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Consider the sum, where $\epsilon > 0$

$$S_N(\epsilon) = \sum_{k=1}^\infty \frac{k^N - (k-1)^N}{k^{N+\epsilon}}$$

  1. Does $S_N(\epsilon)$ converge for all $\epsilon > 0$ for a fixed N?
  2. 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.

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up vote 2 down vote accepted

For fixed $N$, we can use the Limit Comparison Test: $$ \lim_{k\to\infty} \frac{(k^N-(k-1)^N)/k^{N+\epsilon}}{1/k^{1+\epsilon}} \;=\; \lim_{k\to\infty} \frac{k^N-(k-1)^N}{k^{N-1}} \;=\; N. $$ Since $\displaystyle\sum_{k=1}^\infty \frac{1}{k^{1+\epsilon}}$ converges, the given series converges as well.

Now, consider what happens if we fix $k$ and $\epsilon$, and vary $N$. We have $$ \frac{k^N-(k-1)^N}{k^{N+\epsilon}} \;=\; \frac{1}{k^\epsilon}\left(1-\left(1-\frac{1}{k}\right)^N\right) $$ As we increase $N$, the quantity on the right increases, approaching $\dfrac{1}{k^\epsilon}$ as $N\to\infty$. Therefore, by the Monotone Convergence Theorem, $$ \lim_{N\to\infty} \sum_{k=1}^\infty \frac{k^N-(k-1)^N}{k^{N+\epsilon}} \;=\; \sum_{k=1}^\infty \frac{1}{k^\epsilon} $$ Thus $S_N(\epsilon)$ diverges as $N\to\infty$ for $\epsilon\leq 1$, and converges for $\epsilon>1$.

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Awesome, I will look into this. I just posted my edit and came back to notice your answer :) – Roupam Ghosh Jun 22 '12 at 3:04

Since $\,a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\ldots +ab^{n-2}+b^{n-1})\,$ , we get:

$$k^N-(k-1)^N=k^{N-1}+k^{N-2}(k-1)+\ldots +k(k-1)^{N-2}+(k-1)^{N-1}\leq N\,k^{N-1}$$ so

$$\frac{k^N-(k-1)^N}{k^{N+\epsilon}}\leq N\frac{1}{k^{1+\epsilon}}$$ and our series converges by the comparison test (for $\,N\,$ fixed, any $\,\epsilon > 0\,$).

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