Question: 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 $c_{n+1} > c_{n}$ for infinitely many values of $n$. Also, show that $\sum_{n=1}^\infty (-1)^{n-1} c_n $ converges.
Attempt: Now I want to do induction for the first part but that proof is throwing me off. With the base of $n = 1$, I get $\frac{1}{16} > 1$ which is not true. I don't know if I am thinking this right or if there is a typo. For the second part, if the first part is true, then we cannot use alternating series test. I was leaning towards the ration test, but I am leading to a problem. Do I have to test the casea when n is even and odd? Please give me hints on how to solve this! Thank you very much guys!!
If $n $ is even then
$$c_{n+1}=\frac{1}{(n+1)^2}$$ and $$c_n=\frac {1}{n^4} $$ but
$$n^4-(n+1)^2=$$ $$(n^2+n+1)(n^2-n-1)= $$ $$(n^2+n+1)((n-1)^2+n-2)>0$$ for $n\ge2$.
for the series, we have
$$|(-1)^nc_n|\le \frac {1}{n^2} $$ and comparison test.