Return to Answer

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 $c$.

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.

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 $a=2$, 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}, \end{align} again matching your result. For $a=3$, 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!}2^k\frac{1+\exp(2\pi ik/3)+\exp(4\pi ik/3)}{3}\\ &=\frac{1}{9}\sum_{k=0}^\infty \left(1+\exp(2\pi ik/3)+\exp(4\pi ik/3)\right)^2\frac{(2x)^k}{k!}. \end{align} Now expand the square to obtain 9 terms that reduce to 3 terms involving $\exp(2cx)$ for various constants $c$.

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 $c$.

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 $a=2$, 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}, \end{align} For $a=3$, 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!}2^k\frac{1+\exp(2\pi ik/3)+\exp(4\pi ik/3)}{3}\\ &=\frac{1}{9}\sum_{k=0}^\infty \left(1+\exp(2\pi ik/3)+\exp(4\pi ik/3)\right)^2\frac{(2x)^k}{k!}. \end{align} Now expand the square to obtain 9 terms that reduce to 3 terms involving $\exp(2cx)$ for various constants $c$.

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 $a=2$, 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}, \end{align} again matching your result. For $a=3$, 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!}2^k\frac{1+\exp(2\pi ik/3)+\exp(4\pi ik/3)}{3}\\ &=\frac{1}{9}\sum_{k=0}^\infty \left(1+\exp(2\pi ik/3)+\exp(4\pi ik/3)\right)^2\frac{(2x)^k}{k!}. \end{align} Now expand the square to obtain 9 terms that reduce to 3 terms involving $\exp(2cx)$ for various constants $c$.