Problem 2

On rearranging

$\large{(1+\frac{1}{9}+\frac{1}{17}+...)-(\frac{1}{7}+\frac{1}{15}+....)}$

$\large{\displaystyle \sum_{r=1}^{\infty}\frac{1}{8r-7}-\displaystyle \sum_{r=1}^{\infty} \frac{1}{8r-1}}$

$\large{\displaystyle \sum_{r=1}^{\infty} \displaystyle \int_{0}^{1} x^{8r-8} dx-\displaystyle \sum_{r=1}^{\infty} \displaystyle \int_{0}^{1} x^{8r-2} dx}$

$\large{ \displaystyle \int_{0}^{1} \displaystyle \sum_{r=1}^{\infty} x^{8(r-1)} dx-\displaystyle \int_{0}^{1} \displaystyle \sum_{r=1}^{\infty} x^{8r-2} dx}$

$\large{\displaystyle \int_{0}^{1} \frac{1}{1-x^{8}} dx-\displaystyle \int_{0}^{1} \frac{x^{6}}{1-x^{8}} dx}$

$\large{\displaystyle \int_{0}^{1} \frac{1-x^{6}}{1-x^8} dx=}$

$\large{\displaystyle \int_{0}^{1} \frac{1}{2(x^2+1)}dx+\frac{x^2+1}{2(x^4+1)}dx}$

$\large{\displaystyle \int_{0}^{1} \frac{1}{2(x^2+1)}dx+\displaystyle \int_{0}^{1}\frac{1}{2}\frac{1+\frac{1}{x^2}}{(x-\frac{1}{x})^2+2}dx}$

we finally get answer as

$\large{\frac{\pi}{8}+\frac{\pi}{4\sqrt{2}}=\boxed{0.948}}$

I will give a general method how to solve "such" problems.

Suppose we have a summation like the following -

$\displaystyle \sum_{n=1}^{\infty}{\frac{{a}_{n}}{{s}_{1,n}{s}_{2,n}{s}_{3,n}}}$

where

${a}_{n}$ is a linear equation in $n$ .

${s}_{1,n} = (n+{\alpha}_{1})(n+{\alpha}_{2})\cdots (n+{\alpha}_{p})$

${s}_{2,n} = {(n+{\beta}_{1})}^{2}{(n+{\beta}_{2})}^{2} \cdots {(n+{\beta}_{q})}^{2}$

${s}_{3,n} = {(n+{\gamma}_{1})}^{3}{(n+{\gamma}_{2})}^{3} \cdots {(n+{\gamma}_{r})}^{3}$

and we can change into partial fractions -

$\displaystyle \sum_{i=1}^{p}{\frac{{c}_{1,i}}{n+{\alpha}_{i}}} + \sum_{j=1}^{q}{\frac{{c}_{2,j}}{n+{\beta}_{j}}} + \sum_{j=1}^{q}{\frac{{c'}_{2,j}}{{(n+{\beta}_{j})}^{2}}} + \sum_{k=1}^{r}{\frac{{c}_{3,k}}{n+{\gamma}_{k}}} + \sum_{k=1}^{r}{\frac{{c'}_{3,k}}{{(n+{\gamma}_{k})}^{2}}} + \sum_{k=1}^{r}{\frac{{c''}_{3,k}}{{(n+{\gamma}_{k})}^{2}}}$

Then, the value of sum is

$\displaystyle -\sum_{i=1}^{p}{{c}_{1,i} \psi(1+{\alpha}_{i})} - \sum_{j=1}^{q}{{c}_{2,j} \psi(1+{\alpha}_{j})} + \sum_{j=1}^{q}{{c'}_{2,j} \psi'(1+{\alpha}_{j})} - \sum_{k=1}^{r}{{c}_{3,k} \psi(1+{\alpha}_{k})} + \sum_{k=1}^{r}{{c'}_{3,k} \psi'(1+{\alpha}_{k})} - \sum_{k=1}^{r}{\frac{{c''}_{3,k}}{2} \psi''(1+{\alpha}_{k})}$