I'm trying to figure out what the infinite sum of $\dfrac 1{2^{n(n+1)/2}}$ is.

That's $\dfrac 12 + \dfrac 1{2^3} + \dfrac 1{2^6} + \dfrac 1{2^{10}} + \cdots $ where the powers $1,3,6,10,\cdots$ are the triangular numbers.

It's convergent as its bounded above by the sum of $\left(\dfrac 12\right)^n = 1$.

I've noticed that the ration of the terms, that is $\dfrac{a_{n+1}}{a_n} = \left(\dfrac 12\right)^n$, and so the ratios form a geometric series, but can't seem to find a way to possibly use this.

I was wondering if anyone could give me a hint or a push in the right direction? Thanks


(I'm settling this cream puff before I retire for today...)


$$\vartheta_2(z,q)=2\sqrt{q}\sum_{k=0}^\infty q^\frac{k(k+1)}{2}\cos((2k+1)z)$$

be the second Jacobi theta function. Letting $z=0$ and $q=\frac12$ gives

$$\vartheta_2\left(0,\frac12\right)=\sqrt{2}\left(1+\sum_{k=1}^\infty \left(\frac12\right)^\frac{k(k+1)}{2}\right)$$

and thus the expression you want is $\frac1{\sqrt{2}}\vartheta_2\left(0,\frac12\right)-1$. I know not of any more elementary functions that can represent your sum...


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.