Prove by induction, that $ \sum_{i=1}^n \frac{i}{(i+1)!}= \frac{1}{(n+1)(n-1)!}$

Question

If I'm not wrong,

$$\sum_{i=1}^n \frac{i}{(i+1)!}= \frac{1}{(n+1)(n-1)!},$$ but I am having trouble proving it by induction...

If $n=1$ the formulas coincide.

Sup $n=k$ is valid: $$\sum_{i=1}^k \frac{i}{(i+1)!}= \frac{1}{(k+1)(k-1)!}.$$

Then if $n=k+1$,

$$\sum_{i=1}^{k+1} \frac{i}{(i+1)!}= \sum_{i=1}^k \frac{i}{(i+1)!} + \frac{k+1}{(k+1)!} $$

$$ = \frac{1}{(k+1)(k-1)!} + \frac{k+1}{(k+1)!} $$ $$ = \frac{k + (k +1)}{(k+1)!} = \frac{2k + 1}{(k+1)!} $$ I really don't see from this how I should arrive to

$$ \sum_{i=1}^{k+1} \frac{i}{(i+1)!} = \frac{1}{(k+2)(k)!} $$

Answer 1

Let's see what that sum actually is, since your result is obviously wrong.

$\begin{array}\\ \sum_{i=1}^n \frac{i}{(i+1)!} &=\sum_{i=1}^n \frac{i+1-1}{(i+1)!}\\ &=\sum_{i=1}^n \frac{i+1}{(i+1)!}-\sum_{i=1}^n \frac{1}{(i+1)!}\\ &=\sum_{i=1}^n \frac{1}{i!}-\sum_{i=2}^{n+1} \frac{1}{i!}\\ &=(1+\sum_{i=2}^n \frac{1}{i!})-(\sum_{i=2}^{n} \frac{1}{i!}+\frac1{(n+1)!})\\ &=1-\frac1{(n+1)!}\\ \end{array} $

And that is what you should try to prove by induction.

Answer 2

Try this in your inductive step: $$\sum_{i=1}^{k+1} \frac{i}{(i+1)!}= \sum_{i=1}^k \frac{i}{(i+1)!} + \frac{k+1}{(k+2)!} $$

Notice the denominator is $(k + 2)!$ instead of $(k+1)!$, which makes simplification easier:

$$ = \frac{1}{(k+1)(k-1)!} + \frac{k+1}{(k+2)!} $$ $$ = \frac{(k + 2)k}{(k+2)!} + \frac{k+1}{(k+2)!} = \frac{k^2 + 3k + 1}{(k+2)!} $$

$$ = \frac{(k + 2)(k + 1) - 1}{(k+2)!} = \frac{1}{k!} - \frac{1}{(k+2)!}$$

It looks like your inductive hypothesis is incorrect.

Answer 3

For $n=2$ we have $$ \frac{1}{2!}+\frac{2}{3!}=\frac{1}{2}+\frac{1}{3}=\frac{5}{6} $$ On the other hand, $$ \frac{1}{(2+1)(2-1)!}=\frac{1}{3} $$ so your conjecture is wrong.

For $n=3$ we have $$ \frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}= \frac{12}{4!}+\frac{8}{4!}+\frac{3}{4!}= \frac{23}{24} $$ and you can make a better conjecture.