Infinite sums: adding terms 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!!!
 A: 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 $1.64393456671489$.
A: Let us look at the truncation error. We have for $k\gt 1$
$$\frac{1}{k(k+1)}\lt \frac{1}{k^2}\lt \frac{1}{(k-1)k}.$$
Using the partial fraction decomposition $\frac{1}{t(t+1)}=\frac{1}{t}-\frac{1}{t+1}$, and telescoping sums, we find that the truncation error $E_N$ when we truncate at the $\frac{1}{N^2}$ term satisfies the inequality
$$\frac{1}{N+1}\lt E_N\lt \frac{1}{N}.$$
Truncating of course underestimates the infinite sum. So the error when we add your suggested $\frac{1}{N}$ correction has absolute value less than $\frac{1}{N}-\frac{1}{N+1}$, that is, $\frac{1}{N(N+1)}$. For large $N$, this is much smaller than $E_N$, since $E_N\gt \frac{1}{N+1}$.
Remark: Nice idea! We used telescoping sums to give bounds on the truncation error. It is more common to use integrals.  For something much more elaborate that uses a similar insight, please look at Euler-Maclaurin summation.
A: First, since the sum is monotonically increasing, we can use the Euler-Maclaurin formula,
$$(1) \quad \lim_{N \to {\infty}} \sum_{n=1}^N {1 \over {n^2}}= \lim_{N \to {\infty}} {{\pi^2} \over {6}}-1+\int_1^N {1 \over {n^2}} \ dn$$
We also know that $(1)$ becomes an equality in the limit. That's why we added a constant after applying the Euler formula.  It's the error term.
However, we know what the integral is,
 $$\int_1^N {1 \over 
{n^2}} \ dn=1-{1 \over N}$$
Therefore,
$$\lim_{N \to {\infty}} \sum_{n=1}^N {1 \over {n^2}} = \lim_{N \to {\infty}}  {{\pi^2} \over {6}} -{1 \over N}$$
Since we'd like to retain non asymptotic approximations, we can't arbitrarily add terms. Rather, I'd suggest adding, ${1 \over N}$ to eliminate the ${1 \over N}$ term from the asymptotic expansion. This speeds up the convergence, as there will no longer be a subtraction term. Instead you'll have,
$$\lim_{n \to {\infty}} \sum_{n=1}^N {1 \over {n^2}}+{1 \over N}={{\pi^2} \over {6}}$$
I don't know if you think this defeats the purpose or not, but this is a more formal way. Below is a graph. Red is the partial summations of $\sum {1 \over n^2}$. Green is the newly adjusted terms. Orange is ${{\pi^2} \over 6}$.

