Nicontio1

Answered

2021-12-23

I was asked to compute this series:

$\sum _{n=1}^{\mathrm{\infty}}\frac{1}{{(2n-1)}^{2}}$

but by using the fact that$\sum _{n=1}^{\mathrm{\infty}}\frac{1}{{n}^{2}}=\frac{{\pi}^{2}}{6}$

but by using the fact that

Answer & Explanation

scomparve5j

Expert

2021-12-24Added 38 answers

By direct calculation, we see that

$\sum _{n=1}^{\mathrm{\infty}}\frac{1}{{n}^{2}}=\sum _{k=1}^{\mathrm{\infty}}\frac{1}{{(2k-1)}^{2}}+\frac{1}{4}\sum _{n=1}^{\mathrm{\infty}}\frac{1}{{n}^{2}}$

which means

$\sum _{k=1}^{\mathrm{\infty}}\frac{1}{{(2k-1)}^{2}}=\frac{3}{4}\sum _{n=1}^{\mathrm{\infty}}\frac{1}{{n}^{2}}=\frac{3}{4}\frac{{\pi}^{2}}{6}=\frac{{\pi}^{2}}{8}$

which means

kalfswors0m

Expert

2021-12-25Added 24 answers

Observe that

$\sum _{n=1}^{\mathrm{\infty}}\frac{1}{{\left(2n\right)}^{2}}=\frac{1}{{(2\cdot 1)}^{2}}+\frac{1}{{(2\cdot 2)}^{2}}+\frac{1}{{(2\cdot 3)}^{2}}\dots =\frac{1}{4}(\frac{1}{{1}^{2}}+\frac{1}{{2}^{2}}+\frac{1}{{3}^{2}}+\dots )=\frac{{\pi}^{2}}{4\left(6\right)}$

But

$\sum _{n=1}^{\mathrm{\infty}}\frac{1}{{\left(2n\right)}^{2}}+\sum _{n=1}^{\mathrm{\infty}}\frac{1}{{(2n-1)}^{2}}=\sum _{n=1}^{\mathrm{\infty}}\frac{1}{{n}^{2}}$

where rearrangement is allowed since we are dealing with non-negative series.

But

where rearrangement is allowed since we are dealing with non-negative series.

user_27qwe

Expert

2021-12-30Added 230 answers

Very detailed answer. Thanks.

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