Some infinite series (just for fun!)

Question

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!

Answer 1

\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.

Answer 2

The tag telescoping is a great hint:

\begin{align*} A_n &:= \frac{2^n}{2^{2^n}-1} &\Rightarrow & \qquad A_n - A_{n+1} = \frac{2^n}{2^{2^n}+1} \\ B_n &:= \frac{4^n \cdot 4^{2^n}}{(4^{2^n}-1)^2} &\Rightarrow & \qquad B_n - B_{n+1} = \frac{4^n \cdot 4^{2^n}}{(4^{2^n}+1)^2} \\ C_n &:= \frac{3^n}{3^{3^n}-1} &\Rightarrow & \qquad C_n - C_{n+1} = \frac{3^n(3^{3^n}+2)}{3^{2\cdot 3^n}+3^{3^n} + 1} \end{align*}

From these, we obtain

\begin{align*} \sum_{n=0}^{\infty} \frac{2^n}{2^{2^n}+1} &= \lim_{N\to\infty} (A_0 - A_{N+1}) = 1 \\ \sum_{n=0}^{\infty} \frac{4^n \cdot 4^{2^n}}{(4^{2^n}+1)^2} &= \lim_{N\to\infty} (B_0 - B_{N+1}) = \frac{4}{9} \\ \sum_{n=0}^{\infty} \frac{3^n(3^{3^n}+2)}{3^{2\cdot 3^n}+3^{3^n} + 1} &= \lim_{N\to\infty} (C_0 - C_{N+1}) = \frac{1}{2} \end{align*}

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For a generalization, let $a \geq 2$ be an integer and consider

$$ P_n = \frac{a^n}{X^{a^n}-1}. $$

Then it is not hard to check that

$$ P_n - P_{n+1} = \frac{a^n \sum_{k=0}^{a-2} (a-1-k) X^{k \cdot a^n}}{\sum_{k=0}^{a-1} X^{k \cdot a^n}} $$

So if $|X| > 1$, we obtain

$$ \sum_{n=0}^{\infty} \frac{a^n \sum_{k=0}^{a-2} (a-1-k) X^{k \cdot a^n}}{\sum_{k=0}^{a-1} X^{k \cdot a^n}} = P_0 = \frac{1}{X-1}. \tag{*}$$

Then

• The 1st series corresponds to $a = 2$ and $X = 2$ applied to $\text{(*)}$.
• The 2nd series corresponds to $a = 2$ and $X = 4$ applied to the derivative of $\text{(*)}$ w.r.t. $X$.
• The 3rd series corresponds to $a = 3$ and $X = 3$ applied to $\text{(*)}$.