Question

$S_1=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...\infty S_2=\frac{1}{3}+\frac{1}{5^2}+\frac{1}{3^3}+\frac{1}{5^4}+\frac{1}{3^5}+\frac{1}{5^6}+...\infty \text{If}{S}_1-2S_2=\frac{a}{6},\text{then}a=$

Answer 1

$S_1=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...\infty =\sum _{n=1}^{\infty }\frac{1}{n(n+1)}=\sum _{n=1}^{\infty }\frac{1}{n}-\frac{1}{n+1}=\sum _{n=1}^{\infty }(\frac{1}{1}-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+...+(\frac{1}{n}-\frac{1}{n+1})=\underset{n\to \infty }{lim}(1-\frac{1}{n+1})S_1=1-\underset{n\to \infty }{lim}\frac{1}{n+1}=1S_2=\frac{1}{3}+\frac{1}{5^2}+\frac{1}{3^3}+\frac{1}{5^4}+\frac{1}{3^5}+\frac{1}{5^6}+...\infty =(\frac{1}{3}+\frac{1}{3^3}+\frac{1}{3^5}+...\infty )+(\frac{1}{5^2}+\frac{1}{5^4}+\frac{1}{5^6}+...\infty )=\frac{\frac{1}{3}}{1-\frac{1}{3^2}}+\frac{\frac{1}{5^2}}{1-\frac{1}{5^2}}=\frac{3}{8}+\frac{1}{24}=\frac{5}{12}{S}_1-2S_2=1-\frac{5}{6}=\frac{1}{6}\Rightarrow a=1$