Equating the coefficients in $\sum_{n=1}^\infty C_{n-1} \frac{t^n}{n!}=\sum_{n=0}^\infty C_n \frac{t^n}{n!} \sum_{m=0}^\infty \frac{t^m}{m!}$

Question

$$\sum_{n=1}^\infty C_{n-1} \frac{t^n}{n!}=\sum_{n=0}^\infty C_n \frac{t^n}{n!} \sum_{m=0}^\infty \frac{t^m}{m!}$$

In the above equation is it possible to equate the coefficients of $\frac{t^k}{k!}$? Or do we need to consider $(\frac{t^k}{k!})^2$? Could you please give me an idea for this?

Answer 1

Equating the coefficients at equal powers of $t$ on both sides one obtains:

$$C_{n-1}=\sum_{k=0}^n\frac{n!}{k!(n-k)!}C_k.$$

The equating process goes as follows:

$$\sum_{k=0}^\infty C_k \frac{t^k}{k!} \sum_{m=0}^\infty \frac{t^m}{m!} =\sum_{k=0}^\infty C_k\frac1{k!}\sum_{m=0}^\infty \frac{t^{k+m}}{m!}\\ \stackrel{n=k+m}=\sum_{k=0}^\infty C_k\frac1{k!} \sum_{n=k}^{\infty} \frac{t^{n}}{(n-k)!} =\sum_{n=0}^\infty t^n\sum_{k=0}^n C_k \frac{1}{k!(n-k)!}\\ =\sum_{n=0}^\infty\frac{t^n}{n!}\sum_{k=0}^n C_k \frac{n!}{k!(n-k)!}.$$

The reason for the possibility of the exchange of the summation order in the second line is that both double sums run over the same set of indices: $0\le k\le n<\infty$. It can be seen also from the fact that the summation in both $\sum_{n=k}^{\infty}$ and $\sum_{k=0}^n$ formally can run from $0$ to infinity due to $(n-k)!$ term.

Answer 2

$\sum_{n=1}^\infty C_{n-1} \frac{t^n}{n!} =\sum_{n=0}^\infty C_n \frac{t^n}{n!} \sum_{m=0}^\infty \frac{t^m}{m!} $

Here's how you get the binomial coefficients.

Note: These are called exponential generating functions because of their similarity to $e^t =\sum_{m=0}^\infty \frac{t^m}{m!} $.

$\begin{array}\\ \sum_{n=0}^\infty C_n \frac{t^n}{n!} \sum_{m=0}^\infty \frac{t^m}{m!} &=\sum_{n=0}^\infty\sum_{m=0}^\infty C_n \frac{t^{n+m}}{n!m!}\\ &=\sum_{k=0}^\infty\sum_{n=0}^k C_n \frac{t^{k}}{n!(k-n)!} \qquad n+m = k, m = k-n\\ &=\sum_{k=0}^\infty\frac{t^k}{k!}\sum_{n=0}^k C_n \frac{k!}{n!(k-n)!}\\ &=\sum_{k=0}^\infty\frac{t^k}{k!}\sum_{n=0}^k C_n \binom{k}{n}\\ \end{array} $