I'm working on the following problem:
Let $\{ c_n \}_{n=1}^{\infty}$ be a sequence such that $c_n = \frac{1}{n^2}$ if $n$ is odd, and $c_n = \frac{1}{n^4}$ if $n$ is even. Prove that (i) $c_{n+1} > c_n$ for infinitely many $n$ and (ii) $\sum_{n=1}^{\infty} (-1)^{n-1} c_n$ converges.
My attempt:
(i) We claim that when $n$ is even, $$c_{n+1} > c_n$$ and proceed via induction. First, note that for $n=2$, $$c_3 = \frac{1}{9} > \frac{1}{16} = c_2.$$ Now, suppose that $c_{k+1} > c_k$ for some even $k$. Then, we prove that \begin{align*} c_{k+3} &> c_{k+2} \\ \frac{1}{(k+3)^2} &> \frac{1}{(k+2)^4} \\ \end{align*} by showing that $$(k+2)^4 > (k+3)^2.$$ Accordingly, \begin{align*} (k+2)^4 &= k^4 + 8k^3 + 24k^2 + 32k + 16 \\ &>^{I.H.} (k^2 + 2k + 1) + 8k^3 + 24k^2 + 32k + 16 \\ &= 8k^3 + 25k^2 + 34k + 17. \\ \end{align*} Now, \begin{align*} 8k^3 + 25k^2 + 34k + 17 &> (k+3)^2 \\ \Rightarrow 8k^3 + 24k^2 + 28k + 8 &> 0 \\ \end{align*} holds as $k>0$ and each coefficient of the polynomial is positive. Thus, $c_{n+1} > c_n$ for each even $n$.
(ii) To see that $\sum_{n=1}^{\infty} (-1)^{n-1} c_n$ converges, we first note that
$$ |(-1)^{n-1} c_n | = c_n \leq \frac{1}{n^2} \qquad \forall n \in \mathbb{N} $$
Further, $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges via $p$-Test with $p = 2 > 1$, so
$$ \sum_{n=1}^{\infty} (-1)^{n-1} c_n $$
converges via the Weierstrass M-test.
Is this sound reasoning? The inductive portion seems long, as the result is somewhat intuitive. Is there an shorter way to prove this?