This question already has an answer here:

How do you calculate the sum of an infinite series like

$$ \sum_{n = 0}^\infty \frac{n}{2^\sqrt{n}}$$



I searched up how to find this with infinite geometric series solution which was


$a$ being first value and $r$ being common ratio. But I could not find common ratio.

Please give me common ratio of this series.


I have also tried limits like

$$ \lim_{a \to \infty} \sum_{n = 0}^a \frac{n}{2^\sqrt{n}} $$

but they don't actually give me a solid answer.


Prove this infinite series is convergent.


marked as duplicate by Cameron Williams, user147263, Gerry Myerson, J. W. Perry, Hanul Jeon Dec 14 '14 at 5:48


  • 1
    $\begingroup$ This is not a geometric series. That is why you could not find $r$. $\endgroup$ – parsiad Dec 14 '14 at 5:06
  • $\begingroup$ How do you find the sum then? $\endgroup$ – Irrational Person Dec 14 '14 at 5:08
  • $\begingroup$ It's not evident that there should be any closed form for this sum; where does it come from? $\endgroup$ – Milo Brandt Dec 14 '14 at 5:09
  • $\begingroup$ I got this from a Math.SE question that I couldn't solve at all. $\endgroup$ – Irrational Person Dec 14 '14 at 5:10
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    $\begingroup$ @Bot Er, what exactly is your goal in posting a new question, when you found the question here? This question isn't any more likely to get an answer than the original one you link to. $\endgroup$ – Milo Brandt Dec 14 '14 at 5:20

While you will be hard-pressed to find a closed form for the sum, you can use the comparison test to show that it converges. Numerical computation shows that the value is approximately $$ \sum_{n=0}^{\infty}\frac{n}{2^{\sqrt{n}}}\approx51.919095230880139. $$ You can get pretty good closed-form estimates for this by integrating $$ \sum_{n=0}^{\infty}\frac{n}{2^{\sqrt{n}}}\approx\int_{0}^{\infty}\frac{n}{2^{\sqrt{n}}}dn=\frac{12}{\left(\log\left(2\right)\right)^{4}}\approx51.985162021107868. $$ If you require accurate results, just use numerical computation.

For example, in MATLAB:

nmax = 1e+12;
tol  = 1e-16;

t = 0;
for n = 1:nmax
    m = n/2^sqrt(n);
    t = t + m;
    if m < tol


You can make tol smaller to increase accuracy or larger to increase speed of computation.

