Convergence or divergence of $\left( \frac{\sum_{j=1}^n (-1)^j j^k}{\sum_{j=1}^n j^k} \right)_{n \in \mathbb N}$ for some $k \in \mathbb N$

This sequence is supposed to be the averaging of the sequence $$ (-1, \underbrace{+1,\dots,+1}_{2^k} , \underbrace{-1,\dots,-1}_{3^k}, \underbrace{+1,\dots,+1}_{4^k},\dots,\underbrace{(-1)^n,\dots,(-1)^n}_{n^k},\dots) $$ My guess is that it doesn't converge.

Proof: Take only even $n = 2 \nu$ where $\nu \in \mathbb N$, so by Leibniz theorem for alternating series the numerator is the sequence (with index $\nu$) $$ \sum_{\mu =1}^{2\nu} ((-1)^{2\mu-1} (2\mu-1)^k + (-1)^{2\mu} (2\mu)^k ) = \sum_{\mu =1}^{2\nu} (-(2\mu-1)^k + (2\mu)^k ) > 0 $$ Taking odd $n = 2\nu + 1$, where $\nu \in \mathbb N_0$ $$ -1 + \sum_{\mu =1}^{2\nu+1} ((-1)^{2\mu} (2\mu)^k + (-1)^{2\mu+1} (2\mu+1)^k ) = -1 + \sum_{\mu =1}^{2\nu+1} ((2\mu)^k - (2\mu+1)^k ) < 0 $$ It can be shown futher that this sequence is bounded in $(-1,1)$.

So there are two subsequences which go in different directions and the given sequenece diverges.

Is my argument correct ?

Solution 1:

The Riemann sum $$\frac 1 n \sum_{j=1}^n \left( \frac j n\right)^k\rightarrow \int_0^1x^kdx=\frac 1 {k+1}$$

Thus, $$\sum_{j=1}^nj^k= \frac{ n^{k+1}}{k+1}\left(1+o(1)\right)\tag{1}$$

Note that this already shows that if the numerator is bounded, then the ratio converges to $0$, and quite fast at that!

For the numerator, as you did, let's split between odd and even indices. $$\begin{split} \sum_{j=1}^n (-1)^j j^k &=\sum_{j=1, \text{ j even}}^n j^k - \sum_{j=1, \text{ j odd}}^n j^k\\ &= \sum_{j=1, \text{ j even}}^n j^k - \left ( \sum_{j=1}^n j^k - \sum_{j=1, \text{ j even}}^n j^k\right)\\ &= 2\sum_{\mu=1}^{\lfloor\frac n 2 \rfloor} (2\mu)^k - \sum_{j=1}^n j^k\\ \end{split}$$ Thus $$\begin{split} \frac{\sum_{j=1}^n (-1)^j j^k}{\sum_{j=1}^n j^k} &= \frac{2\sum_{\mu=1}^{\lfloor\frac n 2 \rfloor} (2\mu)^k}{\sum_{j=1}^n j^k} -1\\ &= \frac{2^{k+1}\lfloor\frac n 2 \rfloor^{k+1}\left(1 + o\left(1\right)\right)}{n^{k+1}\left(1 + o\left(1\right)\right)} -1\\ &\leq \left(1+\frac2 n\right)^{k+1}\left(1+o(1)\right)-1\\ \end{split}$$ where we have used the fact that $\frac n 2 \leq \lfloor\frac n 2\rfloor<\frac n 2 +1$.

Clearly, the right-hand side converges to $0$, and so does your ratio.

Note: As an exercise, you can push things further and replace those $o(1)$ with second order terms in $(1)$ and obtain an asymptotic approximation of the ratio.