Convergence of $\sum_{n=1}^∞ a_n/n^2 $ where $a_n = 1+1/2^a + 1/3^a + ... + 1/n^a$ and $a \gt 0$ Given $a$ > $0$, define the sequence:
$a_n = 1+1/2^a + 1/3^a + ... + 1/n^a$
I have to show the following series converges:
$\sum_{n=1}^∞ a_n/n^2 $
I was thinking about using the Integral of $1/x^a$, So we will get: $a_n \lt \int_1^n 1/x^a =F(n)$.
But I could not go further with my proof.
Any tips?
Thanks!
 A: The statement you are trying to prove:
$$\sum_{n=1}^{\infty} \sum_{k=1}^{n} \frac{k^{-a}}{n^2} < \infty$$
I'm not sure ift this is the easiest way to prove this statement, but here is a proof:
First we change the order of the summations and calculate the first term:
$$\sum_{n=1}^{\infty} \sum_{k=1}^{n} \frac{k^{-a}}{n^2} = \sum_{k=1}^{\infty}\sum_{n=k}^{\infty} \frac{k^{-a}}{n^2} = \frac{\pi^2}{6} 
+ \sum_{k=2}^{\infty}\sum_{n=k}^{\infty} \frac{k^{-a}}{n^2} $$
Then we can use the  fact that $\frac{1}{n^2}$ is strictly decreasing to give an upperbound with  the integal
$$\sum_{k=2}^{\infty}\sum_{n=k}^{\infty} \frac{k^{-a}}{n^2} \le \sum_{k=2}^{\infty} k^{-a} \int_{k-1}^{\infty} \frac{1}{x^2} dx =
 \sum_{k=2}^{\infty} \frac{k^{-a}}{k-1}$$
And for the last step you can prove that for every $a>0$  this is convergent (I leavethis part to you).
A: If $0<a<1$, then $a_n<F(n)=\frac {n^{1-a}-1} {1-a}<\frac {n^{1-a}} {1-a}$ and $\sum \frac 1 {n^{2}} n^{1-a}=\sum \frac 1  {n^{1+a}} <\infty$. The same method works for $a \geq 1$ and I will let you write out  a proof.
[In fact $a_n$ becomes smaller when you increase $a$!].
