Check if a series $\sum^{\infty}_{n=0} \frac{1}{(2n-1)^2}$ converges Actually a series seems to be easy:
$$\sum^{\infty}_{n=0} \frac{1}{(2n-1)^2}$$
ratio test gives 1, which is not ok, I thought about integral test, but it seems to be too complicated for such a series, please direct me right way
 A: $(2n-1)^2 >n(n+1)$
for $n \ge 2$
so $\frac1{(2n-1)^2}<\frac1{n(n+1)}
=\frac1{n}-\frac1{n+1}$.
This telescopes,
giving a direct proof of convergence.
A: The easiest way to prove it is a convergent series is probably to exploit a telescopic series: the general term is positive and 
$$ \sum_{n\geq 0}\frac{1}{(2n-1)^2} = 2+\sum_{n\geq 1}\frac{1}{(2n+1)^2} \leq 2+\sum_{n\geq 1}\frac{1}{(2n-1)(2n+1)} \tag{A}$$
where:
$$ \sum_{n\geq 1}\frac{1}{(2n-1)(2n+1)} = \frac{1}{2}\sum_{n\geq 1}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right) = \frac{1}{2} \tag{B}$$
leading to $\sum_{n\geq 0}\frac{1}{(2n-1)^2}\leq\frac{5}{2}$. Another chance is given by the fact that $\frac{1}{(2x+1)^2}$ is a decreasing, convex and integrable function over $\mathbb{R}^+$, from which:
$$\sum_{n\geq 1}\frac{1}{(2n+1)^2}\leq \int_{0}^{+\infty}\frac{dx}{(2x+1)^2}=\frac{1}{2}.\tag{C}$$

We may also compute a closed form for the given series: we just need to pick our favourite proof from this historic thread about Basel problem to derive:
$$ \sum_{n\geq 0}\frac{1}{(2n-1)^2} = 1 + \eta(2) = 1+\frac{3}{4}\zeta(2) = \color{blue}{1+\frac{\pi^2}{8}}\approx 2.2337.\tag{D} $$
We may also use series acceleration techniques and/or the Taylor series of the squared arcsine function to show the following identity:
$$ \sum_{n\geq 0}\frac{1}{(2n-1)^2} = 1 +\frac{9}{2}\sum_{n\geq 1}\frac{1}{n^2\binom{2n}{n}}.\tag{E}$$
A: By the limit comparasion test,
$$\sum_{n=1}^\infty \frac{1}{(2n-1)^2} \ \text{converges iff the series} \sum_{n=1}^\infty \frac{1}{n^2} \ \text{converges} $$
Again the last series converges by the integral test. 
Note- I am talking about the sums from $n=1$ (different from your question), but it doesn't matter as the difference is just one term.
