It is given that -r - 1-12 r - 12 - -28- then -r - 1-1r2 is equal to

The correct option is D π^26 ^∞ _r=1Σ1 ( 2r 1 )^2=π^28 11^2+13^2+15^2+17^2+........=π^28 I= 11^2+12^2+13^2+14^2+15+16^2+17^2 I= ( ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Summation by Sigma Method

It is given t...

Question

It is given that $\sum_{r = 1}^{\infty} \frac{1}{(2 r - 1)^{2}} = \frac{π^{2}}{8}$ , then $\sum_{r = 1}^{\infty} \frac{1}{r^{2}}$ is equal to

\frac{π^{2}}{24}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{π^{2}}{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{π^{2}}{6}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

None of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

Suggest Corrections

Similar questions

\sum_{r = 1}^{\infty} \frac{1}{(2 r - 1)^{2}} = \frac{π^{2}}{8}

\sum_{r = 1}^{\infty} \frac{1}{r^{2}}

\frac{π^{2}}{2 k}

I f \sum_{r = 1}^{\infty} \frac{1}{(2 r - 1)^{2}} = \frac{π^{2}}{8}

\sum_{r = 1}^{\infty} \frac{1}{r^{2}}

\sum_{r = 1}^{\infty} \frac{1}{{(2 r - 1)}^{2}} = \frac{π^{2}}{8}

\sum_{r = 1}^{\infty} \frac{1}{r^{2}}

\sum_{r = 1}^{\infty} \frac{1}{{(2 r - 1)}^{2}} = \frac{π^{2}}{8}

\sum_{r = 1}^{\infty} \frac{1}{r^{2}}

\infty \sum r = 1 \frac{1}{{(2 r - 1)}^{2}} = \frac{π^{2}}{8}

\infty \sum r = 1 \frac{1}{r^{2}}

Join BYJU'S Learning Program