Two cases involving Maclaurin Series

Question

Could you help me to prove it?

I'm working hard in it, but I got nothing.

Answer 1

Examine the series for $\cos x$ and $\cosh x$ for the first one

$$\cos x =\sum_{n=0}^\infty {(-1)^nx^{2n}\over (2n)!}$$

if $x<0$ then we can substitute in $\sqrt{-x}$ for $x$ and get the series in question, so that covers that case.

Next say $x>0$, then we note that $\cosh x={1\over 2}\left(e^x+e^{-x}\right)$, adding the two series we get:

$${1\over 2}\sum_{n=0}^\infty {x^n+(-1)^nx^n\over n!}=\sum_{n=0}^\infty {x^{2n}\over (2n)!}$$

Since $x\ge 0$, we can substitute in $\sqrt{x}$ for $x$ and we get the desired series.

Now for the second one, we use absolute convergence and the log functional equation to get

$${1\over 2}\log\left({1+x\over 1-x}\right)={1\over 2}\left(\log(1+x)-\log(1-x)\right)=-{1\over 2}\left(\sum_{n=1}^\infty {(-1)^{n}x^n\over n}-\sum_{n=1}^\infty {x^n\over n}\right)$$

which combine together to

$$\sum_{k=1}^\infty {x^{2k+1}\over 2k+1}$$

since the even terms cancel and only numbers of the form $n=2k+1$ survive and are counted twice, which cancels with the ${1\over 2}$ factor to give the final result.