes to zero.
– Did
Apr 7 '12 at 8:23
@Andre, can you pls throw some light on why the denominator has to be n^n, any article.
– Vikram
Apr 7 '12 at 11:37
1
No article needed here, you simply made the mistake of replacing $(n-1)/n^2$ by $(1-1/n)/n^2$ instead of $(1-1/n)/n$.
– Did
Apr 7 '12 at 11:42
1
1
If you got $e^{-1}$ as $ntoinfty$ you did something wrong.
– anon
Apr 7 '12 at 6:21
If you got $e^{-1}$ as $ntoinfty$ you did something wrong.
– anon
Apr 7 '12 at 6:21
2
2
Not that it matters, but it should be $n^n$ in the denominator, not $n^{2n}$.
– André Nicolas
Apr 7 '12 at 7:17
Not that it matters, but it should be $n^n$ in the denominator, not $n^{2n}$.
– André Nicolas
Apr 7 '12 at 7:17
1
1
Your $e^{-1}$ is the limit of the $n$th root of the numerator only, you should be computing the $n$th root of the fraction. The denominator should be $n^n$ (and not your $n^{2n}$) and its $n$th root is $n$, hence the $n$th root of the fraction is $sim1/(en)$, which goes to zero.
– Did
Apr 7 '12 at 8:23
Your $e^{-1}$ is the limit of the $n$th root of the numerator only, you should be computing the $n$th root of the fraction. The denominator should be $n^n$ (and not your $n^{2n}$) and its $n$th root is $n$, hence the $n$th root of the fraction is $sim1/(en)$, which goes to zero.
– Did
Apr 7 '12 at 8:23
@Andre, can you pls throw some light on why the denominator has to be n^n, any article.
– Vikram
Apr 7 '12 at 11:37
@Andre, can you pls throw some light on why the denominator has to be n^n, any article.
– Vikram
Apr 7 '12 at 11:37
1
1
No article needed here, you simply made the mistake of replacing $(n-1)/n^2$ by $(1-1/n)/n^2$ instead of $(1-1/n)/n$.
– Did
Apr 7 '12 at 11:42
No article needed here, you simply made the mistake of replacing $(n-1)/n^2$ by $(1-1/n)/n^2$ instead of $(1-1/n)/n$.
– Did
Apr 7 '12 at 11:42
|
show 2 more comments
1 Answer
1
active
oldest
votes
up vote
6
down vote
accepted
You are right that $limlimits_{ntoinfty} left(1-frac1nright)^n=e^{-1}$. But this is not the limit you use in the root test. You are looking for the value of $limlimits_{ntoinfty} sqrt[n]{a_n}$. (You worked only with the denominator of your expression for $a_n$, but you have to work with the whole expression and take the $n$-th root.)
The limits $limlimits_{ntoinfty} sqrt[n]{a_n}$ is exactly $limlimits_{ntoinfty} left(frac1n-frac1{n^2}right)=0$.
You will get the same value from
$$sqrt[n]{a_n}=sqrt[n]{frac{(1+(-1)/n)^n}{n^{n}}} = frac{1-frac1n}{n}$$
which tends to $0$ as $ntoinfty$.
This implies that the series $sum a_n$ converges.
answered Apr 7 '12 at 6:59
Martin Sleziak
44.4k7115268
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
You are right that $limlimits_{ntoinfty} left(1-frac1nright)^n=e^{-1}$. But this is not the limit you use in the root test. You are looking for the value of $limlimits_{ntoinfty} sqrt[n]{a_n}$. (You worked only with the denominator of your expression for $a_n$, but you have to work with the whole expression and take the $n$-th root.)
The limits $limlimits_{ntoinfty} sqrt[n]{a_n}$ is exactly $limlimits_{ntoinfty} left(frac1n-frac1{n^2}right)=0$.
You will get the same value from
$$sqrt[n]{a_n}=sqrt[n]{frac{(1+(-1)/n)^n}{n^{n}}} = frac{1-frac1n}{n}$$
which tends to $0$ as $ntoinfty$.
This implies that the series $sum a_n$ converges.
answered Apr 7 '12 at 6:59
Martin Sleziak
44.4k7115268
add a comment |
up vote
6
down vote
accepted
You are right that $limlimits_{ntoinfty} left(1-frac1nright)^n=e^{-1}$. But this is not the limit you use in the root test. You are looking for the value of $limlimits_{ntoinfty} sqrt[n]{a_n}$. (You worked only with the denominator of your expression for $a_n$, but you have to work with the whole expression and take the $n$-th root.)
The limits $limlimits_{ntoinfty} sqrt[n]{a_n}$ is exactly $limlimits_{ntoinfty} left(frac1n-frac1{n^2}right)=0$.
You will get the same value from
$$sqrt[n]{a_n}=sqrt[n]{frac{(1+(-1)/n)^n}{n^{n}}} = frac{1-frac1n}{n}$$
which tends to $0$ as $ntoinfty$.
This implies that the series $sum a_n$ converges.
answered Apr 7 '12 at 6:59
Martin Sleziak
44.4k7115268
add a comment |
up vote
6
down vote
accepted
up vote
6
down vote
accepted
You are right that $limlimits_{ntoinfty} left(1-frac1nright)^n=e^{-1}$. But this is not the limit you use in the root test. You are looking for the value of $limlimits_{ntoinfty} sqrt[n]{a_n}$. (You worked only with the denominator of your expression for $a_n$, but you have to work with the whole expression and take the $n$-th root.)
The limits $limlimits_{ntoinfty} sqrt[n]{a_n}$ is exactly $limlimits_{ntoinfty} left(frac1n-frac1{n^2}right)=0$.
You will get the same value from
$$sqrt[n]{a_n}=sqrt[n]{frac{(1+(-1)/n)^n}{n^{n}}} = frac{1-frac1n}{n}$$
which tends to $0$ as $ntoinfty$.
This implies that the series $sum a_n$ converges.
answered Apr 7 '12 at 6:59
Martin Sleziak
44.4k7115268
You are right that $limlimits_{ntoinfty} left(1-frac1nright)^n=e^{-1}$. But this is not the limit you use in the root test. You are looking for the value of $limlimits_{ntoinfty} sqrt[n]{a_n}$. (You worked only with the denominator of your expression for $a_n$, but you have to work with the whole expression and take the $n$-th root.)
The limits $limlimits_{ntoinfty} sqrt[n]{a_n}$ is exactly $limlimits_{ntoinfty} left(frac1n-frac1{n^2}right)=0$.
You will get the same value from
$$sqrt[n]{a_n}=sqrt[n]{frac{(1+(-1)/n)^n}{n^{n}}} = frac{1-frac1n}{n}$$
which tends to $0$ as $ntoinfty$.
This implies that the series $sum a_n$ converges.
answered Apr 7 '12 at 6:59
Martin Sleziak
44.4k7115268
edited Apr 7 '12 at 7:20
answered Apr 7 '12 at 6:59
Martin Sleziak
44.4k7115268
answered Apr 7 '12 at 6:59
Martin Sleziak
44.4k7115268
answered Apr 7 '12 at 6:59
Martin Sleziak
44.4k7115268
44.4k7115268
add a comment |
add a comment |
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1
If you got $e^{-1}$ as $ntoinfty$ you did something wrong.
– anon
Apr 7 '12 at 6:21
2
Not that it matters, but it should be $n^n$ in the denominator, not $n^{2n}$.
– André Nicolas
Apr 7 '12 at 7:17
1
Your $e^{-1}$ is the limit of the $n$th root of the numerator only, you should be computing the $n$th root of the fraction. The denominator should be $n^n$ (and not your $n^{2n}$) and its $n$th root is $n$, hence the $n$th root of the fraction is $sim1/(en)$, which goes to zero.
– Did
Apr 7 '12 at 8:23
@Andre, can you pls throw some light on why the denominator has to be n^n, any article.
– Vikram
Apr 7 '12 at 11:37
1
No article needed here, you simply made the mistake of replacing $(n-1)/n^2$ by $(1-1/n)/n^2$ instead of $(1-1/n)/n$.
– Did
Apr 7 '12 at 11:42