I am looking for a closed-form solution to this infinite (converging) series containing a ceiling function:

\begin{equation}\sum_{k=1}^{\infty}\frac{1}{2^{2k+\lceil 2\sqrt{k} \rceil}} \approx 0.0715726 \end{equation}

I found some similar problem...

Infinite sum of floor functions

...but I was not able to use Markus Scheuer's approach featuring in this link.

The idea was to write (with $[\cdot]$ the Iverson bracket and $m$ running over all integers):

\begin{align} \sum_{k=1}^{\infty}\frac{1}{2^{2k+\lceil 2\sqrt{k} \rceil}} & = \sum_{k=1}^{\infty}\sum_{m}\frac{1}{2^{2k+m}}\Big[m=\lceil 2\sqrt{k} \rceil\Big] \\ & = \sum_{k=1}^{\infty}\sum_{m}\frac{1}{2^{2k+m}}\Big[m-1 < 2\sqrt{k} \leq m\Big] \\ & = \sum_{k=1}^{\infty}\sum_{m}\frac{1}{2^{2k+m}}\Big[(m-1)^2/4 < k \leq m^2/4\Big] \\ \end{align}

The next step would be (as far as I understood it) to remove the Iverson bracket by including its interval limits to the bounds for the $j$ sum, switching the sums thereby somehow, then shift the limits and get rid of the $m$ in the sum's bounds, split the sums, and calculate their values relatively straighforward. However, in my case, the first problem is that my interval limits are fractional. One could remedy this, I think, by writing...

\begin{align} \sum_{k=1}^{\infty}\frac{1}{2^{2k+\lceil 2\sqrt{k} \rceil}} & = \sum_{k=1}^{\infty}\sum_{m}\frac{1}{2^{2k+m}}\Big[\lfloor (m-1)^2/4 \rfloor < k \leq \lfloor m^2/4 \rfloor \Big] \\ & = \sum_{k=1}^{\infty}\sum_{m}\frac{1}{2^{2(k+\lfloor (m-1)^2/4 \rfloor)+m}} \Big[0 < k \leq \lfloor m/2 \rfloor \Big] \\ \end{align}

..., which still evaluates to the same result, but anyhow I will not get rid of the $m$ in the interval's/sum's bound, while shifting the limits, rather ironically, results in introducing another floor function in the summand.

I also tried $2k+\lceil 2 \sqrt{k}\rceil = 2k+1+\lfloor \sqrt{4k-3} \rfloor$ instead, adjusting the Iverson bracket term to $\Big[m \leq \sqrt{4k-3} < m+1\Big]$ and proceeding from there, but without success.

I also noted a paper...


...which gives an expression for the sum $\sum_{k=1}^{n}\lfloor \sqrt{k} \rfloor$, but I think I cannot simply adjust the arguments in their derivation to my case.

Another identity, but possibly not helpful either, is $2k+\lceil 2 \sqrt{k}\rceil = 2k+1+\lfloor \sqrt{k}+\sqrt{k-1} \rfloor$

Any suggestions?

[Added on 2017-09-18]

I have thought again about the summation and considered the first few terms searching for a pattern. I have splitted the fraction according to (see above for the floor part):

\begin{equation} \frac{1}{2^{2k+1}}\cdot\frac{1}{2^{\lfloor\sqrt{4k-3}\rfloor}} \end{equation}

If one considers only the exponents, written as tuples, one gets something like


If one now writes down the difference between the two values one gets

$(2), (3), (4, 6), (7, 9), (10, 12, 14), (15, 17, 19)$

Here the values are collected in parentheses for the second coordinate being $1,2,3,4,5,6$. One sees that odd second coordinates have even differences, and vice versa. Moreover, the difference between values in the brackets is always 2, with their number constantly increasing. One can use this to determine boundaries for these values, and it helps to treat the odd and even integers separately (thanks Ahmad!). Anyhow, I fear that I am not very clear in the deduction (it was not straightforward either for me), one can write the sum eventually as

\begin{equation} \sum_{k=1}^{\infty}\sum_{j=0}^{k-1}\left(2^{-2(1+j+k^2)}+2^{-2(\frac{3}{2}+j+k+k^2)} \right)\approx 0.0715726 \end{equation}

The first term accounts for the even, the second one for the odd cases. Using Mathematica the first term on its own evaluates to

\begin{equation} \frac{1}{6} \left(1-\sqrt{2}\,\vartheta_2\!\left(0,\frac{1}{4}\right)+\vartheta _3\!\left(0,\frac{1}{4}\right)\right) \approx 0.063722 \end{equation}

with Jacobi theta functions involved. However, the second term, numerically $\approx 0.00785066$, does not evaluate so nicely. But, intuitively, I would think, since it is not looking much more complicated, it should also have a nice solution? Anyone having an idea how to proceed? It somehow looks like 'simply' calculating the sum, but how?

[Added on 2017-09-19 morning]

I was able to split the second sum more and now I got this partial result:

\begin{equation} \frac{1}{6} \left(1-\sqrt{2}\,\vartheta_2\!\left(0,\frac{1}{4}\right)+\vartheta _3\!\left(0,\frac{1}{4}\right)\right) + \frac{1}{6} \left(-1+\frac{\vartheta_2\!\left(0,\frac{1}{4}\right)}{\sqrt{2}}-\sum_{k=1}^{\infty}2^{-2(2k+k^2)}\right) \end{equation}

So it is the last sum which still resists.

[Added on 2017-09-19 evening]

I think I solved it by discovering a pattern again in the exponents of the second sum. I found them on the On-Line Encyclopedia of Integer Sequences as A054000 with an offset of one. A sum with the generating function stated in the title yields a Mathematica solution in terms of Jacobi theta function again, and (by chance) seems to be the sought-for number plus one (I checked it with Mathematica up to 500 decimals with the topmost sum of this entry evaluated up to 1000). Simplification of the sums yields finally:

\begin{equation} \sum_{k=1}^{\infty}\frac{1}{2^{2k+\lceil 2\sqrt{k} \rceil}} = \frac{1}{12} \left(6-\sqrt{2}\, \vartheta _2\left(0,\frac{1}{4}\right)-2 \,\vartheta _3\left(0,\frac{1}{4}\right)\right) \end{equation}

Of course, one should check the 'by chance' step again.

If anyone would independently arrive at the same (conjectured) answer, I would be very interested in that.

P.S. I would also be interested in some general context, e.g. some qualitative interpretation of what it means that the Jacobi theta function is appearing here. Does it tell you something about the constant it describes or the process how it was generated? Does it tell you if the constant is transcendent (possibly not, I guess)? Are there other constants known, in which such values of the theta function feature prominently?

[Added on 2017-09-21]

I think a general formula for $m = 2k+\big\lceil 2\sqrt{k} \big\rceil$ and any base $b$ is (without proof, but conjectured because of numerical evidence):

\begin{equation} \sum_{k=1}^{\infty}b^{-m}=\frac{1}{2(b^{2}-1)}\!\left\{(b+1)-(b-1)\left[\frac{\vartheta_{2}(0,b^{-2})}{\sqrt{b}}+\vartheta_{3}(0,b^{-2})\right]\right\} \end{equation}

  • $\begingroup$ This is a very different question than the question on $\sum \lfloor \sqrt k \rfloor$. I would be very surprised if this had a closed form expression. $\endgroup$
    – davidlowryduda
    Sep 17, 2017 at 18:33

1 Answer 1


separating odd cases from even cases for $k$ and then because of the ceiling function property there is $t$ such that all $t < 2\sqrt{k} \leq t+1$ have there ceiling function equal to $t+1$ , and in last adding all these term together one can arrive at this summation (which don't include ceiling function) :

$\sum _{t=1}^{\infty } \frac{1}{3} 4^{-2-t (3+2 t)} \left(-1+4^{1+t+t^2} \left(-1+2^{1+2 t}\right)\right)+\frac{1}{16} \approx 0.0716337$

The problem is that the integration of the function in the summation contain Erf function and its so complicated that there is a good chance you will not find a closed form for the summation.

  • $\begingroup$ Thank you for the tip of splitting the sum into odd and even integer parts. Regarding the point about integration, I am not sure if I understand you right. Do you mean that one approximates the sum given above by an integral and then thinks about any residual error term separately? Is this, were the error function arises? $\endgroup$
    – Iridium
    Sep 18, 2017 at 8:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.