<s>[INST] How to prove $\exp(x)/(\exp(x)+1)^2$ is even? I've spent some time trying to prove that the function:
$$f(x)=\frac{\exp x}{(\exp x+1)^2}$$
is even. 
I tried expanding the different $\exp x$ as power series, but I had a very difficult time trying to track the different indices. Is that the correct way to proceed or is there some other property that I am no taking into account ? 
Greetings.
 [/INST] Just for fun:
The power series of the given function is awful to process. But that of its inverse is straightforward:
$$\frac1{f(x)}=\frac{(e^x+1)^2}{e^x}=\frac{e^{2x}+2e^x+1}{e^x}=e^x+2+e^{-x}=2\cosh(x)+2.$$
By cancellation of the odd powers, the series is $$4+x^2+2\frac{x^4}{4!}+2\frac{x^6}{6!}\cdots$$

Now, by an act of courage, let us compute the terms up to sixth degree:
$$f(x)=\frac14\frac1{1+(\frac{x^2}4+\frac{x^4}{2\cdot4!}+\frac{x^6}{2\cdot6!}\cdots)}\\
=\frac14\left(1-\left(\frac{x^2}4+\frac{x^4}{2\cdot4!}+\frac{x^6}{2\cdot6!}\cdots\right)+\left(\frac{x^2}4+\frac{x^4}{2\cdot4!}+\frac{x^6}{2\cdot6!}\cdots\right)^2-\left(\frac{x^2}4+\frac{x^4}{2\cdot4!}+\frac{x^6}{2\cdot6!}\cdots\right)^3\cdots\right)\\
=\frac14\left(1-\left(\frac{x^2}4+\frac{x^4}{2\cdot4!}+\frac{x^6}{2\cdot6!}\cdots\right)+\left(\frac{x^4}{4^2}+2\frac{x^6}{4\cdot2\cdot4!}\cdots\right)-\left(\frac{x^6}{4^3}\cdots\right)\right)\\
=\frac14-\frac{x^2}4+\frac{13x^4}{48}-\frac{299x^6}{1440}\cdots$$
Obviously there can't be odd terms.
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