# Solution of problem T12/363 in M&Y

Let $f:\mathbb{N}\to\mathbb{R}$ be a function such that $f(1)=\dfrac{2007}{6}$ and $\dfrac{f(1)}{1}+\dfrac{f(2)}{2}+\cdots+\dfrac{f(n)}{n}=\dfrac{n+1}{2}\cdot f(n)\forall n\in\mathbb{N}$. Find the limit $\lim_{n\to\infty} (2008+n)f(n)$.

My solution.

Induction on $n:f(n)>0\forall n$.

From hypothesis we have $\sum_{i=1}^n\dfrac{f(i)}{i}=\dfrac{n+1}{2}\cdot f(n)$ and $\sum_{i=1}^n\dfrac{f(i)}{i}+\dfrac{f(n+1)}{n+1}=\dfrac{n+2}{2}\cdot f(n+1)$ therefore $\dfrac{f(n+1)}{f(n)}=\dfrac{(n+1)^2}{n(n+3)}\forall n$(*).

In (*), use telescopic product : $f(n)=\dfrac{2007n}{(n+1)(n+2)}$ therefore that limit is equal to $2007.\Box$.

## One thought on “Solution of problem T12/363 in M&Y”

1. fatih says:

nice solution..