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Compute the sum of the series ∑ (x-1)n / (n+2)!

by RoRi

Find all x \in \mathbb{R} such that the series

    \[ \sum_{n=0}^{\infty} \frac{(x-1)^n}{(n+2)!} \]

converges and compute the sum.

First, we apply the ratio test,

    \begin{align*} \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| &= \lim_{n \to \infty} \left( \frac{|x-1|^{n+1}}{(n+3)!} \right) \left( \frac{(n+2)!}{|x-1|^n} \right) \\[9pt] &= \lim_{n \to \infty} \frac{|x-1|}{n+3} \\[9pt] &= 0 \end{align*}

for all x. Therefore, the series converges for all x \in \mathbb{R}. Next, we compute the sum for x \neq 1,

    \begin{align*} \sum_{n=0}^{\infty} \frac{(x-1)^n}{(n+2)!} &= \sum_{n=2}^{\infty} \frac{(x-1)^{n-2}}{n!} \\[9pt] &= \frac{1}{(x-1)^2} \sum_{n=2}^{\infty} \frac{(x-1)^n}{n!} \\[9pt] &= \frac{1}{(x-1)^2} \left( \sum_{n=0}^{\infty} \frac{(x-1)^n}{n!} - 1 - x + 1 \right) \\[9pt] &= \frac{1}{(x-1)^2} \left( e^{x-1} - x \right). \end{align*}

In the case x = 1 then all of the terms except the n = 0 term are 0, and that term is \frac{1}{2}; hence, the sum is \frac{1}{2}.

3 comments

  2. Giovanni says:

    When you re-open the sum from n=2 back to n=0, iget that you are subtracting the terms corresponding no n=2 and n=1, in that case, shouldnt you have a (x^2)/2 term instead of a x term? Just wondering.

    • RoRi says:

      In the third line of the second equation block, right? I’m adding and subtracting the terms corresponding to n = 0 and n =1 (adding the terms inside the sum, and the subtracting them outside the sum), not the n =2 term. That’s already in the sum. For the benefit of anyone looking at this later, I’ll expand on what happened in that line (it looks a bit terse in retrospect). The n = 0 term is

          \[ \frac{(x-1)^n}{n!} = \frac{(x-1)^0}{0!} = 1 \]

      and the n = 1 term is

          \[ \frac{(x-1)^1}{1!} = (x-1). \]

      So then we have

          \begin{align*} \sum_{n=2}^{\infty} \frac{(x-1)^n}{n!} &= \sum_{n=2}^{\infty} \frac{(x-1)^n}{n!} + (1+(x-1)) - (1+(x-1)) \\[9pt] &= \sum_{n=0}^{\infty} \frac{(x-1)^n}{n!} - x. \end{align*}

      Hopefully this makes sense.

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