Sum's of powers I'm trying to show $$\sum_{n \leq x}\sigma_{\alpha}=\frac{\zeta(\alpha+1)}{\alpha + 1}x^{\alpha+1}+O(x^{\alpha})$$
I have come to a step where I need to prove that $$\sum_{n \leq x}=\frac{x^{\alpha +1}}{\alpha+1}+O(x^{\alpha})$$
I'm unsure how to show this part, the most I have done is saw that, 
$$1+2^{\alpha}+ \cdots + x^{\alpha} \leq xx^{\alpha}=x^{\alpha+1}$$
Any help with this would be greatly appreciated thanks!
 A: Since $\sigma_{\alpha}(n)=\sum_{d\mid n}d^{\alpha}$, for any $s$ large enough we have
$$ \sum_{n\geq 1}\frac{\sigma_\alpha(n)}{n^s} = \zeta(s)\zeta(s-\alpha)$$
and the LHS is actually convergent for any $s>\alpha+1$. In a right neighbourhood of $s=\alpha+1$ the LHS behaves like $\left(\zeta(\alpha+1)+o(1)\right)\left(\frac{1}{s-1-\alpha}+\gamma+o(1)\right)$. By invoking the Hardy-Littlewood tauberian theorem we have that 
$$ \sum_{n\leq N}\frac{\sigma_\alpha(n)}{n^{\alpha}}\sim \zeta(\alpha+1) N $$
and the claim follows by summation by parts.
A: the key to the approximation you wish to derive is an elementary but powerful summation formula due to Euler. this states that for a function $f \in C^1([y,x])$ 
$$
\sum_{y \lt n \le x} f(n) = \int_y^x f(t)dt +\int_y^x (t-\lfloor t \rfloor)f'(t)dt
+f(x)(\lfloor x\rfloor -x) - f(y)(\lfloor y \rfloor-y) \tag{E}
$$
i will append the short proof, but for the moment, assuming this statement (E), and considering a particular application with $y=1$ and $f(t)=t^{\alpha}$ this gives, for $\alpha \gt 0$:
$$
\sum_{n \le x} n^{\alpha} =1+\sum_{1 \lt n \le x} n^{\alpha}  \\
= \int_1^x t^{\alpha} dt + \bigg(1 +\int_1^x(t-\lfloor t \rfloor)\alpha t^{\alpha-1}dt +x^{\alpha}(\lfloor x \rfloor - x) \bigg)\\
= \frac{x^{\alpha+1}}{\alpha+1} + O(x^{\alpha})
$$
which is the result you need. (because $\alpha \gt 0$ the constant term $1-\frac1{\alpha+1}$ may be absorbed into the $O(x^{\alpha})$)
to prove Euler's summation formula, we note that for all $n$ satisfying $y \le n-1$ and also $n \le x$ we have:
$$
\begin{align}
\int_{n-1}^n \lfloor t \rfloor f'(t)dt &= (n-1)(f(n)-f(n-1) \\
                                       &=nf(n) -(n-1)f(n-1) -f(n) \\
\end{align}
$$
summing from $n=\lfloor y \rfloor +1$ to $\lfloor x \rfloor$ we obtain:
$$
\int_{\lfloor y \rfloor }^{\lfloor x \rfloor} \lfloor t \rfloor f'(t)dt = \lfloor x \rfloor f(\lfloor x \rfloor) -\lfloor y \rfloor f(\lfloor y \rfloor) - \sum_{y \lt n \le x}f(n)
$$
and, therefore,
$$
\int_{ y}^{ x} \lfloor t \rfloor f'(t)dt = \lfloor x \rfloor f( x ) -\lfloor y \rfloor f( y) - \sum_{y \lt n \le x}f(n) \tag{1}
$$
but integration by parts also gives us:
$$
\int_y^x tf'(t)dt = xf(x|) -yf(y) - \int_y^x f(t)dt \tag{2}
$$
now subtracting (1) from (2) and rearranging gives us (E)
