Limit of $\frac1{n^{50}}\sum\limits_{k=1}^{n}(-1)^{k}k^{50}$ when $n\to\infty$ Evaluate $\lim_{n\to\infty}\dfrac{\sum_{k=1}^{n}(-1)^{k}k^{50}}{n^{50}}$.
Or can we get some formula when $50$ is replaced with $m$?
 A: Let $S_n=n^{-a}\sum\limits_{k=1}^n(-1)^kk^{a}$ with $a=50$ or with any $a\geqslant1$, then, regrouping every $(2k)$th and $(2k-1)$th terms, one sees that
$$
S_{2n}=(2n)^{-a}\sum\limits_{k=1}^n((2k)^{a}-(2k-1)^{a}).
$$
Using the fact that the derivative of $t\mapsto t^{a}$ is $t\mapsto at^{a-1}$, one sees that for every $k$ there exists some $t_k$ in $(0,1)$ such that
$$
(2k)^{a}-(2k-1)^{a}=a(2k-t_k)^{a-1}.
$$
Thus,
$$
S_{2n}=\frac1{2n}\sum\limits_{k=1}^na\left(\frac{k-\frac12t_k}n\right)^{a-1}.
$$
This is a multiple of a Riemann sum of the function $t\mapsto at^{a-1}$ on $[0,1]$ hence
$$
\lim S_{2n}=\frac12\int_0^1at^{a-1}\mathrm dt=\frac12\left[t^{a}\right]_0^1=\frac12.
$$
Furthermore, note that
$$
S_{2n+1}=\left(\frac{2n}{2n+1}\right)^aS_{2n}-1,
$$
and that $\left(\frac{2n}{2n+1}\right)^a\to1$ hence 
$$\lim S_{2n+1}=-\frac12.$$
Finally, for every $a\geqslant1$,
$$
\lim \left(\frac1{n^a}\sum\limits_{k=1}^n(-1)^kk^{a}\right)-(-1)^n\frac12=0.
$$
A: Denote $T_m(n) = \sum\limits_{k=1}^{n} (-1)^m k^m$;$\qquad$ 
$S_m(n) = \sum\limits_{k=1}^{n} k^m$ $\qquad (m\in\mathbb{N})$.
Then $T_m(2n) = -S_m(2n)+2\cdot 2^m\cdot S_m(n)$, $\qquad$
 $T_m(2n+1) = T_m(2n)-(2n+1)^m$.
Since $S_m(n) = \dfrac{n^{m+1}}{m+1}+\dfrac{n^m}{2}+Q_{m-1}(n) $, (see  Sum of powers), 
$\quad$ where $Q_{m-1}(n)$ $-$ some polynomial of $(m-1)$-th degree, $\quad$ then 
$T_m(2n) = \dfrac{(2n)^m}{2}+R_{m-1}(n)$, $\quad$where $R_{m-1}(n)$ $-$ some polynomial of $(m-1)$-th degree.
$T_m(2n+1) = \dfrac{(2n)^m-2(2n+1)^m}{2}+R_{m-1}(n)$.
So,
$$
\lim_{n\to\infty} \dfrac{T_m(2n)}{(2n)^m} = \dfrac{1}{2};
$$
$$
\lim_{n\to\infty} \dfrac{T_m(2n+1)}{(2n+1)^m} = -\dfrac{1}{2}.
$$
