I need help with proving the partial sum formula for the following series with induction:
$$ \sum^\infty_{n=1}\frac{n}{(n+1)!}$$
I found a partial sum formula for the series: $$S_n=1-\dfrac{1}{(n+1)!}$$
I am stuck on proving the partial sums with induction.
My attempt is:
For the base case:
$S_1=\dfrac{1}{2}$
Next assume true for k:
$$S_k=\dfrac{k}{(k+1)!}$$
However, at this point I am not sure how to proceed with proving $k+1$, if someone could give me some pointers that would be awesome.
Best Answer
Let $a_k = \frac{k}{(k+1)!}$ be the $k^{\mathrm{th}}$ term. Your hypothesis is that $S_n = 1-\frac{1}{(n+1)!}$ for all $n$. Given your inductive hypothesis, it is enough to show that $$ S_{k+1} = S_k + a_{k+1} = 1-\frac{1}{(k+1)!} + \frac{k+1}{(k+2)!} = 1-\frac{1}{(k+2)!}.$$ This is a calculation that I suspect you can perform.