For a function $f:\Bbb N\to\Bbb R$, the following are satisfied, (i) $f(1)=1$ and (ii) $f(1)+2f(2)+3f(3)+....+nf(n)=n(n+1)f(n).$ Find $f(n)$

Question

I found the value of $f(2), f(3),f(4)$ and observed $f(n)=\frac{1}{2n}$ for $n \gt 1$. I am able to prove this by Mathematical induction. Another approach I tried is $f(1)+2f(2)+3f(3)+....+nf(n)=n(n+1)f(n)$. Now subtract $n*f(n)$ from both sides sides, we get $f(1)+2* f(2)+3f(3)+....+(n-1)f(n-1)=(n^2)f(n)$. now replacing $n$ by $n+1$ we get $f(1)+2f(2)+3f(3)+....+nf(n)=(n+1)^2 f(n+1)$. Now if we try to obtain value of $f(2),f(3)......$ from this equation, then I get same values as from previous equations. As LHS of this equation and other equation are same, we may equate them and solve. We get, $\frac{f(n+1)}{f(n)}=\frac{n}{(n+1)}$. On doing a telescoping product, we get $f(n)=\frac{1}{n}$ which does not give correct values of $f(2),f(3)......$ Where is the fallacy?

Answer 1

We have $$f(1) + 2f(2) + \ldots + nf(n) = n(n+1)f(n)$$ Replace $n$ by $n-1$ to get $$f(1) + 2f(2) + \ldots + (n-1)f(n-1) = n(n-1)f(n-1)$$ Subtracting the second equation from the first, you'll get $$nf(n) = n(n+1)f(n) - n(n-1)f(n-1)$$ $$\frac{f(n)}{f(n-1)} = \frac{n(n-1)}{n^2} = 1 - \frac{1}{n}$$ This holds only for $n\ge 2$, because $n$ starts from $1$. When we replace $n$ by $n-1$, we're assuming $n\ge 2$, because $f(0)$ does not make sense. The idea is simple, we compute $$\frac{f(n)}{f(2)} = \frac{f(n)}{f(n-1)}\cdot \frac{f(n-1)}{f(n-2)} \ldots \frac{f(3)}{f(2)}\cdot$$ Using $$\frac{f(n)}{f(n-1)} = 1 - \frac{1}{n} \quad (n\ge 2)$$ and $f(1) = 1$ we get $$f(n) =f(2)\left(1 - \frac{1}{3}\right)\left(1 - \frac{1}{4}\right)\ldots\left(1 - \frac{1}{n}\right)$$ or $$f(n) = f(2)\prod_{k=3}^n \left(\frac{k-1}{k}\right) = \frac 1 4 \cdot \frac {2}{n} = \frac{1}{2n}$$ since $f(2) = 1/4$. Finally, $$f(n) = \begin{cases} 1 & n = 1\\ \frac{1}{2n} & n \ge 2 \end{cases}$$

Answer 2

The fallacy is to extend the product of the telescope to $n=1$. For $n=1$ the definition is separate, and is independent of the recursive equation. Instead you should take the telescopic product upto $n=2$, answer will match.

Answer 3

The problem that arises here is the fact that $f(1)$ has been explicitly defined, and does not satisfy the functional equation given. Hence any telescoping product must not include the $f(1)$ term as it risks being wrong. A must safer approach would be this: $$f(1)+2f(2)+…+nf(n)=n(n+1)f(n)$$ Replace $n$ by $n−1$ to get $$f(1)+2f(2)+…+(n−1)f(n−1)=n(n−1)f(n−1)$$ Subtracting the second equation from the first, we get $$nf(n)=n(n+1)f(n)−n(n−1)f(n−1) \\ \frac{f(n)}{f(n-1)}=\frac{n-1}{n}$$ Now, we telescope from $f(2)$ onwards:$$\frac{f(3)}{f(2)}.\frac{f(4)}{f(3)}...\frac{f(n)}{f(n-1)}=\frac{2}{3}.\frac{3}{4}...\frac{n-1}{n} \\ f(n)=\frac{2f(2)}{n}$$ Now, $$f(1)+2f(2)=(2)(3)f(2) \\ f(1)=4f(2)\\ f(2)=\frac{1}{4}$$ Therefore, $$f(n)=\frac{1}{2n}$$