Is the information sufficient 5?

Relevant wiki: Telescoping Series - Product

First we are going to prove by induction that $f(n) = \frac{2f(1)}{n(n+1)}$ . This is obviously clear for $n=1$ .

Considering the case $n=2$ we obtain

$f(2)=\frac{f(1) + f(2)}{2^2}$

$3f(2)=f(1)$

$f(2)=\frac{f(1)}{3}$ ,

so our statement is true for $n=2$ too. Assuming that the induction hypothesis is proven for all positive integers $k<n$ with $n>2$ we have from the definition

$(n-1)(n+1)f(n) = (n^2-1)f(n) = \displaystyle \sum_{i=1}^{n-1} f(i) = \displaystyle \sum_{i=1}^{n-1} \frac{2f(1)}{i(i+1)} = 2f(1) \displaystyle \sum_{i=1}^{n-1} \frac{1}{i(i+1)} = 2f(1) \displaystyle \sum_{i=1}^{n-1} (\frac{1}{i} - \frac{1}{i+1}) = 2f(1)(1 - \frac{1}{n}) = 2f(1) \frac{n-1}{n}$ ,

since the terms of the telescoping series cancel out. Thus we obtain $f(n) = 2f(1) \frac{n-1}{n(n-1)(n+1)} = \frac{2f(1)}{n(n+1)}$ hence our induction is complete.

Using this result for $n>1$ we get

$\lim_{n \to \infty} n^2f(n) = \lim_{n \to \infty} 2f(1) \frac{n^2}{n(n+1)} = 2f(1) \lim_{n \to \infty} \frac{1}{1+ \frac{1}{n}} = 2f(1) = 4034$ .

Thus ultimately our answer is $\frac{4034 + 2016}{11} = 550$ .

$\begin{aligned} f(n) & = \frac {f(1)+f(2)+f(3)+ \cdots + f(n)}{n^2} \\ n^2 f(n) & = \sum_{k=1}^n f(k) \\ (n+1)^2 f(n+1) & = \sum_{k=1}^{n+1} f(k) \\ (n+1)^2 f(n+1) - n^2 f(n) & = \sum_{k=1}^{n+1} f(k) - \sum_{k=1}^n f(k) \\ (n^2+2n+1) f(n+1) - n^2 f(n) & = f(n+1) \\ n(n+2)f(n+1) & = n^2 f(n) \\ \implies f(n+1) & = \frac n{n+2}f(n) \\ f(2) & = \frac 13 f(1) \\ f(3) & = \frac 24 \cdot \frac 13 f(1) \\ f(4) & = \frac 35 \cdot \frac 24 \cdot \frac 13 f(1) \\ \cdots \ & = \ \cdots \\ \implies f(n) & = \prod_{k=1}^{n-1} \frac k{k+2}f(1) = \prod_{k=1}^{n-1} \frac {2017k}{k+2} = \frac {4034(n-1)!}{(n+1)!} \\ & = \frac {4034}{n(n+1)} = \frac {4034}{n^2+n} \end{aligned}$

$\begin{aligned} \implies \lim_{n \to \infty} n^2f(n) & = \lim_{n \to \infty} \frac {4034n^2}{n^2+n} = \lim_{n \to \infty} \frac {4034}{1+\frac 1n} = 4034 \\ \implies \frac {\displaystyle \lim_{n \to \infty} n^2f(n)+2016}{11} & = \frac {4034+2016}{11} = \boxed{550} \end{aligned}$