Question

Given $\sum _{n=1}^{\infty }\frac{1}{n^2}=\frac{{\pi }^2}{6}$ : the value of $\sum _{n=1}^{\infty }\frac{1+3+5+......+(2n-1)}{1^3+2^3+3^3+........n^3}$ is:

• A. $2{\pi }^2/3$
• B. $4{\pi }^2/3$
• C. $4(\frac{{\pi }^2}{6}-1)$
• D. ${\pi }^2/6$

Answer 1

Answer: (A)

$2{\pi }^2/3$

${\sum }_{n=1}^{\infty }\frac{1+3+5+....+(2n-1)}{(1^3+2^3+3^3+....+n^3)}={\sum }_{n=1}^{\infty }\frac{n^2}{{\left[\frac{n(n+1)}{2}\right]}^2}{\sum }_{n=1}^{\infty }\frac{4}{{(n+1)}^2}={\sum }_{n=1}^{\infty }\frac{4}{(n^2+1+2n)}={\sum }_{n=1}^{\infty }\frac{4}{n^2[1+\frac{1}{n^2}+\frac{2}{n^2}]}$

as the value of $n$ increases

the term $1+\frac{1}{n^2}+\frac{2}{n^2}$ will be as equal as $1$

${\sum }_{n=1}^{\infty }\frac{4}{n^2}=4\times \frac{{\pi }^2}{6}=\frac{2{\pi }^2}{3}$