Consider the sum, where $\epsilon > 0$
$S_N(\epsilon) = \sum_{k=1}^\infty \frac{k^N - (k-1)^N}{k^{N+\epsilon}}$
- Does $S_N(\epsilon)$ converge for all $\epsilon > 0$ for a fixed N?
- Does $S_N(\epsilon)$ coverge or diverge for a fixed $\epsilon$ as $N \to \infty$?
I have been unsuccessfully trying to apply convergence tests to solve this.
EDIT: A bit too late for me to edit. As I have already found two nice answers only for question 1 from DonAntonio and Jim. Here's what I came up with for question 1.
$\sum_{k=1}^\infty \frac{k^N - (k-1)^N}{k^{N+\epsilon}} = (\epsilon+N)\int_1^\infty \frac{\lfloor x \rfloor^N}{x^{\epsilon + N + 1}} dx$ And since, $(\epsilon+N)\int_1^\infty \frac{\lfloor x \rfloor^N}{x^{\epsilon + N + 1}} dx < (\epsilon+N)\int_1^\infty \frac{1}{x^{\epsilon + 1}} dx = \frac{\epsilon + N}{\epsilon}$ and since the sum is growing for fixed N hence, it converges for fixed N.
Question 2. I still can't resolve. Edit 2: Thanks Jim. :) Now solved.