Determine whether the series $frac{e^frac{1}{n}}{n^2}$ converges or diverges
The problem states
Determine whether the series converges or diverges: $$sum_{n=1}^inftyfrac{e^{1/n}}{n^2}$$
Immediately I always like to try the test for divergence:
$$lim_{ntoinfty} sum_{n=1}^inftyfrac{e^{1/n}}{n^2} = 0$$ I judge this to be true by looking at the numerator and denominator of the series. The numerator will tend to 1 as $nto infty$ and the denominator will grow bigger and bigger, so eventually you will have $frac{1}{infty}$ which will equal to zero. This tells us absolutely nothing about the convergence or divergence of this series.
Next I'll try the integral test because i see in the denominator a term resulting from the derivative of the numerator. To apply the integral test the sequence $a_n$ of our series $sum_{n=1}^infty a_n$ should be continuous, positive and decreasing. The summation begins at one and not zero so we have no issues with continuity here. Both numerator and denominator will yield positive terms as $n to infty$ and sequential terms will get smaller as the numerator shrinks whilst the denominator grows as $nto infty$.
$$text{let } f(x) = frac{e^{x^{-1}}}{x^2}$$
$$int_1^infty f(x) dx = int_1^infty frac{e^{x^{-1}}}{x^2}dx$$
$$= -e^{x{^{-1}}}|_1^infty$$
$$= lim_{xto infty}- e^{frac{1}{x}} +e$$
$$=-1 + e$$
because $int_1^infty f(x)$ converges then by the integral test so must our series.
I have two questions:
- How can i write the conditions for the integral test (continuous,positive,decreasing) in a mathematical way. What would be the mathematically accepted way to show these conditions?
- Would it be possible to determine whether this series converges or diverges using the limit comparison test?
sequences-and-series convergence
edited Nov 25 at 11:35
amWhy
191k28224439
asked Oct 1 '17 at 16:16
Blargian
161314
add a comment |
The problem states
Determine whether the series converges or diverges: $$sum_{n=1}^inftyfrac{e^{1/n}}{n^2}$$
Immediately I always like to try the test for divergence:
$$lim_{ntoinfty} sum_{n=1}^inftyfrac{e^{1/n}}{n^2} = 0$$ I judge this to be true by looking at the numerator and denominator of the series. The numerator will tend to 1 as $nto infty$ and the denominator will grow bigger and bigger, so eventually you will have $frac{1}{infty}$ which will equal to zero. This tells us absolutely nothing about the convergence or divergence of this series.
Next I'll try the integral test because i see in the denominator a term resulting from the derivative of the numerator. To apply the integral test the sequence $a_n$ of our series $sum_{n=1}^infty a_n$ should be continuous, positive and decreasing. The summation begins at one and not zero so we have no issues with continuity here. Both numerator and denominator will yield positive terms as $n to infty$ and sequential terms will get smaller as the numerator shrinks whilst the denominator grows as $nto infty$.
$$text{let } f(x) = frac{e^{x^{-1}}}{x^2}$$
$$int_1^infty f(x) dx = int_1^infty frac{e^{x^{-1}}}{x^2}dx$$
$$= -e^{x{^{-1}}}|_1^infty$$
$$= lim_{xto infty}- e^{frac{1}{x}} +e$$
$$=-1 + e$$
because $int_1^infty f(x)$ converges then by the integral test so must our series.
I have two questions:
- How can i write the conditions for the integral test (continuous,positive,decreasing) in a mathematical way. What would be the mathematically accepted way to show these conditions?
- Would it be possible to determine whether this series converges or diverges using the limit comparison test?
sequences-and-series convergence
edited Nov 25 at 11:35
amWhy
191k28224439
asked Oct 1 '17 at 16:16
Blargian
161314
3
Compare, to say $e/n^2$.
– Lord Shark the Unknown
Oct 1 '17 at 16:19
This makes no sense: $$lim_{ntoinfty} sum_{n=1}^inftyfrac{e^{frac{1}{n}}}{n^2} = 0$$
– Simply Beautiful Art
Oct 1 '17 at 16:27
The test for divergence is applied to the sequence and not the series? I.e) if i had written $$ lim_{ntoinfty} frac{e^{frac{1}{n}}}{n^2} = 0$$ that would be fine?
– Blargian
Oct 1 '17 at 16:49
Yes, that is correct.
– Simply Beautiful Art
Oct 1 '17 at 16:53
Don’t forget the $dx$ on your integral
– Chase Ryan Taylor
Oct 1 '17 at 18:24
add a comment |
The problem states
Determine whether the series converges or diverges: $$sum_{n=1}^inftyfrac{e^{1/n}}{n^2}$$
Immediately I always like to try the test for divergence:
$$lim_{ntoinfty} sum_{n=1}^inftyfrac{e^{1/n}}{n^2} = 0$$ I judge this to be true by looking at the numerator and denominator of the series. The numerator will tend to 1 as $nto infty$ and the denominator will grow bigger and bigger, so eventually you will have $frac{1}{infty}$ which will equal to zero. This tells us absolutely nothing about the convergence or divergence of this series.
Next I'll try the integral test because i see in the denominator a term resulting from the derivative of the numerator. To apply the integral test the sequence $a_n$ of our series $sum_{n=1}^infty a_n$ should be continuous, positive and decreasing. The summation begins at one and not zero so we have no issues with continuity here. Both numerator and denominator will yield positive terms as $n to infty$ and sequential terms will get smaller as the numerator shrinks whilst the denominator grows as $nto infty$.
$$text{let } f(x) = frac{e^{x^{-1}}}{x^2}$$
$$int_1^infty f(x) dx = int_1^infty frac{e^{x^{-1}}}{x^2}dx$$
$$= -e^{x{^{-1}}}|_1^infty$$
$$= lim_{xto infty}- e^{frac{1}{x}} +e$$
$$=-1 + e$$
because $int_1^infty f(x)$ converges then by the integral test so must our series.
I have two questions:
- How can i write the conditions for the integral test (continuous,positive,decreasing) in a mathematical way. What would be the mathematically accepted way to show these conditions?
- Would it be possible to determine whether this series converges or diverges using the limit comparison test?
sequences-and-series convergence
edited Nov 25 at 11:35
amWhy
191k28224439
asked Oct 1 '17 at 16:16
Blargian
161314
The problem states
Determine whether the series converges or diverges: $$sum_{n=1}^inftyfrac{e^{1/n}}{n^2}$$
Immediately I always like to try the test for divergence:
$$lim_{ntoinfty} sum_{n=1}^inftyfrac{e^{1/n}}{n^2} = 0$$ I judge this to be true by looking at the numerator and denominator of the series. The numerator will tend to 1 as $nto infty$ and the denominator will grow bigger and bigger, so eventually you will have $frac{1}{infty}$ which will equal to zero. This tells us absolutely nothing about the convergence or divergence of this series.
Next I'll try the integral test because i see in the denominator a term resulting from the derivative of the numerator. To apply the integral test the sequence $a_n$ of our series $sum_{n=1}^infty a_n$ should be continuous, positive and decreasing. The summation begins at one and not zero so we have no issues with continuity here. Both numerator and denominator will yield positive terms as $n to infty$ and sequential terms will get smaller as the numerator shrinks whilst the denominator grows as $nto infty$.
$$text{let } f(x) = frac{e^{x^{-1}}}{x^2}$$
$$int_1^infty f(x) dx = int_1^infty frac{e^{x^{-1}}}{x^2}dx$$
$$= -e^{x{^{-1}}}|_1^infty$$
$$= lim_{xto infty}- e^{frac{1}{x}} +e$$
$$=-1 + e$$
because $int_1^infty f(x)$ converges then by the integral test so must our series.
I have two questions:
- How can i write the conditions for the integral test (continuous,positive,decreasing) in a mathematical way. What would be the mathematically accepted way to show these conditions?
- Would it be possible to determine whether this series converges or diverges using the limit comparison test?
sequences-and-series convergence
sequences-and-series convergence
edited Nov 25 at 11:35
amWhy
191k28224439
asked Oct 1 '17 at 16:16
Blargian
161314
edited Nov 25 at 11:35
amWhy
191k28224439
asked Oct 1 '17 at 16:16
Blargian
161314
edited Nov 25 at 11:35
amWhy
191k28224439
edited Nov 25 at 11:35
amWhy
191k28224439
edited Nov 25 at 11:35
amWhy
191k28224439
191k28224439
asked Oct 1 '17 at 16:16
Blargian
161314
asked Oct 1 '17 at 16:16
Blargian
161314
asked Oct 1 '17 at 16:16
Blargian
161314
161314
3
Compare, to say $e/n^2$.
– Lord Shark the Unknown
Oct 1 '17 at 16:19
This makes no sense: $$lim_{ntoinfty} sum_{n=1}^inftyfrac{e^{frac{1}{n}}}{n^2} = 0$$
– Simply Beautiful Art
Oct 1 '17 at 16:27
The test for divergence is applied to the sequence and not the series? I.e) if i had written $$ lim_{ntoinfty} frac{e^{frac{1}{n}}}{n^2} = 0$$ that would be fine?
– Blargian
Oct 1 '17 at 16:49
Yes, that is correct.
– Simply Beautiful Art
Oct 1 '17 at 16:53
Don’t forget the $dx$ on your integral
– Chase Ryan Taylor
Oct 1 '17 at 18:24
add a comment |
3
Compare, to say $e/n^2$.
– Lord Shark the Unknown
Oct 1 '17 at 16:19
This makes no sense: $$lim_{ntoinfty} sum_{n=1}^inftyfrac{e^{frac{1}{n}}}{n^2} = 0$$
– Simply Beautiful Art
Oct 1 '17 at 16:27
The test for divergence is applied to the sequence and not the series? I.e) if i had written $$ lim_{ntoinfty} frac{e^{frac{1}{n}}}{n^2} = 0$$ that would be fine?
– Blargian
Oct 1 '17 at 16:49
Yes, that is correct.
– Simply Beautiful Art
Oct 1 '17 at 16:53
Don’t forget the $dx$ on your integral
– Chase Ryan Taylor
Oct 1 '17 at 18:24
3
3
Compare, to say $e/n^2$.
– Lord Shark the Unknown
Oct 1 '17 at 16:19
Compare, to say $e/n^2$.
– Lord Shark the Unknown
Oct 1 '17 at 16:19
This makes no sense: $$lim_{ntoinfty} sum_{n=1}^inftyfrac{e^{frac{1}{n}}}{n^2} = 0$$
– Simply Beautiful Art
Oct 1 '17 at 16:27
This makes no sense: $$lim_{ntoinfty} sum_{n=1}^inftyfrac{e^{frac{1}{n}}}{n^2} = 0$$
– Simply Beautiful Art
Oct 1 '17 at 16:27
The test for divergence is applied to the sequence and not the series? I.e) if i had written $$ lim_{ntoinfty} frac{e^{frac{1}{n}}}{n^2} = 0$$ that would be fine?
– Blargian
Oct 1 '17 at 16:49
The test for divergence is applied to the sequence and not the series? I.e) if i had written $$ lim_{ntoinfty} frac{e^{frac{1}{n}}}{n^2} = 0$$ that would be fine?
– Blargian
Oct 1 '17 at 16:49
Yes, that is correct.
– Simply Beautiful Art
Oct 1 '17 at 16:53
Yes, that is correct.
– Simply Beautiful Art
Oct 1 '17 at 16:53
Don’t forget the $dx$ on your integral
– Chase Ryan Taylor
Oct 1 '17 at 18:24
Don’t forget the $dx$ on your integral
– Chase Ryan Taylor
Oct 1 '17 at 18:24
add a comment |
1 Answer
1
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oldest
votes
As Lord Shark the Unknown comments, simply note that
$$e^{1/n}<e,quad n>1$$
And so,
$$0<sum_{n=1}^inftyfrac{e^{1/n}}{n^2}<sum_{n=1}^inftyfrac e{n^2}=frac{epi^2}6<infty$$
How can i write the conditions for the integral test (continuous,positive,decreasing) in a mathematical way. What would be the mathematically accepted way to show these conditions?
If $f(x)$ is continuous, $f(x)ge0$, and $f'(x)le0$, then...
Would it be possible to determine whether this series converges or diverges using the limit comparison test?
Sure. You could apply the limit comparison test against $1/n^2$, since $e^{1/n}to1$ as $ntoinfty$.
answered Oct 1 '17 at 16:31
Simply Beautiful Art
50.2k578181
Could you please explain how $$sum_{n=1}^inftyfrac e{n^2}=frac{epi^2}6$$
– Blargian
Oct 1 '17 at 16:50
1
See the Basel problem.
– Simply Beautiful Art
Oct 1 '17 at 16:52
1
You probably don't want this, but another interpretation of the conditions for the integral test would be$$(fin C^1)landforall xge0[(f(x)ge0)land f'(x)le0)]impliesleft(sum_{n=0}^infty f(n)<inftyiffint_0^infty f(x)~mathrm dx<inftyright)$$
– Simply Beautiful Art
Oct 1 '17 at 16:56
1
If you don't want to restrict this to differentiable functions, then use $forall xge0forall yge0[x<yimplies f(x)>f(y)]$
– Simply Beautiful Art
Oct 1 '17 at 16:57
add a comment |
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As Lord Shark the Unknown comments, simply note that
$$e^{1/n}<e,quad n>1$$
And so,
$$0<sum_{n=1}^inftyfrac{e^{1/n}}{n^2}<sum_{n=1}^inftyfrac e{n^2}=frac{epi^2}6<infty$$
How can i write the conditions for the integral test (continuous,positive,decreasing) in a mathematical way. What would be the mathematically accepted way to show these conditions?
If $f(x)$ is continuous, $f(x)ge0$, and $f'(x)le0$, then...
Would it be possible to determine whether this series converges or diverges using the limit comparison test?
Sure. You could apply the limit comparison test against $1/n^2$, since $e^{1/n}to1$ as $ntoinfty$.
answered Oct 1 '17 at 16:31
Simply Beautiful Art
50.2k578181
Could you please explain how $$sum_{n=1}^inftyfrac e{n^2}=frac{epi^2}6$$
– Blargian
Oct 1 '17 at 16:50
1
See the Basel problem.
– Simply Beautiful Art
Oct 1 '17 at 16:52
1
You probably don't want this, but another interpretation of the conditions for the integral test would be$$(fin C^1)landforall xge0[(f(x)ge0)land f'(x)le0)]impliesleft(sum_{n=0}^infty f(n)<inftyiffint_0^infty f(x)~mathrm dx<inftyright)$$
– Simply Beautiful Art
Oct 1 '17 at 16:56
1
If you don't want to restrict this to differentiable functions, then use $forall xge0forall yge0[x<yimplies f(x)>f(y)]$
– Simply Beautiful Art
Oct 1 '17 at 16:57
add a comment |
As Lord Shark the Unknown comments, simply note that
$$e^{1/n}<e,quad n>1$$
And so,
$$0<sum_{n=1}^inftyfrac{e^{1/n}}{n^2}<sum_{n=1}^inftyfrac e{n^2}=frac{epi^2}6<infty$$
How can i write the conditions for the integral test (continuous,positive,decreasing) in a mathematical way. What would be the mathematically accepted way to show these conditions?
If $f(x)$ is continuous, $f(x)ge0$, and $f'(x)le0$, then...
Would it be possible to determine whether this series converges or diverges using the limit comparison test?
Sure. You could apply the limit comparison test against $1/n^2$, since $e^{1/n}to1$ as $ntoinfty$.
answered Oct 1 '17 at 16:31
Simply Beautiful Art
50.2k578181
Could you please explain how $$sum_{n=1}^inftyfrac e{n^2}=frac{epi^2}6$$
– Blargian
Oct 1 '17 at 16:50
1
See the Basel problem.
– Simply Beautiful Art
Oct 1 '17 at 16:52
1
You probably don't want this, but another interpretation of the conditions for the integral test would be$$(fin C^1)landforall xge0[(f(x)ge0)land f'(x)le0)]impliesleft(sum_{n=0}^infty f(n)<inftyiffint_0^infty f(x)~mathrm dx<inftyright)$$
– Simply Beautiful Art
Oct 1 '17 at 16:56
1
If you don't want to restrict this to differentiable functions, then use $forall xge0forall yge0[x<yimplies f(x)>f(y)]$
– Simply Beautiful Art
Oct 1 '17 at 16:57
add a comment |
As Lord Shark the Unknown comments, simply note that
$$e^{1/n}<e,quad n>1$$
And so,
$$0<sum_{n=1}^inftyfrac{e^{1/n}}{n^2}<sum_{n=1}^inftyfrac e{n^2}=frac{epi^2}6<infty$$
How can i write the conditions for the integral test (continuous,positive,decreasing) in a mathematical way. What would be the mathematically accepted way to show these conditions?
If $f(x)$ is continuous, $f(x)ge0$, and $f'(x)le0$, then...
Would it be possible to determine whether this series converges or diverges using the limit comparison test?
Sure. You could apply the limit comparison test against $1/n^2$, since $e^{1/n}to1$ as $ntoinfty$.
answered Oct 1 '17 at 16:31
Simply Beautiful Art
50.2k578181
As Lord Shark the Unknown comments, simply note that
$$e^{1/n}<e,quad n>1$$
And so,
$$0<sum_{n=1}^inftyfrac{e^{1/n}}{n^2}<sum_{n=1}^inftyfrac e{n^2}=frac{epi^2}6<infty$$
How can i write the conditions for the integral test (continuous,positive,decreasing) in a mathematical way. What would be the mathematically accepted way to show these conditions?
If $f(x)$ is continuous, $f(x)ge0$, and $f'(x)le0$, then...
Would it be possible to determine whether this series converges or diverges using the limit comparison test?
Sure. You could apply the limit comparison test against $1/n^2$, since $e^{1/n}to1$ as $ntoinfty$.
answered Oct 1 '17 at 16:31
Simply Beautiful Art
50.2k578181
answered Oct 1 '17 at 16:31
Simply Beautiful Art
50.2k578181
answered Oct 1 '17 at 16:31
Simply Beautiful Art
50.2k578181
answered Oct 1 '17 at 16:31
Simply Beautiful Art
50.2k578181
50.2k578181
Could you please explain how $$sum_{n=1}^inftyfrac e{n^2}=frac{epi^2}6$$
– Blargian
Oct 1 '17 at 16:50
1
See the Basel problem.
– Simply Beautiful Art
Oct 1 '17 at 16:52
1
You probably don't want this, but another interpretation of the conditions for the integral test would be$$(fin C^1)landforall xge0[(f(x)ge0)land f'(x)le0)]impliesleft(sum_{n=0}^infty f(n)<inftyiffint_0^infty f(x)~mathrm dx<inftyright)$$
– Simply Beautiful Art
Oct 1 '17 at 16:56
1
If you don't want to restrict this to differentiable functions, then use $forall xge0forall yge0[x<yimplies f(x)>f(y)]$
– Simply Beautiful Art
Oct 1 '17 at 16:57
add a comment |
Could you please explain how $$sum_{n=1}^inftyfrac e{n^2}=frac{epi^2}6$$
– Blargian
Oct 1 '17 at 16:50
1
See the Basel problem.
– Simply Beautiful Art
Oct 1 '17 at 16:52
1
You probably don't want this, but another interpretation of the conditions for the integral test would be$$(fin C^1)landforall xge0[(f(x)ge0)land f'(x)le0)]impliesleft(sum_{n=0}^infty f(n)<inftyiffint_0^infty f(x)~mathrm dx<inftyright)$$
– Simply Beautiful Art
Oct 1 '17 at 16:56
1
If you don't want to restrict this to differentiable functions, then use $forall xge0forall yge0[x<yimplies f(x)>f(y)]$
– Simply Beautiful Art
Oct 1 '17 at 16:57
Could you please explain how $$sum_{n=1}^inftyfrac e{n^2}=frac{epi^2}6$$
– Blargian
Oct 1 '17 at 16:50
Could you please explain how $$sum_{n=1}^inftyfrac e{n^2}=frac{epi^2}6$$
– Blargian
Oct 1 '17 at 16:50
1
1
See the Basel problem.
– Simply Beautiful Art
Oct 1 '17 at 16:52
See the Basel problem.
– Simply Beautiful Art
Oct 1 '17 at 16:52
1
1
You probably don't want this, but another interpretation of the conditions for the integral test would be$$(fin C^1)landforall xge0[(f(x)ge0)land f'(x)le0)]impliesleft(sum_{n=0}^infty f(n)<inftyiffint_0^infty f(x)~mathrm dx<inftyright)$$
– Simply Beautiful Art
Oct 1 '17 at 16:56
You probably don't want this, but another interpretation of the conditions for the integral test would be$$(fin C^1)landforall xge0[(f(x)ge0)land f'(x)le0)]impliesleft(sum_{n=0}^infty f(n)<inftyiffint_0^infty f(x)~mathrm dx<inftyright)$$
– Simply Beautiful Art
Oct 1 '17 at 16:56
1
1
If you don't want to restrict this to differentiable functions, then use $forall xge0forall yge0[x<yimplies f(x)>f(y)]$
– Simply Beautiful Art
Oct 1 '17 at 16:57
If you don't want to restrict this to differentiable functions, then use $forall xge0forall yge0[x<yimplies f(x)>f(y)]$
– Simply Beautiful Art
Oct 1 '17 at 16:57
add a comment |
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3
Compare, to say $e/n^2$.
– Lord Shark the Unknown
Oct 1 '17 at 16:19
This makes no sense: $$lim_{ntoinfty} sum_{n=1}^inftyfrac{e^{frac{1}{n}}}{n^2} = 0$$
– Simply Beautiful Art
Oct 1 '17 at 16:27
The test for divergence is applied to the sequence and not the series? I.e) if i had written $$ lim_{ntoinfty} frac{e^{frac{1}{n}}}{n^2} = 0$$ that would be fine?
– Blargian
Oct 1 '17 at 16:49
Yes, that is correct.
– Simply Beautiful Art
Oct 1 '17 at 16:53
Don’t forget the $dx$ on your integral
– Chase Ryan Taylor
Oct 1 '17 at 18:24