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Is there a closed form for the following sum?

$$\sum_{n=0}^\infty e^{-\sqrt n}$$

Watson
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sigma
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    Wolfram Alpha does not show an expression, so probably there is no closed form. – Peter Jul 18 '16 at 15:31
  • Various inverse symbolic calculators come up with nothing either. – George V. Williams Jul 18 '16 at 17:04
  • Depends upon the meaning of "closed form"; for instance, one could probably add/subtract cosh(), sinh() and do some symmetry touch-up. – rrogers Jul 19 '16 at 21:15
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    $$\int_{0}^{+\infty}e^{-\sqrt{x}},dx = 2$$ and the difference between the sum and the above integral can be estimated through Abel's or Euler-MacLaurin summation formulas. – Jack D'Aurizio Sep 14 '16 at 23:45
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    Based on my answer to 1, we can write the answer as a more interesting sum $$ \sum_{n=0}^\infty e^{-\sqrt{n}}=\frac{5}{2} - \sum_{s=1}^\infty \frac{\Gamma(1+\frac{s}{2})\zeta(1+\frac{s}{2})\sin(\frac{\pi s}{4})}{(-1)^s \pi(2\pi)^{s/2} \Gamma(s+1)} $$ – Benedict W. J. Irwin Jun 29 '17 at 17:26
  • Also related. – metamorphy Apr 04 '23 at 08:03
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    With Abel-Plana: $$\sum _{n=0}^{\infty } \exp \left(-\sqrt{n}\right)=\frac{5}{2}+\int_0^{\infty } \frac{i \left(-e^{-\sqrt{-i x}}+e^{-\sqrt{i x}}\right)}{-1+e^{2 \pi x}} , dx$$ – Mariusz Iwaniuk Jan 18 '25 at 12:00
  • Use Inverse Laplace transform of $e^{-\sqrt{n}}$ $$\sum_{n=0}^{\infty}e^{-\sqrt{n}}=\frac{1}{2\sqrt{\pi}}\int_{0}^{\infty}\frac{x^{-3/2}e^{-1/{(4x)}}}{1-e^{-x}}dx$$ – Quý Nhân Jan 18 '25 at 15:17

1 Answers1

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Since we know that $$\int e^{-\sqrt{n}}\,dn=-2 e^{-\sqrt{n}} \left(1+\sqrt{n}\right)$$ we can use Euler-MacLaurin summations.

For example, it will give $$S_n=\sum_{m=0}^n e^{-\sqrt m}=1+\frac{97305549483745747}{21424936845312000\, e}-$$ $$2 \sqrt{n}\,e^{-\sqrt{n}}\Bigg(1+\frac{3}{4 n^{1/2}}+\frac{1}{48 n}-\frac{1}{11520 n^2}-\frac{1}{3840 n^{5/2}} -\frac{503}{1935360 n^3}+O\left(\frac{1}{n^{7/2}}\right) \Bigg)$$ The value of the constant is $\color{red}{2.670}80$

Edit

If we write $$S_n=C_n-2 \sqrt{n}\,e^{-\sqrt{n}}\Bigg(1+\frac{3}{4 n^{1/2}}+\cdots-\frac{503}{1935360 n^3}+O\left(\frac{1}{n^{7/2}}\right) \Bigg)$$ $$T_n=\sum_{m=0}^n e^{-\sqrt m}$$ and adjust the constant $C_n$ in order to have $S_n=T_n$ for small values of $n$

$$ \left( \begin{array}{cc} n & C_n \\ 1 & 2.670339100 \\ 2 & 2.670403221 \\ 3 & 2.670406242 \\ 4 & 2.670406667 \\ 5 & 2.670406766 \\ 6 & 2.670406797 \\ 7 & 2.670406808 \\ 8 & 2.670406813 \\ 9 & 2.670406815 \\ 10 & 2.670406816 \\ \end{array} \right)$$

Claude Leibovici
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    Do you really mean exact equality? This does not seem to be correct. Can you explain how you get this fraction? (Euler-Maclaurin formula only gives a formula for the partial sum). – Martin Jan 18 '25 at 10:30
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    $\sum_{n=0}^\infty e^{-\sqrt n}=1+\frac{703008193}{154828800, e}$ is of course false. If you are going to use an approximation you should write $\sum_{n=0}^\infty e^{-\sqrt n}\color{red}{\approx} 1+\frac{703008193}{154828800, e}$ – jjagmath Jan 18 '25 at 12:29
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    The LHS of your equation is a real number, while the RHS is a function of $n$, so there's obviously something wrong here. – jjagmath Jan 18 '25 at 17:10
  • @jjagmath. Typo's fixed. Thanks for pointing – Claude Leibovici Jan 18 '25 at 18:48
  • This is an approximate equality for the sum, right? Otherwise we'd be able to take $n\rightarrow\infty$ and get an exact answer. – John Barber Jan 18 '25 at 20:12