Find the sum \(\displaystyle \frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} + \frac{4}{5!} + \dotsb\)
+9519 We notice that the n-th term is \(\dfrac{n}{(n + 1)!}\)
If we did some sort of arithmetic trick on it:
\(\dfrac{n}{(n + 1)!} = \dfrac{(n + 1) - 1}{(n + 1)!} = \dfrac{(n + 1)}{(n + 1)(n!)} - \dfrac{1}{(n + 1)!} = \dfrac1{n!} - \dfrac1{(n + 1)!}\)
That means, the sum is \(\displaystyle\sum_{n = 1}^\infty \left(\dfrac1{n!} - \dfrac1{(n + 1)!}\right)\).
To evaluate general infinite sums, we consider the partial sum first, then take the limit.
So first, we consider \(\displaystyle\sum_{n = 1}^N \left(\dfrac1{n!} - \dfrac1{(n + 1)!}\right)\).
Listing the terms,
\(\displaystyle\sum_{n = 1}^N \left(\dfrac1{n!} - \dfrac1{(n + 1)!}\right) = \left(\dfrac1{1!} - \dfrac1{2!}\right) + \left(\dfrac1{2!} - \dfrac1{3!}\right) + \left(\dfrac1{3!} - \dfrac1{4!}\right) + \cdots + \left(\dfrac1{N!} - \dfrac1{(N+1)!}\right)\)
As you may have noticed, the terms in the middle eliminates each other nicely. The only terms left are the first, and the last.
\(\displaystyle\sum_{n = 1}^N \left(\dfrac1{n!} - \dfrac1{(n + 1)!}\right) = \dfrac1{1!} - \dfrac1{(N + 1)!} = 1 - \dfrac1{(N + 1)!}\)
Next, we will take \(\displaystyle\lim_{N\to \infty}\) on both sides.
(If you don't know limits, don't panic. The above notation just means "N is a really large number".)
\(\displaystyle\lim_{N\to \infty}\displaystyle\sum_{n = 1}^N \left(\dfrac1{n!} - \dfrac1{(n + 1)!}\right)= \displaystyle\lim_{N\to \infty}\left(1 - \dfrac1{(N + 1)!}\right)\\ \displaystyle\sum_{n=1}^\infty \dfrac{n}{(n + 1)!} = 1 - \dfrac1{\text{Really big number}} = 1\)
The required sum is 1.
