Question Number 163435 by amin96 last updated on 06/Jan/22
Answered by Mathspace last updated on 07/Jan/22
$${S}=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{4}{n}+\mathrm{1}}\left(\frac{\mathrm{1}}{\mathrm{4}{n}+\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{4}{n}+\mathrm{4}}\right) \\ $$$$=\sum_{{m}=\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\left(\mathrm{4}{n}+\mathrm{1}\right)\left(\mathrm{4}{n}+\mathrm{3}\right)} \\ $$$$−\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{4}{n}+\mathrm{1}\right)\left(\mathrm{4}{n}+\mathrm{4}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{16}}\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left({n}+\frac{\mathrm{1}}{\mathrm{4}}\right)\left({n}+\frac{\mathrm{3}}{\mathrm{4}}\right)} \\ $$$$−\frac{\mathrm{1}}{\mathrm{16}}\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\left({n}+\frac{\mathrm{1}}{\mathrm{4}}\right)\left({n}+\mathrm{1}\right)} \\ $$$${we}\:{have}\:\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\left({n}+{a}\right)\left({n}+{b}\right)} \\ $$$$=\frac{\psi\left({a}\right)−\psi\left({b}\right)}{{a}−{b}}\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\left({n}+\frac{\mathrm{1}}{\mathrm{4}}\right)\left({n}+\frac{\mathrm{3}}{\mathrm{4}}\right)} \\ $$$$=\frac{\psi\left(\frac{\mathrm{1}}{\mathrm{4}}\right)−\psi\left(\frac{\mathrm{3}}{\mathrm{4}}\right)}{\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{3}}{\mathrm{4}}}=\frac{\mathrm{1}}{\mathrm{2}}\left(\psi\left(\frac{\mathrm{3}}{\mathrm{4}}\right)−\psi\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\right\} \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left({n}+\frac{\mathrm{1}}{\mathrm{4}}\right)\left({n}+\mathrm{1}\right)} \\ $$$$=\frac{\psi\left(\frac{\mathrm{1}}{\mathrm{4}}\right)−\psi\left(\mathrm{1}\right)}{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}}=\frac{\mathrm{4}}{\mathrm{3}}\left\{\psi\left(\mathrm{1}\right)−\psi\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\right\}\:\Rightarrow \\ $$$${S}=\frac{\mathrm{1}}{\mathrm{32}}\left\{\psi\left(\frac{\mathrm{3}}{\mathrm{4}}\right)−\psi\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\right\} \\ $$$$−\frac{\mathrm{1}}{\mathrm{12}}\left\{\psi\left(\mathrm{1}\right)−\psi\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\right\} \\ $$$$ \\ $$
Commented by Tawa11 last updated on 07/Jan/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$