# $\sum_{n=0}^{\infty}{e^{\frac{1}{f\left(n\right)}}-1}$ convergence

Let $f\left(n\right)$ be a polynomial in $n$ of degree $m$. What can we say about the convergence of the series

$$S = \sum_{n=0}^{\infty}{e^{\frac{1}{f\left(n\right)}}-1}.$$

Obviously $\lim_{n \rightarrow \infty }S = 0$, but I can't figure out what to do after that. I have been thinking about comparing it with

$$\sum_{n=0}^{\infty}{e^{\frac{1}{n^{m}}}-1}\quad \text{or}\quad \sum_{n=0}^{\infty}{e^{\frac{1}{n}}-1},$$

but this has not made things simpler for me, since I don't know ho w to show that either of those two converges/diverges.

Being a positive terms series, I could rearrange it obtaining

$$\lim_{k\rightarrow\infty}\left(\sum_{n=0}^{k}{e^{\frac{1}{n}}}\right)-k$$

but once again, I'm stuck after this. I'm quite sure that the solution is in front of my eyes, can't anybody please point me in the right direction?

• Isn't that the other way around? that statement is false even for 1.... – kaharas Aug 26 '13 at 13:16
• Duh, yes, sorry, $\lvert e^x - 1\rvert \leqslant 2\lvert x\rvert$ for $\lvert x\rvert \leqslant 1$. – Daniel Fischer Aug 26 '13 at 13:18

Note that $$e^{1/n} - 1 = \frac{1}{n} + \frac{1}{2n^2} + \frac{1}{6n^3} + \cdots \ge \frac 1n$$ so that $$\sum_{n=1}^\infty e^{1/n} - 1$$ diverges.
On the other hand, $$e^{1/n^2} - 1 = \frac{1}{n^2} + \frac{1}{2n^4} + \frac{1}{6n^6} + \cdots \le \frac{1}{n^2} \left(1 + \frac 12 + \frac 16 + \cdots\right) \le \frac{e}{n^2}$$ so that $$\sum_{n=1}^\infty e^{1/n^2} - 1$$ converges.