Prove the Euler-Maclaurin summation formula:
If $m, n \in \mathbb{Z}$ and $f$ has a continuous derivative for $m \leq x \leq n $ then
$\displaystyle \sum_{k=m} ^{n} f(k) = \int_m ^n f(x)$ $dx + \frac{1}{2} f(m) + \frac{1}{2} f(n) + \int_m ^n f'(x) n(x)$ $dx$,
where $n(x) = x - [x] - \frac{1}{2}$, where $[x]$ is the greatest integer less than or equal to $x$.
Hint: For $m \leq k < n$ integrate by parts with $\int$ $dx = n(x) $to obtain
$\displaystyle \int _{k} ^{k + \frac{1}{2}} f(x)$ $\displaystyle dx = \frac{1}{2} f(k) - \int _{k} ^{k + \frac{1}{2}} f'(x) n(x)$ $dx$ and
$\displaystyle \int _{k + \frac{1}{2}} ^{k+1} f(x)$ $dx = \frac{1}{2} f(k+1) - \int _{k + \frac{1}{2}} ^{k+1} f'(x) n(x)$ $dx$,
and sum over $k$. From this show that $\displaystyle \sum_{k=1}^n k^{- \alpha } = \frac{n^{1- \alpha}}{1- \alpha } + c_{\alpha} + O(n^{- \alpha})$ if $0< \alpha < 1 $ for suitable constant $c_{\alpha}$.
This is a problem from Leveque's Number Theory. (Chapter 6, #3). All the section discussed is the following:
$f(x) = o(g(x))$ if $\displaystyle \lim_{x \to \infty} \frac{f(x)}{g(x)} = 0$.
$f(x) = O(g(x))$ if $\displaystyle \frac{f(x)}{g(x)} < M \in \mathbb{R}$ for $x$ sufficiently large.
$f(x)$ is asymptotically equal to $g(x)$ if $\displaystyle \lim_{x \to \infty} \frac{f(x)}{g(x)} = 1$.
There is a constant such that $\displaystyle \sum_{k=1} ^n \frac{1}{k} = \log n + \gamma + O(\frac{1}{n})$.
I am not sure how to start/proceed, even with the hint. Any suggestions?
Just follow the suggestions and take the sum (for $k = 1, \ldots , n-1$) of the combined result
$$\int_k^{k+1} f(x) \, dx = \frac1{2} f(k+1) + \frac1{2} f(k) - \int_k^{k+1} f'(x) n(x) \, dx$$ You should get
$$\sum_{k=1}^n f(k) = \int_1^n f(x) \, dx + \frac{1}{2}(f(n) + f(1)) + \int_1^n \left(x - \lfloor x \rfloor - \frac1{2}\right) f'(x) \, dx.$$
With $f(x) = x^{-\alpha}$ and $f'(x) = -\alpha x^{-(\alpha+1)}$, everything should fall into place for the indicated result after you provide the estimate
$$\left|\int_1^n \left(x - \lfloor x \rfloor - \frac1{2}\right) \frac{\alpha}{x^{\alpha+1}} \, dx \right| \leqslant \int_1^n \left|x - \lfloor x \rfloor - \frac1{2}\right|\frac{\alpha}{x^{\alpha+1}} \, dx \\ \leqslant\int_1^n \frac{\alpha}{x^{\alpha+1}} \,dx \\= 1 - \frac{1}{n^\alpha}\\ = 1 +O(n^{-\alpha})$$