\hspace{-16}$If $\mathbf{f:\mathbb{N}\rightarrow \mathbb{R}}$ be a function such that $\mathbf{f(1)=\frac{2007}{6}}$ and \\\\\\ $\mathbf{\frac{f(1)}{1}+\frac{f(2)}{2}+\frac{f(3)}{3}+.........+\frac{f(n)}{n}=\frac{n+1}{2}.f(n)\forall n\in \mathbb{N}}$\\\\\\ Then Find $\mathbf{\lim_{n\rightarrow \infty}(2008+n).f(n)=}$
thanks for pointing vivek
now i have edited it......
71@man111,
I have a confusion. We have f : N → N, And f(1) = 2007/6 ≠Natural Number?
It is also evident from setting n = 1 in second expression.
1\texttt{one can easily figure out that} \\ f(n+1)=\frac{\left(n+1 \right)^2}{n\left(n+3 \right)}f(n) \\ \texttt{Hence , we get} \\ f(n)=\frac{6n}{\left(n+1 \right)\left(n+2 \right)}f(1)\\ \lim_{n\rightarrow \infty}\left(2008+n \right)f(n)=6f(1)\\ \boxed{Ans.=2007}
1@Aditya
\underline{\frac{f(1)}{1}+\frac{f(2)}{2}+\cdots+\frac{f(n)}{n}}+\frac{f(n+1)}{n+1}=\frac{n+2}{2}f(n+1) \\ \\ \\ \frac{n+1}{2}f(n)+\frac{f(n+1)}{n+1}=\frac{(n+2)}{2}f(n+1)
71See, from the given relation, Set n =2 , You would get f(1) = f(2)
Now Set n = 3 and using the derived relation we find out f(3) as
f(3) = 3 2 . f(2)2 . 5
Set n = 4 and we get f(4) = 4 2 . f(3)3 . 6
Inductively, we have f(n+1) = (n+1) 2 . f(n)n . (n+3)
Which boils down to what xYz has done.
EDIT: Sorry I didn't see your reply, was typing this.. Anyways, I didn't find it useful to delete.
