TITLE: Ramanujan's q function
QUESTION [6 upvotes]: I stuck at the following problem:
Let
\begin{equation}
Q(n) := \sum_{k \geq 0}\frac{(n-1)_k}{n^k}
\end{equation}
where $(n)_k = n (n-1) \ldots (n-k + 1)$.
I want to show the following equation:
\begin{equation}
1 + Q(n) = \int^{\infty}_{0}{e^{-x}\left(1 + \frac{x}{n} \right)^n dx}
\end{equation}
By induction for $n \geq 1$.  For $n = 1$ this is clear.
For the induction step I got
\begin{equation}
Q(n+1) = \sum_{k \geq 0}{\frac{(n)_k}{(n+1)^k}} = \sum_{k \geq 0}{\frac{(n-1)_k}{(n+1)^k}\frac{n}{n-k}}  
\end{equation}
and
\begin{equation}
\int^{\infty}_{0}{e^{-x}\left(1 + \frac{x}{n+1} \right)^{n+1} dx} = 1 + \int^{\infty}_{0}{e^{-x} \left( 1 + \frac{x}{n+1}\right)^n dx}
\end{equation}
I want to connect $Q(n+1)$ and the integral expression above by the induction requirement. I used substitution for the integral but this wasn't helpful. I would be thankful if anyone can give me a hint.

REPLY [3 votes]: There is no need of induction, it actually increases our effort. Simplification of the integral will directly give us the summation.
$$\int_{0}^{\infty}e^{-x}\left(1+\frac{x}{n}\right)^{n}dx$$
Using the binomial theorem we can expand $\left(1+\frac{x}{n}\right)^{n}=\sum_{k=0}^{n}\binom{n}{k}\left(\frac{x}{n}\right)^{k}$.
$$\sum_{k=0}^{n}\binom{n}{k}\frac{1}{n^{k}}\int_{0}^{\infty} x^{k}e^{-x}dx$$
The expression inside the integral is gamma function or factorial function.
$$\sum_{k=0}^{n}\binom{n}{k}\frac{k!}{n^{k}}$$
Note that, $k! \binom{n}{k}=\frac{n!}{(n-k)!}=(n-k+1)(n-k+2)..(n)$.
So summation is,
$$\sum_{k=0}^{n}\frac{(n)...(n-k+2)(n-k+1)}{n^{k}}$$
Which proves the required claim