The sum of $\sum_{n=1}^{\infty} e^{-2^{n}}$

Asked Oct 11 '20 at 08:03

Active Oct 11 '20 at 10:32

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I would like to calculate the following sum:

$\sum_{n=1}^{\infty} e^{-2^{n}}$.

I know that numerically the sum is about 0.154.

Is there anyway to expand this analytically?

**Edit:

Following the comments, is there a rather "tight" analytic bound? the obvious (not so tight) bound I can think of is the geometric sum:

$\sum_{n=1}^{\infty} e^{-2^{n}} \leq \sum_{n=1}^{\infty} e^{-2n}$

edited Oct 11 '20 at 10:32

asked Oct 11 '20 at 08:03

Amit

8

No. ${}{}{}{}{}$ – reuns Oct 11 '20 at 08:07
2

Most probably not, though interestingly enough, $\approx \frac{\tan(57°)}{10}$ to 8 digits. – Graviton Oct 11 '20 at 08:10
1

Such sums can usually only be calculated numerically, this is a very fast converging sum. – Peter Oct 11 '20 at 09:58
Related (an alternative expression in my answer, and useful information in the other one), but, basically, the answer is above. – metamorphy Oct 11 '20 at 13:35

0 Answers0