I have a power series nestled inside a power series, and the inside power series is taken to a power. I need to isolate the term that would go like $t^{2mn}$ so that I could integrate it.

I'm not looking to sum up the series, since that's where I'm coming from.

$$ \sum_{n=0}^{\infty} (-1)^n \left(\sum_{m=0}^{\infty} \; \frac{(-1)^m}{(2m)!} \; \beta^{2m} t^{2m} \right)^n $$

Any ideas are appreciated.

  • 2
    $\begingroup$ The inner sum is just the cosine, no? $\endgroup$ Commented Mar 28, 2019 at 5:09
  • $\begingroup$ @J.M.isnotamathematician. Using your comment, the outer sum is simple too. Cheers $\endgroup$ Commented Mar 28, 2019 at 5:31

2 Answers 2


Here are two approaches to extract the coefficient of $t^{2MN}$. One is based upon multinomial expansion, the other is based upon the representation $\cos t=\frac{1}{2}\left(e^{it}+e^{-it}\right)$. Regrettably, none of them seems really promising. Nevertheless, here we go!

It is convenient to use the coefficient of operator $[t^n]$ to denote the coefficient of $t^n$.

First approach:

We obtain \begin{align*} \color{blue}{[t^{2MN}]}&\color{blue}{\sum_{n=0}^\infty(-1)^n\left(\sum_{m=0}^\infty\frac{(-1)^m}{(2m)!}\beta^{2m}t^{2m}\right)^n}\\ &\sum_{n=0}^\infty(-1)^n[t^{2MN}]\left(\sum_{m=0}^{MN}\frac{(-1)^m}{(2m)!}\beta^{2m}t^{2m}\right)^n\\ &=\sum_{n=0}^\infty(-1)^n[t^{2MN}]\sum_{{k_0+\cdots+k_{MN}=n}\atop{k_j\geq 0,\ ,0\leq j\leq MN}} \binom{n}{k_0,\ldots,k_{MN}}\prod_{j=0}^{MN}\left(\frac{\left(-\beta^2\right)^j}{(2j)!}t^{2j}\right)^{k_j}\\ &\,\,\color{blue}{=\sum_{n=0}^\infty(-1)^n[t^{2MN}]\sum_{{k_0+\cdots+k_{MN}=n}\atop{k_j\geq 0,\ ,0\leq j\leq MN}}\binom{n}{k_0,\ldots,k_{MN}}}\\ &\qquad\quad\color{blue}{\cdot\prod_{j=0}^{MN}\left((2j)!\right)^{-k_j}\left(-\beta^2\right)^{\sum_{j=1}^{N}jk_j}t^{2\sum_{j=1}^{N}jk_j}}\tag{1}\\ \end{align*}

We observe in order to find the coefficient of $t^{2MN}$ the expression (1) boils down to find the solutions of \begin{align*} k_0+k_1+\cdots+k_{MN}&=n\\ k_1+2k_2+\cdots+MNk_{MN}&=MN\qquad\qquad k_j\geq 0, \,\, 0\leq j\leq MN \end{align*} which indicates connections with Bell polynomials.

The second approach is somewhat more promising, since we can completely extract the coefficient of $t^{2MN}$ . But, we will see the final expression is far from being simple.

Second approach:

\begin{align*} \color{blue}{[t^{2MN}]}&\color{blue}{\sum_{n=0}^\infty(-1)^n\left(\sum_{m=0}^\infty\frac{(-1)^m}{(2m)!}\beta^{2m}t^{2m}\right)^n}\\ &=[t^{2MN}]\sum_{n=0}^\infty(-1)^n\cos^n(\beta t)\tag{2}\\ &=[t^{2MN}]\sum_{n=0}^\infty(-1)^n\frac{1}{2^n}\left(e^{i\beta t}+e^{-i\beta t}\right)^n\tag{3}\\ &=[t^{2MN}]\sum_{n=0}^\infty\left(-\frac{1}{2}\right)^ne^{-in\beta t}\left(1+e^{2i\beta t}\right)^n\tag{4}\\ &=[t^{2MN}]\sum_{n=0}^\infty\left(-\frac{1}{2}\right)^n\sum_{j=0}^\infty\frac{\left(-in\beta\right)^j}{j!}t^j\left(1+e^{2i\beta t}\right)^n\tag{5}\\ &=\sum_{n=0}^\infty\left(-\frac{1}{2}\right)^n\sum_{j=0}^{2MN}\frac{\left(-in\beta\right)^j}{j!}[t^{2MN-j}]\left(1+e^{2i\beta t}\right)^n\tag{6}\\ &=\sum_{n=0}^\infty\left(-\frac{1}{2}\right)^n\sum_{j=0}^{2MN}\frac{\left(-in\beta\right)^{2MN-j}}{(2MN-j)!}[t^{j}]\sum_{k=0}^n\binom{n}{k}e^{2ik\beta t}\tag{7}\\ &=\sum_{n=0}^\infty\left(-\frac{1}{2}\right)^n\sum_{j=0}^{2MN}\frac{\left(-in\beta\right)^{2MN-j}}{(2MN-j)!} [t^{j}]\sum_{k=0}^n\binom{n}{k}\sum_{l=0}^\infty \frac{\left(2ik\beta\right)^l}{l!}t^l\tag{8}\\ &=\sum_{n=0}^\infty\left(-\frac{1}{2}\right)^n\sum_{j=0}^{2MN}\frac{\left(-in\beta\right)^{2MN-j}}{(2MN-j)!}\sum_{k=0}^n\binom{n}{k}\frac{(2ik\beta)^j}{j!}\tag{9}\\ &\,\,\color{blue}{=\frac{\left(-\beta^2\right)^{MN}}{(MN)!}\sum_{n=0}^\infty\left(-\frac{1}{2}\right)^nn^{2MN} \sum_{j=0}^{2MN}\binom{2MN}{j}\left(-\frac{2}{n}\right)^j\sum_{k=0}^n\binom{n}{k}k^j}\tag{10} \end{align*}


  • In (2) we use the series expansion $\cos t=\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}t^{2n}$.

  • In (3) we use the representation $\cos t=\frac{1}{2i}\left(e^{it}+e^{-it}\right)$.

  • In (4) we factor out $e^{-in\beta t}$.

  • In (5) we expand $e^{-in\beta t}$.

  • In (6) we apply the rule $[t^{p-q}]A(t)=[t^p]t^qA(t)$ and we set the upper limit of the sum with index $j$ to $2MN$, since other terms do not contribute.

  • In (7) we change the order of the inner sum $j\to 2MN-j$.

  • In (8) we expand $e^{2ik\beta t}$.

  • In (9) select the coefficient of $t^j$.

  • In (10) we do some final rearrangements.


Seems like a strange approach but the only thing I think would help would be this: https://en.m.wikipedia.org/wiki/Cauchy_product Although you'd need to establish a general form for the coefficients when multiplying more than 2 series, which would include several nested finite sums, making the result really ugly


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