$S=\sum_{j=1}^{\infty}\frac{1}{j^2(j+1)^2}$
How can I find the lower and upper bound of the above sum? I have tried to write the sum as $\sum_{j=1}^{\infty}\frac{1}{2j+1}[\frac{1}{j^2}-\frac{1}{(j+1)^2}]$ but of no help to get an idea about the lower and upper limits.
On
Since $f(x) = 1/(x^2 (x+1)^2)$ is a decreasing function of $x$ on $[1,\infty)$, you can use the integral bounds $$ \int_{N+1}^\infty f(x)\; dx \le \sum_{j=N+1}^\infty f(j) \le \int_{N}^\infty f(x)\; dx $$ so that $$\sum_{j=1}^N f(j) + \int_{N+1}^\infty f(x)\; dx\le S \le \sum_{j=1}^N f(j) + \int_N^\infty f(x)\; dx$$ which will give you bounds as close as you wish, at the cost of computing a partial sum. Note that $$ \int_N^\infty \frac{dx}{x^2(x+1)^2} = 2 \ln\left(\frac{N}{1+N}\right) + \frac{2N+1}{N(1+N)}$$
On
Using what @Robert Israel wrote in his answer and comments, I think that we could also baound the partial sums.
$$S_n=\sum_{j=1}^{n}\frac{1}{j^2(j+1)^2}$$ $$S_n=2\sum_{j=1}^{n}\left(\frac 1 {j+1}-\frac 1 {j}\right)+\sum_{j=1}^{n}\frac 1 {(j+1)^2}+\sum_{j=1}^{n}\frac 1 {j^2}$$ $$S_n=-\frac{2 n}{n+1}+\left(\frac{\pi ^2}{6}-1-\psi ^{(1)}(n+2)\right)+H_n^{(2)}$$
Using the asymptotics $$S_n=\left(\frac{\pi ^2}{3}-3\right)-\frac{1}{3 n^3}+\frac{1}{n^4}+O\left(\frac{1}{n^5}\right)$$
Depending on how close a bound you need, you can say $$S=\sum_{j=1}^{\infty}\frac{1}{j^2(j+1)^2} \lt \sum_{j=1}^{\infty}\frac 1{j^4} = \frac {\pi^4}{90}\approx 1.08323$$ and $$S=\sum_{j=1}^{\infty}\frac{1}{j^2(j+1)^2} \gt \sum_{j=1}^{\infty}\frac 1{(j+1)^4} = \frac {\pi^4}{90}-1$$ which are fine upper and lower bounds. This lower bound is barely positive, so it might not satisfy what you need to do with it.