<s>[INST] Some infinite series (just for fun!) I have a few infinite series problems that I think MSE might enjoy, whose answers I already know:

$$\sum_{n=0}^\infty \frac{2^{n-2^n}}{1+2^{-2^n}}=\text{?}$$
  $$\sum_{n=0}^\infty \frac{4^{n+2^n}}{(1+4^{2^n})^2}=\text{?}$$
  $$\sum_{n=0}^\infty \frac{3^n(1+2\cdot3^{-3^n})}{2\cosh(3^n\ln 3)+1}=\text{?}$$

After these all get answered, I will post explanations of how I came up with answers for them. Have fun!
 [/INST] \begin{align}
\frac{1}{1-2}+S_1&=\lim_{N\to\infty}(\frac{1}{1-2}+\sum_{n=0}^N \frac{2^{n}}{1+2^{2^n}})\\
&=\lim_{N\to\infty}(\frac{1}{1-2}+\frac{1}{1+2}+\sum_{n=1}^N \frac{2^{n}}{1+2^{2^n}})\\
&=\lim_{N\to\infty}(\frac{2}{1-2^2}+\sum_{n=1}^N \frac{2^{n}}{1+2^{2^n}})\\
&=\lim_{N\to\infty}\frac{2^{N+1}}{1-2^{2^{N+1}}}\\
&=0
\end{align}
\begin{align}
\frac{1}{(4^{-2^{-1}}-4^{2^{-1}})^2}-S_2&=\frac{1}{(4^{-2^{-1}}-4^{2^{-1}})^2}-\sum_{n=0}^\infty \frac{4^{n+2^n}}{(1+4^{2^n})^2}\\
&=\lim_{N\to\infty}(\frac{1}{(4^{-2^{-1}}-4^{2^{-1}})^2}-\frac{1}{(4^{-2^{-1}}+4^{2^{-1}})^2}-\sum_{n=1}^N \frac{4^{n}}{(4^{-2^{n-1}}+4^{2^{n-1}})^2})\\
&=\lim_{N\to\infty}(\frac{4}{(4^{-2^{0}}-4^{2^{0}})^2}-\sum_{n=1}^N \frac{4^{n}}{(4^{-2^{n-1}}+4^{2^{n-1}})^2})\\
&=\lim_{N\to\infty}\frac{4^{N+1}}{(4^{-2^{N}}-4^{2^{N}})^2}\\
&=0
\end{align}
Same thing for $S_3$. Note that $\cosh(x)=\frac{e^x+e^{-x}}{2}$. Just some calculations.
</s>