Good approximation of a partial sum Let $a \in (0,1)$ and consider the partial sum
\begin{equation}
S_{n,a} = \sum_{i=1}^n \frac{1}{(i+1) i^{1+a}}.
\end{equation}
Is there a good approximation of $S_{n,a}$ in terms of $n$ and $a$? This sum is roughly $\sum_{i=1}^n 1 / i^2$ for small $a$. I am trying to use integral-test like approximations but cannot find simple integral approximations. E.g., I can write
\begin{equation}
S_{n,a} = \frac{1}{2} + \sum_{i=2}^n \frac{1}{(i+1) i^{1+a}} \leq \frac{1}{2} + \int_1^n\frac{\mathrm{d} x}{(x+1) x^{(1+a)}}.
\end{equation}
This latter integral does not have a simple closed-form formula. A similar lower bound can derived similarly. In particular, if $a = a_n := 1 / n$, then what is the limit of $S_{n,a}$ as $n \to \infty$? Anyone has an idea? Thanks very much.
 A: I think i have a pretty got approximation. As $n \to \infty$ your series is about equal to $-\frac {a}{a+1}+\frac 32 +H_{\frac a2}-H_{\frac {a+1}{2}}+c$ where c is a very small number, in fact it is less than $0.0005$ for big $a$ and is less than $0.2$ for smaller $a$. Here, $H_n$ is the nth harmonic number. I know that this is not quite what you were asking for but i hope this can help you.
Edit 1:
First you need to use the integral version of summation by parts to get
$$\sum_{i=1}^x \frac 1{(i+1)i^{1+a}} = \frac {\lfloor x\rfloor}{(x+1)x^{1+a}} + \int_1^x \lfloor t \rfloor \frac {at+2t+a+1}{(t+1)^2t^{a+1}} dt$$
Then use the approximation $\lfloor x \rfloor = x$ to get
$$\sum_{i=1}^x \frac 1{(i+1)i^{1+a}} \approx \frac {1}{(x+1)x^{a}} + \int_1^x \frac {at+2t+a+1}{(t+1)^2t^{a}} dt$$
Now let $x \to \infty$
$$\sum_{i=1}^\infty \frac 1{(i+1)i^{1+a}} \approx \int_1^\infty \frac {at+2t+a+1}{(t+1)^2t^{a+1}} dt$$
The next step is complicated, but i transformed the integral, evaluated the it using wolfram alpha and transformated it again to get to
$$-\frac {a}{a+1}+\frac 32 +H_{\frac a2}-H_{\frac {a+1}{2}}$$
If you want to try it yourself use the fact that
$$\int_1^\infty \frac {at+2t+a+1}{(t+1)^2t^{a+1}} dt = \int_0^\infty \frac {a+2}{(t+1)^2t^{a}} dt + \int_0^\infty \frac {a+1}{(t+1)^2t^{a+1}} dt$$
For $H_{x}$ i would recommend using the extension to all the positive real numbers with $H_{x}=\int_0^1 \frac {1-t^x}{1-t} dt$ or $H_x=\psi(x+1)+\gamma$
A: Note that we have
$$\frac1{k+1}=\frac1k\frac1{1+\frac1k}=\sum_{m=1}^\infty\frac{(-1)^{m+1}}{k^m}\tag{$|k|>1$}$$
and hence we can derive bounds such as
$$S_{n,a}-\frac12<\sum_{k=2}^n\frac1{k^{a+2}}<\int_1^n\frac{\mathrm dx}{x^{a+2}}=\frac{1-n^{-(a+1)}}{a+1}\underset{{a=1/n\\n\to\infty}}\longrightarrow1$$
and likewise,
$$S_{n,a}-\frac12>\sum_{k=2}^n\left(\frac1{k^{a+2}}-\frac1{k^{a+3}}\right)>\int_2^{n+1}\frac1{x^{a+2}}-\frac1{x^{a+3}}~\mathrm dx\underset{{a=1/n\\n\to\infty}}\longrightarrow\frac38$$
as it would seem, bounding your limit between $7/8$ and $3/2$.
One may note though that
$$\lim_{(n,a)\to(\infty,0^+)}S_{n,a}=\sum_{k=1}^\infty\frac1{k(k+1)}=1$$
