Determine whether the series is absolutely convergent, conditionally convergent, or divergent.

[HardlyPuzzled]

I've attached a picture of my work but I am not sure if it's completely correct. Does the integral test automatically prove absolute convergence?

 

[MarkFL]

After the substitution, I would write:

[MATH]\lim_{t\to\infty}\left(\int_{\frac{1}{t}}^1 e^u\,du\right)=\lim_{t\to\infty}\left(e-e^{\frac{1}{t}}\right)=e-1[/MATH]
This is of course equivalent to what you wrote, however, I needed to verify your result. Now, consider the following diagram:



This demonstrates visually that:

[MATH]\sum_{k=2}^{\infty}\frac{e^{\frac{1}{k}}}{k^2}<\int_1^{\infty} \frac{e^{\frac{1}{x}}}{x^2}\,dx=e-1[/MATH]
And so we may conclude that:

[MATH]e+\sum_{k=2}^{\infty}\frac{e^{\frac{1}{k}}}{k^2}=\sum_{k=1}^{\infty}\frac{e^{\frac{1}{k}}}{k^2}[/MATH]
converges.

 

[lookagain]

After the substitution, I would write:

[MATH]\lim_{t\to\infty}\left(\int_{\frac{1}{t}}^1 e^u\,du\right)=\lim_{t\to\infty}\left(e-e^{\frac{1}{2}}\right)=e-1[/MATH]

You have a typo. The exponent for e should be (1/t), not (1/2). However, you
worked it as if it were (1/t).

 

[Steven G]

After the substitution, I would write:

[MATH]\lim_{t\to\infty}\left(\int_{\frac{1}{t}}^1 e^u\,du\right)=\lim_{t\to\infty}\left(e-e^{\frac{1}{2}}\right)=e-1[/MATH]
This is of course equivalent to what you wrote, however, I needed to verify your result. Now, consider the following diagram:

View attachment 11826

This demonstrates visually that:

[MATH]\sum_{k=2}^{\infty}\frac{e^{\frac{1}{n}}}{n^2}<\int_1^{\infty} \frac{e^{\frac{1}{x}}}{x^2}\,dx=e-1[/MATH]
And so we may conclude that:

[MATH]e+\sum_{k=2}^{\infty}\frac{e^{\frac{1}{n}}}{n^2}=\sum_{k=1}^{\infty}\frac{e^{\frac{1}{n}}}{n^2}[/MATH]
converges.

Mark, unless I am completely lost I suspect that the index for the summation should be n=2 to infinity rather than k=2 to infinity. My real question is why is it not n=1 to infinity?
Thanks for reading this and for your anticipated reply.

 

[MarkFL]

You have a typo. The exponent for e should be (1/t), not (1/2). However, you
worked it as if it were (1/t).

Yes, just a typo.

Mark, unless I am completely lost I suspect that the index for the summation should be n=2 to infinity rather than k=2 to infinity. My real question is why is it not n=1 to infinity?
Thanks for reading this and for your anticipated reply.

Another typo.

 

[Steven G]

Yes, just a typo.



Another typo.

Mark, yes you meant to use n instead of k but did you mean to start at n=2 or n=1. If you meant n=2 my question is why not n=1?

 

[MarkFL]

The leftmost rectangle representing the series is not under the curve where the definite integral is defined. It's like a right Riemann sum.

 

[Steven G]

Yes, I am not fully awake yet. You are using the rectangle that starts at x=1, but it crosses the curve at x=2. Thanks for pointing out the obvious to me.

 

[MarkFL]

Hehe...after all the typos I made last night I can't say a word.