I'm stumped with the convergence/divergence of this unusual looking series $$\sum_{n=1}^\infty\left(\left(1+\frac{1}{n}\right)^n - e\right)^\sqrt2$$ The summand clearly goes to 0 as n goes to infinity, but I'm totally lost. I've tried the ratio test, looking at it I can't think of a comparison to use that test. My guess would be that it does converge, but I have no idea how to go about proving it or disproving it. Please help me guys this is driving me nuts. What am I missing here?
 A: $$\left(1+\frac1n\right)^n = \exp\left(n \ln\left(1+\frac1n\right)\right)$$  Now $$\ln\left(1+\frac1n\right) = \frac1n - \frac{1}{2n^2} + O\left(\frac{1}{n^3}\right)$$
so
$$\left(1+\frac{1}{n}\right)^n = \exp\left(1 - \frac{1}{2n} + O\left(\frac{1}{n^2}\right)\right) = e \; \left(1 - \frac{1}{2n} + O\left(\frac{1}{n^2}\right)\right)$$
$$\left(\left(1+\frac{1}{n}\right)^n - e\right)^{\sqrt{2}} =\left(-\frac{1}{2n} + O\left(\frac{1}{n^2}\right)\right)^\sqrt{2} $$
Note that this is an irrational power of a  negative number, so some care is needed in its definition. If you don't want to allow complex numbers, you'll just have to say it's undefined. The general definition is $a^b = \exp(b \ln a)$, where $\ln a$ is any branch of the (multivalued) complex logarithm of $a$.
Now I'll assume you're using the principal branch, so
$$ \ln\left(-\frac{1}{2n} +  O\left(\frac{1}{n^2}\right)\right) = -\ln n + i \pi  - \ln 2 + O\left(\frac1n\right)$$
and thus
$$\eqalign{ \left(\left(1+\frac{1}{n}\right)^n - e\right)^{\sqrt{2}} &=
\exp\left(- \sqrt{2} \ln n + \sqrt{2}(i \pi \sqrt{2} - \ln 2) + O\left(\frac{1}{n}\right)\right)\cr 
&= n^{-\sqrt{2}} \exp\left( \sqrt{2}(i \pi \sqrt{2} - \ln 2)\right) + O\left(n^{-1-\sqrt{2}}\right) }$$
and since $\sum_n n^p$ converges for any $p > 1$, your series converges.
