# How can I simplify this sum any further?

Recently I was playing around with some numbers and I stumbled across the following formal power series:

$$\sum_{k=0}^\infty\frac{x^{ak}}{(ak)!}\biggl(\sum_{l=0}^k\binom{ak}{al}\biggr)$$

I was able to "simplify" the above expression for $$a=1$$:

$$\sum_{k=0}^\infty\frac{x^k}{k!}\cdot2^k=e^{2x}$$ I also managed to simplify the expression for $$a=2$$ with the identity $$\sum_{i=0}^\infty\frac{x^{2k}}{(2k)!}=\cosh(x)$$:

$$\sum_{k=0}^\infty\frac{x^{2k}}{(2k)!}\biggl(\sum_{l=0}^k\binom{2k}{2l}\biggr)=\mathbf[\cdots\mathbf]=\frac{1}{4}\cdot(e^{2x}+e^{-2x})+\frac{1}{2}=\frac{1}{2}\cdot(\cosh(2x)+1)$$

However, I couldn't come up with a general method for all $$a\in\Bbb{N}$$. I would be very thankful if someone could either guide me towards simplifying this expression or post his solution here.

You might be able to use the fact that $$\sum_{k=0}^\infty b_{ak}=\sum_{k=0}^\infty \left(\frac{1}{a}\sum_{j=0}^{a-1} \exp\left(2\pi ijk/a\right)\right)b_k.$$ For example, when $$a=1$$, taking $$b_k = \frac{x^k}{k!}\sum_{\ell \ge 0} \binom{k}{\ell}$$ yields $$\sum_{k=0}^\infty b_{k}=\sum_{k=0}^\infty \frac{x^k}{k!}\sum_{\ell \ge 0} \binom{k}{\ell}=\sum_{k=0}^\infty \frac{x^k}{k!}2^k=\exp(2x),$$ as you already obtained. For $$a=2$$, first note that \begin{align} \sum_{\ell \ge 0}\binom{k}{2 \ell} &= \sum_{\ell\ge 0} \left(\frac{1}{2}\sum_{j=0}^1 \exp\left(2\pi ij\ell/2\right)\right)\binom{k}{\ell}\\ &= \sum_{\ell\ge 0} \frac{1+(-1)^\ell}{2}\binom{k}{\ell}\\ &= \frac{1}{2}\sum_{\ell\ge 0} \binom{k}{\ell}+ \frac{1}{2}\sum_{\ell\ge 0} (-1)^\ell\binom{k}{\ell}\\ &= \frac{2^k+0^k}{2}. \end{align} Now taking $$b_k = \frac{x^k}{k!}\sum_{\ell \ge 0}\binom{k}{2 \ell}$$ yields \begin{align}\sum_{k=0}^\infty b_{2k}&=\sum_{k=0}^\infty \left( \frac{1}{2}\sum_{j=0}^1 \exp\left(\pi ijk\right)\right)\frac{x^k}{k!}\sum_{\ell \ge 0}\binom{k}{2 \ell}\\ &=\frac{1}{2}\sum_{k=0}^\infty \left(1+(-1)^k\right)\frac{x^k}{k!}\left(2^{k-1}+\frac{1}{2}[k=0]\right)\\ &=\frac{1}{4}\sum_{k=0}^\infty \frac{x^k}{k!}2^k+\frac{1}{4}\sum_{k=0}^\infty (-1)^k\frac{x^k}{k!}2^k+\frac{1}{2}\\ &=\frac{\exp(2x)+\exp(-2x)}{4} +\frac{1}{2}\\ &=\cosh^2(x), \end{align} again matching your result. For $$a=3$$, first note that \begin{align} \sum_{\ell \ge 0}\binom{k}{3 \ell} &= \sum_{\ell\ge 0} \left(\frac{1}{3}\sum_{j=0}^2 \exp\left(2\pi ij\ell/3\right)\right)\binom{k}{\ell}\\ &= \sum_{\ell\ge 0} \frac{1+\exp(2\pi i\ell/3)+\exp(4\pi i\ell/3)}{3}\binom{k}{\ell}\\ &= \frac{1}{3}\sum_{\ell\ge 0} \binom{k}{\ell}+ \frac{1}{3}\sum_{\ell\ge 0} \exp(2\pi i/3)^\ell\binom{k}{\ell}+ \frac{1}{3}\sum_{\ell\ge 0} \exp(4\pi i/3)^\ell\binom{k}{\ell}\\ &= \frac{2^k+(1+\exp(2\pi i/3))^k+(1+\exp(4\pi i/3))^k}{3}\\ &= \frac{2^k+\exp(\pi i/3)^k+\exp(-\pi i/3)^k}{3}. \end{align} Now taking $$b_k = \frac{x^k}{k!}\sum_{\ell \ge 0}\binom{k}{3 \ell}$$ yields \begin{align}\sum_{k=0}^\infty b_{3k}&=\sum_{k=0}^\infty \left( \frac{1}{3}\sum_{j=0}^2 \exp\left(2\pi ijk/3\right)\right)\frac{x^k}{k!}\sum_{\ell \ge 0}\binom{k}{3 \ell}\\ &=\frac{1}{3}\sum_{k=0}^\infty \left(1+\exp(2\pi ik/3)+\exp(4\pi ik/3)\right)\frac{x^k}{k!}\frac{2^k+\exp(\pi i/3)^k+\exp(-\pi i/3)^k}{3}\\ &=\frac{1}{9}\sum_{k=0}^\infty (1+\exp(2\pi i/3)^k+\exp(4\pi i/3)^k)(2^k+\exp(\pi i/3)^k+\exp(-\pi i/3)^k)\frac{x^k}{k!}. \end{align} Now expand the product of trinomials to obtain 9 sums that reduce to $$\exp(cx)$$ for various constants $$c$$.

Alternatively, note that: $$\sum_{k=0}^\infty \frac{x^{ak}}{(ak)!}\sum_{\ell \ge 0}\binom{ak}{a\ell} = \left(\sum_{k=0}^\infty \frac{x^{ak}}{(ak)!}\right)^2,$$ so you might as well just compute \begin{align} \sum_{k=0}^\infty \frac{x^{ak}}{(ak)!} &= \sum_{k=0}^\infty \left( \frac{1}{a}\sum_{j=0}^{a-1} \exp\left(2\pi ijk/a\right)\right)\frac{x^k}{k!} \\ &= \frac{1}{a}\sum_{j=0}^{a-1} \sum_{k=0}^\infty \frac{(\exp(2\pi ij/a)x)^k}{k!} \\ &= \frac{1}{a}\sum_{j=0}^{a-1} \exp(\exp(2\pi ij/a)x), \end{align} and then square the result.

• Thank you very much for your answer! Is there an online platform where I can find identities such as the one you mentioned? Or is it just something one has to know from experience? – Susp1cious Nov 16 at 22:06
• Wilf's generatingfunctionology and Graham, Knuth, et al.'s Concrete Mathematics are good sources for manipulating sums. – Rob Pratt Nov 16 at 23:26
• My result for $a=2$ is slightly different. Is it possible that you forgot to eliminate the $2$ from the Binomial? Shouldn't the expression be $\sum_{k=0}^\infty b_{2k}=\sum_{k=0}^\infty\biggl(\frac{1}{2}\sum_{j=0}^1\exp(\pi\ ijk)\biggr)\frac{x^k}{k!}\sum_{l\geq0}\binom{k}{l}$ – Susp1cious Nov 17 at 8:57
• @Susp1cious, why is the bottom of your binomial coefficient $\ell$ when it is independent of $k$? – J. M. is not a mathematician Nov 17 at 15:08
• No, I made a different error, which I should have noticed because evaluating at $x=0$ should yield 1. I’ll fix it. – Rob Pratt Nov 17 at 15:20

The explicit formula is as follows: $$S_a=\frac{1}{a^2}\left(\sum_{z^a=2^a}+2\sum_{p_a(z)=0}\right)e^{az}$$ where the polynomials $$p_a$$ are given by A244608. For example, \begin{align} p_9(z)&= 1 - 13604 z^9 - 13359 z^{18} + 247 z^{27} + z^{36}\\ p_{10}(z)&=3125-383750 z^{10}-73749 z^{20}+502 z^{30}+z^{40} \end{align} whose roots in the complex plane look like follows:  The first few solutions are \begin{align} S_1&=e^{2 x}\\ S_2&=\frac{e^{-2 x}}{4}+\frac{e^{2 x}}{4}\\ S_3&=\frac{2 e^{-x}}{9}+\frac{e^{2 x}}{9}+\frac{2}{9} e^{\frac{x}{2}-\frac{1}{2} i \sqrt{3} x}+\frac{2}{9} e^{\frac{x}{2}+\frac{1}{2} i \sqrt{3} x}+\frac{1}{9} e^{-x-i \sqrt{3} x}+\frac{1}{9} e^{-x+i \sqrt{3} x}\\ S_4&=\frac{e^{-2 x}}{16}+\frac{1}{8} e^{(-1-i) x}+\frac{1}{8} e^{(-1+i) x}+\frac{1}{16} e^{-2 i x}+\frac{1}{16} e^{2 i x}+\frac{1}{8} e^{(1-i) x}+\frac{1}{8} e^{(1+i) x}+\frac{e^{2 x}}{16} \end{align} as given by \begin{align} p_1(z)&=0\\ p_2(z)&=0\\ p_3(z)&=1+z^3\\ p_4(z)&=4+z^4\\ p_5(z)&=-1+11z^5+z^{10}\\ p_6(z)&=-27+26z^6+z^{12} \end{align} etc. Quoting the OEIS entry, the coefficients are found as follows:

Let $$\omega$$ be a primitive $$j$$-th root of unity. Let $$L(k)=\sum_{p=0}^{j-1} c(p)\omega^{kp}$$ with $$c(0)=2$$ and $$c(i)=C(j,i)$$ if $$i>0$$. Then $$p(j,X)=(X-L(1))(X-L(2))\dots(X-L([(n-1)/2]))$$.

Actually the true generalisation for your $$a=1$$, $$a=2$$ cases is this expression for your sum: $$\frac{1}{a^2}\sum_{s=0}^{a-1}\sum_{r=0}^{a-1}e^{x\omega^r(1+\omega^s)}.$$ where $$\omega$$ is an $$a$$-th root of unity ($$\omega^a=1$$, $$a \neq 1$$).

When $$a=1$$ we have $$\omega=1$$ and your sum is $$e^{2x}$$. For $$a=2$$ we have $$\omega=-1$$ and your sum is $$\frac{1}{4}(e^{2x}+e^{-2x}+e^0+e^0)=\frac{1}{4}(e^{2x}+e^{-2x}+2)$$. Both agreeing with your observations.

Proof: Using the fact that $$\sum_{l=0}^{a-1}\omega^{lr}=0$$ unless $$a|l$$ we can show: $$\sum_{l=0}^k\binom{ak}{al}=\frac{1}{a}\sum_{s=0}^{a-1}(1+\omega^s)^{ak}$$ and$$\sum_{k=0}^\infty\frac{x^{ak}}{(ak)!}=\frac{1}{a}\sum_{r=0}^{a-1}e^{\omega^r x}.$$

Substituting the above expressions into your sum we obtain:

$$\sum_{k=0}^\infty\frac{x^{ak}}{(ak)!}\biggl(\sum_{l=0}^k\binom{ak}{al}\biggr)=\sum_{k=0}^\infty\frac{x^{ak}}{(ak)!}\frac{1}{a}\sum_{s=0}^{a-1}(1+\omega^s)^{ak}=\frac{1}{a}\sum_{s=0}^{a-1}\sum_{k=0}^\infty\frac{(x(1+\omega^s))^{ak}}{(ak)!}$$$$=\frac{1}{a}\sum_{s=0}^{a-1}(\frac{1}{a}\sum_{r=0}^{a-1}e^{x\omega^r(1+\omega^s)})=\frac{1}{a^2}\sum_{s=0}^{a-1}\sum_{r=0}^{a-1}e^{x\omega^r(1+\omega^s)}.$$

• "Actually the true...": Are you suggesting that something is untrue in my solution? I did a sanity check via WolframAlpha that your double sum matches my square of a sum for $a=3$.. – Rob Pratt Nov 17 at 19:05
• @RobPratt No but your answer was not a finite expression for the sum in question in terms of exponential functions which was the case for a=1 and a=2 in the OP's question. – Ivan Meir Nov 17 at 19:17
• Right, I didn't complete the computation for $a=3$ and instead gave a result for general $a$ in terms of the square of a finite sum of exponentials. – Rob Pratt Nov 17 at 19:19