Partial sum of $\sum \frac {1} {k^2}$ It is pretty well-known that $\sum_{k = 1}^{\infty} \frac {1} {k^2} = \frac {\pi^2} {6}$. I am interested in evaluating the partial sum $\sum_{k = 1}^{N} \frac {1} {k^2}$. Here is what I have done so far. Since we have
$$\sum_{k = 1}^{N} \frac {1} {k^2} = 1 + \sum_{k = 2}^{N} \frac {1} {k^2}
< 1 + \sum_{k = 2}^{N} \frac {1} {k^2 - 1} =\\= 1 + \frac {1} {2} \left ( \sum_{k = 2}^{N} \frac {1} {k - 1} - \sum_{k = 2}^{N} \frac {1} {k + 1} \right )
= 1 + \frac {1} {2} \left ( \frac {3} {2} - \frac {2N + 1} {N^2 + N} \right )
= \frac {7} {4} - \frac {2N + 1} {2 N^2 + 2 N}$$
and
$$\sum_{k = 2}^{N} \frac {1} {k^2 - 1} - \sum_{k = 2}^{N} \frac {1} {k^2} < \int_{2}^{N} \left ( \frac {1} {x^2 - 1} - \frac {1} {x^2} \right ) \textrm {d}x = \int_{2}^{N} \frac {\textrm {d}x} {x^2 - 1} - \int_{2}^{N} \frac {\textrm {d}x} {x^2} < \frac {1} {N^2 - N} - \frac {1} {2} + \log \sqrt {3},$$
we have
$$\frac {7} {4} + \frac {1} {2} - \log \sqrt {3} - \frac {2 N^2 + N + 1} {2N^3 - 2N} < \sum_{k = 1}^{N} \frac {1} {k^2} < \frac {7} {4} - \frac {2N + 1} {2 N^2 + 2 N}$$
But how to obtain a more precise expression without using analytic methods such as Euler summation?

The Euler summation formula gives, for all $x\ge 1$,
$$
\sum_{n\le x}\frac{1}{n^2}=\frac{\pi^2}{6}-\frac{1}{x}+O(x^{-2}),
$$
by taking $f(t)=t^{-2},\, f'(t)=-2t^{-3}$.
