Infinite sum with two parameters Evaluate the convergence of the following infinite sum. I am so bad when it comes to parameters. I don't know where to start.
$\sum_{n=1}^{\infty}\dfrac{(n+1)^a-n^a}{n^b}$, where $a,b\in\mathbb{R}$.
 A: Hint. As $n \to \infty$, by using a Taylor series expansion, one may write
$$
\left(1+\frac1{n}\right)^a=1+\frac{a}n+O\left(\frac{1}{n^{2}}\right)
$$ giving
$$
\frac{(n+1)^a-n^a}{n^b}=\frac{1}{n^{b-a}}\cdot\left(\left(1+\frac1{n}\right)^a-1\right)=\frac{a}{n^{b-a+1}}+O\left(\frac{1}{n^{b-a+2}} \right).
$$
A: According to binomial expansion, we have
$$(n+1)^a=n^a+an^{a-1}+\frac{a(a-1)}2n^{a-2}+\dots$$
So, we have
$$\frac{(n+1)^a-n^a}{n^b}=an^{a-1-b}+\frac{a(a-1)}2n^{a-2-b}+\dots$$
Which is equivalent to Oliver's answer.
Summating this for $n\ge1$, we get
$$=a\zeta(1+b-a)+\frac{a(a-1)}2\zeta(2+b-a)+\dots$$
Assuming we have $b>a$ so that it converges.  For $a\in\mathbb N$, we get a closed form solution in terms of the zeta function:
$$\sum_{n\ge1}{(n+1)^a-n^a\over n^b}=a\zeta(1+b-a)+\frac{a(a-1)}2\zeta(2+b-a)+\dots+\frac{a(a-1)(a-2)\dots3\cdot2\cdot1}{1\cdot2\cdot3\cdot\ldots(a-1)a}\zeta(b)$$
Though I can't expect much better out of this.

The special case is $a=0$, whereupon the whole sum is $0$.
