1
$\begingroup$

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?

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

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.

The extension of this idea to the general case should be straightforward.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.