Speed of Convergence for some series (double sum) For $\alpha>2$, i want to find the speed of convergence of $$S_n=\sum_{i=1}^n\sum_{j=n-i+1}^n i^{-\alpha}j^{-\alpha}.$$
In particular, i want $\sum_{i=1}^n\sum_{j=n-i+1}^n i^{-\alpha}j^{-\alpha}\leq Cn^{-\alpha+1-\epsilon}$ for some $C,\epsilon>0$.
My first attempt was to estimate by integrals
$$\sum_{i=1}^n\sum_{j=n-i+1}^n i^{-\alpha}j^{-\alpha}\leq \int_{x=1}^n\int_{y=n-x+1}^n x^{-\alpha}y^{-\alpha}dxdy\\=\int_{x=1}^nx^{-\alpha}(n^{-\alpha+1})dx+\int_{x=1}^nx^{-\alpha}(n-x+1)^{-\alpha+1}dx$$
but the second integral on the left is not an easy one.
My second attempt was to show that $$n^{\alpha-1+\epsilon}\sum_{i=1}^n\sum_{j=n-i+1}^n i^{-\alpha}j^{-\alpha}$$
is bounded.
 A: I'm not an expert at this, but looking at your first attempt's integration,
$$I = \int_1^n x^{-\alpha} (n-x+1)^{-\alpha+1} dx = \int_1^n (n - x + 1)^{-\alpha} x^{-\alpha + 1} dx$$
$$\begin{align*} I &= \frac{1}{2}\int_1^n x^{-\alpha} (n - x + 1)^{-\alpha} (n + 1) dx \\ &= \frac{n + 1}{2}\int_{-\frac{n - 1}{2}}^{\frac{n - 1}{2}} \left(\frac{(n + 1)^2}{4} - x^2\right)^{-\alpha} dx & x' = x - \frac{n + 1}{2} \\ &= v\int_{1 - v}^{v - 1} \frac{1}{(v^2 - x^2)^{\alpha}} dx\end{align*}$$
Where $v = \frac{n + 1}{2}$ is a constant. Perhaps someone else can fill in the integration steps (which should either be a trig-sub recurrence or partial fractions), but by using Sage, I get that
$$I \sim C(\alpha) v^{-\alpha}$$
Where $C(\alpha)$ is a constant that depends only on $\alpha$ and not $n$. In particular, your $\epsilon$ seems to be $1$ but should be $\left(\frac{n}{2}\right)^{-\alpha}$ instead.
a = 5
v = var('v')
it = integrate(1/(v^2-x^2)^a,x,1-v,v-1)
print([it.subs(v=10^k).log(10).n() for k in range(1, 8)])

Hope this helps at least slightly.
A: Starting from @Gareth Ma's answer
$$I_a=v\int_{1 - v}^{v - 1} \frac{1}{(v^2 - x^2)^{\alpha}}\, dx$$
$$I_a=2\, (v-1)\, v^{1-2 a} \,\,
   _2F_1\left(\frac{1}{2},a;\frac{3}{2};\frac{(v-1)^2}{v^2}\right)$$
What it seems is that $v^aI_a$ is almost a straight line with positive slope.
A: $$I_a=\int_{1}^n\int_{n-x+1}^n x^{-a}y^{-a}\,dy\,dx$$
$$I_a=\frac 1{a-1}\int_{1}^n \left(n^a (n-x+1)-n (n-x+1)^a\right) (n x (n-x+1))^{-a}\,dx$$
$$(a-1)n^aI_a=\frac{n \left(n^{1-a}-1\right)}{a-1}+$$
$$\frac{(n+1)^{1-a} \left(n^a \, _2F_1\left(1-a,a;2-a;\frac{1}{n+1}\right)-n \,
   _2F_1\left(1-a,a;2-a;\frac{n}{n+1}\right)\right)}{a-1}+$$
$$\frac{(n+1)^{-a} \left(n^2 \, _2F_1\left(2-a,a;3-a;\frac{n}{n+1}\right)-n^a \,
   _2F_1\left(2-a,a;3-a;\frac{1}{n+1}\right)\right)}{a-2}$$
Now, for large values of $n$
$$\, _2F_1\left(1-a,a;2-a;\frac{1}{n+1}\right) \quad \to \quad1$$
$$\, _2F_1\left(1-a,a;2-a;\frac{n}{n+1}\right)\quad \to\quad \frac{\sqrt{\pi } 2^{2 a-1} \Gamma (2-a)}{\Gamma \left(\frac{3}{2}-a\right)}$$
$$\, _2F_1\left(2-a,a;3-a;\frac{1}{n+1}\right) \quad \to \quad1$$
$$\, _2F_1\left(2-a,a;3-a;\frac{n}{n+1}\right)\quad \to\quad-\frac{\sqrt{\pi } 4^{a-1} (a-2) \Gamma (1-a)}{\Gamma \left(\frac{3}{2}-a\right)}$$ which, hoping no mistakes, makes
$$I_a=\frac{a-1}{4}  (n+1)^{-a}\Bigg[4\frac{(a-2) n-1}{(a-2) (a-1)} +\frac{\sqrt{\pi } 4^a (n+2) n^{1-a} \Gamma (1-a)}{\Gamma
   \left(\frac{3}{2}-a\right)}\Bigg]$$
For $n=100$ and $a=2\pi$, this gives $2.4463\times 10^{-14}$ while the exact value would be the same
A: We will show that $S_n=o(n^{-(\alpha-1)})$. We claim
$$n^{(\alpha-1)}S_n \approx  n^{(\alpha-1)} \sum_{i=1}^n i^{-\alpha}\left[(n-i+1)^{-(\alpha-1)}-n^{-(\alpha-1)}\right] =\sum_{i=1}^n i^{-\alpha}\left[\left(1+\frac{1-i}{n}\right)^{-(\alpha-1)}-1\right]$$ converges to $0$ where we replace the inner sum (over $j$) with difference of the tails.
Let $k=i-1$. Then
\begin{align*}
\sum_{i=1}^n i^{-\alpha}\left[\left(1+\frac{1-i}{n}\right)^{-(\alpha-1)}-1\right]&=\sum_{k=0}^{n-1} (k+1)^{-\alpha}\left[\left(1-\frac{k}{n}\right)^{-(\alpha-1)}-1\right]\\
&=0+\sum_{k=1}^{n-1} (k+1)^{-\alpha}\left[\left(1-\frac{k}{n}\right)^{-(\alpha-1)}-1\right]\\
&\leq \sum_{k=1}^{n-1} k^{-\alpha}\left[\left(1-\frac{k}{n}\right)^{-(\alpha-1)}-1\right].
\end{align*}
Let $a_k(n)=k^{-\alpha}\left[\left(1-\frac{k}{n}\right)^{-(\alpha-1)}-1\right]$. Applying the generalized binomial theorem we have that $$a_k(n)\leq \frac{1}{n}C_{\alpha}k^{-(\alpha-1)}\left(1-\frac{k}{n}\right)^{-(\alpha-1)}.$$
Together with the fact that $$\left(\frac{n}{k(n-k)}\right)^{(\alpha-1)}=2^{(\alpha-1))}\left(\frac{n}{2k(n-k)}\right)^{(\alpha-1)}<2^{(\alpha-1)}\frac{n}{2k(n-k)}$$
we have that
$$n^{(\alpha-1)}S_n\leq \frac{C_{\alpha}}{n}\sum_{k=1}^{n-1}\frac{n}{k(n-k)}\leq \frac{C_{\alpha}}{n}\int_{1}^{n-1}\frac{n}{x(n-x)}dx=\frac{C_{\alpha}}{n}2\log{(n-1)}\to 0.$$
