An honouring divisibility(self-made)

[Fm Jakaria]

Prove that for all k$\in \mathbb N$, k > 1; they're exists n$\in \mathbb N$ , n < 2k such that
$\frac {\sigma(n)}{n} = \frac{k+1}{k}$. [Here $\sigma(n)$ denotes the sum of positive divisors of n; as usual.]

[mutasimmim]

What about $k=4$? And

they're

should be there.

[Fm Jakaria]

hmm....... If,we give n freedom, then ?(edited problem)

[Nirjhor]

The assertion is clearly true for prime \(k\) since you can just take \(n=k\). But not for composite \(k\).

Notice that for composite \(k\) you must take composite \(n\). So \(\tau(n)\ge 3\). Let the divisors of \(n\) be \(1<d_1<d_2<\cdots<d_i<n\). Then \(\sigma(n)=1+n+\sum d_j\). Make this simple observation that \(d_j=n/d_{i-j}\). This implies \[1+\dfrac{1}{k}=\dfrac{\sigma(n)}{n}=\dfrac{1+n+\sum d_j}{n}=1+\dfrac{1}{n}+\sum\dfrac{d_j}{n}=1+\dfrac{1}{n}+\sum \dfrac{1}{d_j}.\] So we are looking for \(n\) so that \(1/k=1/n+\sum 1/d_j\) for a fixed \(k\) where \(d_j\mid n\). Now notice that \(k\mid n\) so we must have \(d_n=k\) for some \(n\). This cancels out the \(1/k\) and leaves us with the equation \(1/n+\sum 1/d_j=0\) which is never possible.

[mutasimmim]

we give n freedom

What do you mean? Please explain.

(edited problem)

I don't see how you've edited the problem.