Evaluate the following infinite sum with parameter So the sum actually originates from the following integral:
$$\int_0^\infty \frac{\left\lfloor x \right\rfloor}{x^{a + 1}}dx$$
where $a$ is a real parameter. I've managed to transform the integral into the following sum (except for the case $a = 0$, which is observed separately), which actually became a lot more interesting for me:
$$\sum_{n=0}^\infty \frac{n}{a}(\frac{(n+1)^a - n^a}{n^a(n+1)^a})$$
But I'm clueless as to how to approach that sum? Any ideas? I'm really keen on figuring this out! Many thanks in advance!
 A: $HINT$
$n\frac{(n+1)^a-n^a}{n^a(n+1)^a}=\frac{n}{n^a}-\frac{n+1-1}{(n+1)^a}=\frac{1}{n^{a-1}}-\frac{1}{(n+1)^{a-1}}+\frac{1}{(n+1)^a}$
Now take cases for $a$
For $a=1$ the series diverge.
A: Notice that $$\frac{\lfloor x\rfloor}{x^{a+1}}\underset{x\rightarrow +\infty}{\sim}\frac{1}{x^a}$$ 
thus the integral converges iff $a>1$. What you did is 
$$ \int_1^{+\infty}{\frac{\lfloor x\rfloor}{x^{a+1}}dx}=\sum_{n=1}^{+\infty}\int_n^{n+1}{\frac{\lfloor x\rfloor}{x^{a+1}}dx}=\sum_{n=1}^{+\infty}{n\int_n^{n+1}{\frac{dx}{x^{a+1}}}}=\sum_{n=1}^{+\infty}\frac{n}{a}\left(\frac{1}{n^a}-\frac{1}{(n+1)^a}\right) $$
Let $N\in\mathbb{N}^*$, we have that
$$ \sum_{n=1}^N{n\left(\frac{1}{n^a}-\frac{1}{(n+1)^a}\right)}=\sum_{n=1}^N{\left(\frac{1}{n^{a-1}}-\frac{n}{(n+1)^a}\right)}=\sum_{n=1}^N\left(\frac{1}{n^{a-1}}-\frac{1}{(n+1)^{a-1}}\right)+\sum_{n=1}^N{\frac{1}{(n+1)^a}}=1-\frac{1}{(N+1)^{a-1}}+\sum_{n=1}^N{\frac{1}{(n+1)^a}}\underset{N\rightarrow +\infty}{\longrightarrow}\zeta(a)$$
This only works if $a>1$, otherwise the integral diverges as it was expected. You finally have
$$ \int_1^{+\infty}{\frac{\lfloor x\rfloor}{x^{a+1}}dx}=\frac{\zeta(a)}{a} $$
