2
$\begingroup$

How do you solve this sum:

$$\sum_{n=0}^{\infty}2^n\left(1-e^{2^{-n-1}k}\right)^2$$

I know the result is $e^k - 1 - k$. Where can I find some lectures or materials about solving such problems?

$\endgroup$

1 Answer 1

6
$\begingroup$

Let $k\in\mathbb{R} $, we have :

\begin{aligned} \sum_{n=0}^{+\infty}{2^{n}\left(1-\mathrm{e}^{\frac{k}{2^{n+1}}}\right)^{2}}&=\sum_{n=0}^{+\infty}{\left(2^{n}-2^{n+1}\,\mathrm{e}^{\frac{k}{2^{n+1}}}+2^{n}\,\mathrm{e}^{\frac{k}{2^{n}}}\right)}\\ &=\sum_{n=0}^{+\infty}{\left(2^{n+1}-2^{n}-2^{n+1}\,\mathrm{e}^{\frac{k}{2^{n+1}}}+2^{n}\,\mathrm{e}^{\frac{k}{2^{n}}}\right)}\\ &=\sum_{n=0}^{+\infty}{\left(2^{n+1}\left(1-\mathrm{e}^{\frac{k}{2^{n+1}}}\right)-2^{n}\left(1-\mathrm{e}^{\frac{k}{2^{n}}}\right)\right)}\\ \sum_{n=0}^{+\infty}{2^{n}\left(1-\mathrm{e}^{\frac{k}{2^{n+1}}}\right)^{2}}&=\sum_{n=0}^{+\infty}{\left(u_{n+1}-u_{n}\right)}\end{aligned}

Where $ \left(u_{n}\right)_{n} $ is a numerical sequence defined as follows : $$ \left(\forall n\in\mathbb{N}\right),\ u_{n}=2^{n}\left(1-\mathrm{e}^{\frac{k}{2^{n}}}\right) $$

Since $ \left(u_{n}\right)_{n} $ does converge to a limit $ \ell $, using the fact that $ \lim\limits_{x\to 0}{\frac{\mathrm{e}^{x}-1}{x}}=0 $, we get that $ \ell =-k $, we have that $ \sum\limits_{n\geq 0}{\left(u_{n+1}-u_{n}\right)} $ is a telescopic series that converges, and $ \sum\limits_{n=0}^{+\infty}{\left(u_{n+1}-u_{n}\right)}=\ell -u_{0}=-k-1+\mathrm{e}^{k}.$

Thus : $$ \sum_{n=0}^{+\infty}{2^{n}\left(1-\mathrm{e}^{\frac{k}{2^{n+1}}}\right)^{2}}=\mathrm{e}^{k}-k-1.$$

$\endgroup$
1
  • 1
    $\begingroup$ Very elegant way of solving the problem! $\endgroup$ Apr 7, 2020 at 15:07

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .