TITLE: Infinite sums: adding terms
QUESTION [9 upvotes]: I would like to know where I can find a formal treatment of an idea I had, assuming it makes sense. Consider the infinite sum
$$\sum\limits_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$$
and define the partial sum
$$S_N=\sum\limits_{n=1}^{N}\frac{1}{n^2}$$
I guess that
$$R_N = S_N + \frac{1}{N}$$
should converge faster as $N\rightarrow\infty$. Does it make sense? Any formal theory? Thanks!!!

REPLY [4 votes]: Let $f(x)=1/x^2$.  Then, using the Euler-Mclaurin Sum Formula we obtain
$$\begin{align}
\sum_{n=1}^N\frac1{n^2}&=\int_1^Nf(x)\,dx+\frac12\left(f(N)+f(1)\right)+\sum_{k=1}^{K}\frac{B_{2k}}{(2k)!}\left(f^{(2k-1)}(N)-f^{(2k-1)}(1)\right)+R_K\\\\
&=1-\frac1N+\frac12\left(\frac1{N^2}+1\right)+\sum_{k=1}^{K}B_{2k}\left(1-N^{-(2k+1)}\right)+R_K\\\\
&=\frac{\pi^2}{6}-\frac1N+\frac{1}{2N^2}-\frac1{6N^3}+\frac1{30N^5}-\frac1{42N^7}+\frac1{30N^9}-\frac5{66N^{11}}+O(N^{-13}) \tag 1
\end{align}$$
Therefore, we see that the modified truncated series
$$\sum_{n=1}^N\frac1{n^2}+\frac1N=\frac{\pi^2}{6}+O\left(\frac1{N^2}\right)$$
which converges faster than the un-modified truncated by an order of $1/N$, the former converging as $1/N$ while the latter converging as $1/N^2$.  
And of course, we can improve even further using more terms in the Euler-Mclaurin Summation Formula.  Using the expansion in $(1)$, we have the following results for both the actual truncated sum and the terms from the Euler-Mclaurin Sum Formula (EMSF):
For $N=10$, $\sum_{n=1}^{10}\frac1{n^2}\approx 1.54976773116654$, while the EMSF gives approximately $1.54976773116652$.
For $N=100$, $\sum_{n=1}^{100}\frac1{n^2}\approx 1.63498390018489$, while the EMSF gives approximately $1.63498390021823$.
For $N=1000$, $\sum_{n=1}^{1000}\frac1{n^2}\approx 1.64393456668156$, while the EMSF gives approximately $1.64393456671489$.