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by **dani.bezerra** on Sat Feb 23, 2013 10:59 am

Hi, I'm trying solve this question $\sum_{n=1}^\infty \frac{e^{\frac{1}{n}}}{n}$ by the comparison test, 'cause is the only test that works. But I'm having difficulties, I can't find other series to compare. Someone could help me with it?

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by **nona.m.nona** on Sun Feb 24, 2013 5:28 pm

dani.bezerra wrote:Hi, I'm trying solve this question $\sum_{n=1}^\infty \frac{e^{\frac{1}{n}}}{n}$ by the comparison test, 'cause is the only test that works. But I'm having difficulties, I can't find other series to compare. Someone could help me with it?

It might be helpful to note that $\frac{e^{\frac{1}{n}}}{n}$ is less than $e^{\frac{1}{n}}$ , that $e^{\frac{1}{n}}$ is equal to $\left(\frac{1}{e}\right)^n$ , and that $\frac{1}{e}$ is less than $1$ .

by **dani.bezerra** on Mon Feb 25, 2013 9:02 am

nona.m.nona wrote:It might be helpful to note that $\frac{e^{\frac{1}{n}}}{n}$ is less than $e^{\frac{1}{n}}$ , that $e^{\frac{1}{n}}$ is equal to $\left(\frac{1}{e}\right)^n$ , and that $\frac{1}{e}$ is less than $1$ .

I don't understand why $e^{\frac{1}{n}}$ is equal to $\left(\frac{1}{e}\right)^n$ . Could you explain better?

by **nona.m.nona** on Mon Feb 25, 2013 12:21 pm

dani.bezerra wrote:I don't understand why $e^{\frac{1}{n}}$ is equal to $\left(\frac{1}{e}\right)^n$ . Could you explain better?

It isn't. My apologies; I must have been tired when I posted this.

However, since $n\, \geq\, 1$ , then $\frac{1}{n}\, \geq\, 0$ and $e^{\frac{1}{n}}\, \geq\, 1$ . As a result, $\frac{e^{\frac{1}{n}}}{n}\, \geq\, \frac{1}{n}$ , and you should have a known result for this last expression.

by **dani.bezerra** on Mon Feb 25, 2013 2:27 pm

Now make sense.. thanks for your explanation..

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