# What is function f :N ->R with f(1)=1,f(n+1)=1/n(n+f(n))?

### 1 Answer | Add Yours

There exists a mistake in the provided relation `f(n+1)=1/(n(n+f(n)))` , since, working with this relation, the result would be inconclusive. If you replace 1 for n in the given expression such that, `f(n+1)=1/(n(1+f(n)))` , the mathematical induction method works for all natural n.

You need to evaluate the function at `n=1` and `n=2` , using the given relation,`f(n+1)=1/(n(1+f(n)))` , such that:

`n = 1 => f(2) = 1/(1*(1 + f(1)))`

Since the problem provides `f(1) = 1` , you need to replace 1 for f(1), such that:

`f(2) = 1/(1*(1 + 1)) = 1/2`

`n = 2 => f(3) = 1/(2*(2 + f(2)))`

`f(3) = 1/(2*(1 + 1/2)) = 1/3`

You need to use the mathematical induction to test if `f(n+1) = 1/(n+1)` , hence, considering `f(n) = 1/n` valid, yields:

`f(n+1) = 1/(n(1 + f(n))) => f(n+1) = 1/(n(1 + 1/n))`

`f(n+1) = 1/(n((n + 1)/n))`

Reducing duplicate factors yields:

`f(n+1) = 1/(n+1)`

**Hence, evaluating the function f(n), under the given conditions, using mathematical induction, yields `f(n) = 1/n, n` in domain N.**