## Comparison Test for Series

Limits, differentiation, related rates, integration, trig integrals, etc.
dani.bezerra
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### Comparison Test for Series

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?

nona.m.nona
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### Re: Comparison Test for Series

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 $1$.

dani.bezerra
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Joined: Sat Feb 23, 2013 10:35 am
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### Re: Comparison Test for Series

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?

nona.m.nona
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### Re: Comparison Test for Series

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.

dani.bezerra
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Joined: Sat Feb 23, 2013 10:35 am
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### Re: Comparison Test for Series

Now make sense.. thanks for your explanation..