3

I'm aware that I have to use the expansion for $e^x$ for this problem, ie: $e^x =\sum_{n=0}^{\infty} \frac{x^n}{n!}$.

However the problem is with the computation, and it's getting too long, I'm trying to find a method which wouldn't take as much time to what I have done: I broke the series into two parts and computed them separately:
$$\sum_{n=0}^{\infty}\frac{n^3((2n)!)+(2n-1)(n!)}{(n!)((2n)!)}=\sum_{n=0}^{\infty}\frac{n^3}{n!}+\sum_{n=0}^{\infty} \frac{2n-1}{(2n)!}$$ Now: $$\sum_{n=0}^{\infty}\frac{n^3}{n!}=0+\sum_{n=1}^{\infty}\frac{n^2}{(n-1)!}=\sum_{n=1}^{\infty}\frac{n^2-1+1}{(n-1)!}=\sum_{n=1}^{\infty}\frac{1}{(n-1)!}+\sum_{n=1}^{\infty}\frac{(n-1)(n+1)}{(n-1)!}$$

$$=\sum_{n=1}^{\infty}\frac{1}{(n-1)!}+\sum_{n=2}^{\infty}\frac{(n+1)}{(n-2)!}=\sum_{n=1}^{\infty}\frac{1}{(n-1)!}+\sum_{n=2}^{\infty}\frac{(n-2)+3}{(n-2)!}=\sum_{n=1}^{\infty}\frac{1}{(n-1)!}+\sum_{n=2}^{\infty}\frac{3}{(n-2)!}+\sum_{n=3}^{\infty}\frac{1}{(n-3)!}$$ $$=e+3e+e=5e$$ This was tiresome, and for the second sum: $$\sum_{n=0}^{\infty} \frac{2n-1}{(2n)!}=\sum_{n=1}^{\infty}\frac{1}{(2n-1)!}-\sum_{n=0}^{\infty} \frac{1}{(2n)!}$$ $$e+\frac{1}{e}=2\cdot{\sum_{n=0}^{\infty} \frac{1}{(2n)!}} ~~,e-\frac{1}{e}=2\cdot\sum_{n=1}^{\infty}\frac{1}{(2n-1)!}$$ Hence our answer is: $5e - \frac{1}{e}$
I'm trying to find short/elegant methods which solves the problem in a significant lower amount of steps. Like maybe if I'm able to convert this into a problem where I'm able to use converse of cessaro- stolz theorem, or something else which I'm unaware about in the moment.

q123LsaB
  • 327
  • 1
  • 12
  • 1
    One potentially nice technique will be to use $x \frac{d}{dx}$ - it has the nice property that if $\sum_{n=0}^\infty a_n x^n = f(x)$, then $\sum_{n=0}^\infty n^m a_n x^n = \left( x \frac{d}{dx} \right)^m f(x)$. For example, $\sum_{n=0}^\infty n^3 \frac{x^n}{n!} = \left( x \frac{d}{dx} \right)^3 e^x = (x+3x^2+x^3)e^x$, so $\sum_{n=0}^\infty n^3 \frac{1}{n!} = 5e$. – user196574 Aug 08 '23 at 06:10
  • what does $ (x \frac{d}{dx})^m f(x) $ mean? What operation are you doing here? Also how do you get the following result: $\sum_{n=0}^{\infty} n^m a_n x^n = (x \frac{d}{dx})^m f(x)$ – q123LsaB Aug 08 '23 at 09:16
  • 1
    For example, $(x \frac{d}{dx})^2 = x \frac{d}{dx} x \frac{d}{dx}$, and you act it as follows: $(x \frac{d}{dx})^2 x^n = x \frac{d}{dx}\left( x \frac{d}{dx}\left( x^n \right) \right) = x \frac{d}{dx}\left( x n x^{n-1} \right) = n x \frac{d}{dx}\left(x^n \right) = n x n x^{n-1} = n^2 x^n$ – user196574 Aug 08 '23 at 20:58

1 Answers1

2

1. The first sum can be computed combinatorially using Dobiński's formula, which states that

$$ S_k := \sum_{n=0}^{\infty} \frac{n^k}{n!} = e B_k, $$

where $B_n$ is the $n$-th Bell number. (This, however, also tells that computing $S_k$ for general $k$ is as hard as computing Bell numbers, which is a challenging problem.)

Alternatively, $S_k$ satisfies the following recurrence relation:

$$ S_0 = e, \qquad S_{k+1} = \sum_{n=1}^{\infty} \frac{n^k}{(n-1)!} = \sum_{n=0}^{\infty} \frac{(n+1)^k}{n!} = \sum_{j=0}^{k} \binom{k}{j} S_j $$

This helps us quickly compute $S_k$'s for small $k$'s. In OP's case, we get

\begin{align*} S_1 &= S_0 = e, \\ S_2 &= S_0 + S_1 = 2e, \\ S_3 &= S_0 + 2S_1 + S_2 = 5e. \end{align*}

2. For the second sum, a slightly faster method is as follows:

\begin{align*} \sum_{n=0}^{\infty} \frac{2n-1}{(2n)!} = \sum_{n=1}^{\infty} \frac{1}{(2n-1)!} - \sum_{n=0}^{\infty} \frac{1}{(2n)!} = -\sum_{k=0}^{\infty} \frac{(-1)^k}{k!} = -e^{-1}. \end{align*}

Sangchul Lee
  • 167,468
  • I understood 1) , thank you for letting me know about the recursive formula. However I'm finding the following transformation in 2) , non trivial. ie: $\sum_{n=1}^{\infty} \frac{1}{(2n-1)!} - \sum_{n=0}^{\infty} \frac{1}{(2n)!} = -\sum_{k=0}^{\infty} \frac{(-1)^k}{k!}$, can you explain the simplification? – q123LsaB Aug 08 '23 at 09:07
  • @q123LsaB, It would be easier to start from the sum $\sum_{k=0}^{\infty}\frac{(-1)^k}{k!}$. Try to collect the even-th terms and odd-th terms separately, and see what you get! :) – Sangchul Lee Aug 08 '23 at 10:57
  • Got it thank you very much, its indeed faster :) – q123LsaB Aug 08 '23 at 12:27