# find an upper bound of the sequence like $a_{n+1}=\frac{na_n+a_n^2}{n+1}$

Given a sequence ${a_n}$, $a_{n+1}=\frac{na_n+a_n^2}{n+1}$, $a_1=\frac{1}{2}$.

i can prove that the $\{a_n\}$ is decreasing, and the limit is 0.

but by using computer i found that the limit of $\{n\cdot a_n\}$ is about 1.3... .

how to prove the $\{n\cdot a_n\}$ has a upper bound.

• What are you trying to show? Are you trying to show that the limit of $a_n$ as $n$ goes to infinity is zero? It sounds like you've already done that. Why do you need an upper bound for $na_n$? – Joshua Ruiter Feb 16 '17 at 5:58
• if it is strictly decreasing, you don't even need an upper bound. – Saketh Malyala Feb 16 '17 at 5:59
• It seems natural to consider $b_n=n\cdot a_n$; then $b_{n+1}=b_n+{b_n^2\over n^2}$, and the problem becomes somewhat more well-known: math.stackexchange.com/questions/1672888/…. – Ivan Neretin Feb 16 '17 at 7:54