Proving two interesting series related with integral When working on series and integrals, I attained two interesting limits:
$$
\lim_{M\to \infty}\left\{\sum_{i=0}^{M-1}\left({i\over M}\right)^\alpha-{M\over \alpha+1}\right\}=-{1\over 2}
$$
$$
\lim_{M\to \infty}\left\{M\cdot \sum_{i=0}^{M-1}\left({i\over M}\right)^\alpha-{M^2\over \alpha+1}+{M\over 2}\right\}=-{\alpha\over 12}
$$
for $\alpha>0$. Now my question is:

How could these limits be found using the definition of integral? One I my friends advised to use a trapezoid approach for that, but I'm suspicious. Would it give me the answer to both of the limits?

Also

what is the value for the following limit?
  $$
\lim_{M\to \infty}\left\{M^2\cdot \sum_{i=0}^{M-1}\left({i\over M}\right)^\alpha-{M^3\over \alpha+1}+{M^2\over 2}+{M\alpha\over 12}\right\}
$$

 A: $$\frac 1M \sum_{i=0}^{M-1}\left(\frac iM\right)^\alpha$$
is an approximation for the integral
$$
 \int_0^1 x^\alpha \, dx = \frac{1}{\alpha+1} \, .
$$
The difference can be computed with Taylor's formula for $f(x) = x^{\alpha+1}$ on each interval $[\frac iM, \frac{i+1}{M}]$:
$$
\int_{i/M}^{(i+1)/M} x^\alpha \, dx = \frac{1}{\alpha+1}  \left(\left(\frac {i+1}M\right)^{\alpha+1} -  \left(\frac iM\right)^{\alpha+1} \right) \\
= \frac 1M \left(\frac iM\right)^\alpha + \frac{\alpha}{2M^2} \left(\frac iM\right)^{\alpha-1} + O\left( \frac{1}{M^3}\right) \, .
$$
Now sum  these identities and multiply by $M$. It follows that
$$
\frac{M}{\alpha+1} - \sum_{i=0}^{M-1}\left(\frac iM\right)^\alpha = \frac{\alpha}{2} \frac 1M \sum_{i=0}^{M-1}\left(\frac iM\right)^{\alpha-1} + O\left( \frac{1}{M}\right)
$$
For $M \to \infty$ the right-hand side has the limit
$$
\frac{\alpha}{2}  \int_0^1 x^{\alpha -1} \, dx = \frac 12 \, ,
$$
that is your first limit.
The second and third limit can be obtained in the same way, by using more terms in Taylor's formula for $f(x) = x^{\alpha+1}$.
A: If  $\alpha$ is a positive integer then your limits can be easily derived from the Faulhaber's formula 
$$\sum _{i=0}^{M-1}i^{\alpha}=\frac{1}{\alpha+1}\sum _{j=0}^{\alpha}{\alpha+1 \choose j}B_{j}(M-1)^{\alpha+1-j}$$
where $B_j$ is the $j$-th Bernoulli number.
