I can approximate the sum of $\frac 1 {n^2}$ using its integral. But what about $(-1)^n\frac 1 {n^2}$? Is it possible to approximate this using integrals?
I want to know if there are other ways than using two terms in the series to get a positive series ie: $\frac 1 {n^2}-\frac 1 {(n+1)^2}$
I could use cos/sin like this:
$$\sum _{n=1}^{\infty } \frac{(-1)^n}{n^2}\sim\sum _{n=1}^m \frac{(-1)^n}{n^2}+\frac{\pi}{2} \int_{m+\frac12}^{\infty } \frac{\cos (\pi x)}{x^2} \, dx\sim\sum _{n=1}^m \frac{(-1)^n}{n^2}-\frac{\sin \left(\pi \left(m+\frac{1}{2}\right)\right)}{2 \left(m+\frac{1}{2}\right)^2}$$
as $m\rightarrow \infty$
but oftentimes it is impractical to find the antiderivatives of such functions.
If you want to work with $\frac{(-1)^n}{n^2}$, lets consider the partial sums: $$ S_n=\sum_{k=0}^m\frac{1}{(2k)^2}-\sum_{k=0}^m\frac{1}{(2k+1)^2}. $$ For the sequel, we will consider $S_m-S_n$ for $m>n$. In particular, those sums would be $$ S_m-S_n=\sum_{k=n+1}^m\frac{1}{(2k)^2}-\sum_{k=n+1}^m\frac{1}{(2k+1)^2}. $$
From the integral test, we know that $$ \int_{n+1}^{m+1}\frac{1}{(2x)^2}dx\leq\sum_{k=n+1}^m\frac{1}{(2n)^2}\leq\int_{n}^m\frac{1}{(2x)^2}dx $$ and $$ \int_{n+1}^{m+1}\frac{1}{(2x+1)^2}dx\leq\sum_{k=n+1}^m\frac{1}{(2n+1)^2}\leq\int_{n}^m\frac{1}{(2x+1)^2}dx $$ Then the difference is bounded by: $$ \int_{2n+1}^{2n}\frac{1}{x^2}dx-\int_{2m+1}^{2m+2}\frac{1}{x^2}dx=\int_{n+1}^{m+1}\frac{1}{(2x)^2}dx-\int_{n}^m\frac{1}{(2x+1)^2}dx\leq\sum_{k=n+1}^m\frac{1}{(2n)^2}-\sum_{k=n+1}^m\frac{1}{(2n+1)^2} \leq\int_{n}^m\frac{1}{(2x)^2}dx-\int_{n+1}^{m+1}\frac{1}{(2x+1)^2}dx=\int_{2n}^{2n+3}\frac{1}{x^2}dx-\int_{2m}^{2m+3}\frac{1}{x^2}dx. $$ Assuming that I did the calculations correctly, this gives that the even partial sums (there are the same number of positive and negative terms) are Cauchy. A similar calculation will work for odd partial sums.
I know that the original statement was about approximations, but you can use the same ideas by letting $m\rightarrow\infty$ to get approximations.