How to solve the recurrence $a_{0}=1 , (n+1)a_{n+1}=a_{n}+(\frac{1}{n!}), n\geq 0$

Question

I know that I should got out the n!, since there is no root for it , if I multiply the both side by n! then it will become like :

$n! (n+1)a_{n+1}=a_{n}n!+1$

And then I have to put $b_n=a_n n!\ \ \ \ $ right ??

Any idea how could I get rid the (n+1) ??

Answer 1

Just multiply each side by $n!$, then you get $$ (n+1)!a_{n+1}=n!a_n+1 $$ or $$ (n+1)!a_{n+1}-n!a_n=1 $$ which is straightforward to solve by telescoping.

Answer 2

You are on the right track. By letting $b_n=a_n n!$ the recursion becomes $b_0=1$ and for $n\geq0$, $$b_{n+1}=b_n+1.$$ Now find $b_n$ and then $a_n=\frac{b_n}{n!}$.

Answer 3

I will write the equation as $a_{n+1}=\frac1{n+1}a_n+\frac1{(n+1)!}$.

The answer is $a_n=\frac{n+1}{n!}$. One can see this by induction. It is clearly true for $a_0$. Assume true for $a_n$. Then we have that

$a_{n+1}=\frac1{n+1}\frac{n+1}{n!}+\frac1{(n+1)!}$

writing this as a single fraction we have

$a_{n+1}=\frac{n+2}{(n+1)!}$ and we're done.