Nicontio1

2021-12-23

I was asked to compute this series:
$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{\left(2n-1\right)}^{2}}$
but by using the fact that $\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{n}^{2}}=\frac{{\pi }^{2}}{6}$

scomparve5j

Expert

By direct calculation, we see that
$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{n}^{2}}=\sum _{k=1}^{\mathrm{\infty }}\frac{1}{{\left(2k-1\right)}^{2}}+\frac{1}{4}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{n}^{2}}$
which means
$\sum _{k=1}^{\mathrm{\infty }}\frac{1}{{\left(2k-1\right)}^{2}}=\frac{3}{4}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{n}^{2}}=\frac{3}{4}\frac{{\pi }^{2}}{6}=\frac{{\pi }^{2}}{8}$

kalfswors0m

Expert

Observe that
$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{\left(2n\right)}^{2}}=\frac{1}{{\left(2\cdot 1\right)}^{2}}+\frac{1}{{\left(2\cdot 2\right)}^{2}}+\frac{1}{{\left(2\cdot 3\right)}^{2}}\dots =\frac{1}{4}\left(\frac{1}{{1}^{2}}+\frac{1}{{2}^{2}}+\frac{1}{{3}^{2}}+\dots \right)=\frac{{\pi }^{2}}{4\left(6\right)}$
But
$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{\left(2n\right)}^{2}}+\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{\left(2n-1\right)}^{2}}=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{n}^{2}}$
where rearrangement is allowed since we are dealing with non-negative series.

user_27qwe

Expert