Does anyone have a proof that
$$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{2n-1}\mbox{sech}\left(\frac{(2n-1)\pi}{2}\right)=\frac{\pi}{8}$$
Subscribe to:
Post Comments (Atom)
-
So if I have a matrix and I put it into RREF and keep track of the row operations, I can then write it as a product of elementary matrices. ...
-
I am asked to prove the density of irrationals in $\mathbb{R}$. I understand how to do this by proving the density of $\mathbb{Q}$ first, na...
-
Can someone just explain to me the basic process of what is going on here? I understand everything until we start adding 1's then after ...
No comments:
Post a Comment