Problem with limit of recursive sequence having the actual index "n" in denominator.

Question

The given sequence is: $$ a_1 = 1; \quad a_{n+1} = \frac{2a_n}{n+1} $$ $n$ is natural. What is the limit if $n\to\infty$? Please help!

Answer 1

I find that, whenever there is an $n$ or $n+1$ in a recurrence, throwing a $n!$ or $(n+1)!$ somewhere usually helps.

You have $a_{n+1} = \frac{2a_n}{n+1} $.

If we multiply by $(n+1)!$, this becomes $(n+1)!a_{n+1} = 2n!a_n $. (Note that multiplying by $n!$ or dividing by $n!$ do not work.)

Letting $b_n =n!a_n $, we get $b_{n+1} =2b_n $, which you should be able to solve.

Then get $a_n$ by $\frac{b_n}{n!} $.

Answer 2

Once you have sorted out the closed form for $a_n$, you can try to determine convergence of yhe sum $\sum_n \frac{2^n}{n!}$. If it convrges, then the limit of the series is obvious.