# Evaluate $S_n=\frac{1}{\log(2)}+\frac{2}{\log(3)}+\frac{3}{\log(4)}+\frac{4}{\log(5)}+\cdots+\frac{n}{\log(n+1)}$

Hi I want to evaluate the following sum :

$$S_n=\frac{1}{\log(2)}+\frac{2}{\log(3)}+\frac{3}{\log(4)}+\frac{4}{\log(5)}+\cdots+\frac{n}{\log(n+1)}=?$$

## My try :

Using a well know trick we have :

$$\int_{0}^{1}2^x+3^x+\cdots+(n+1)^xdx=S_n$$

Now we see a link between $$S$$ and the truncated function Zeta .

As I'm stuck now so I propose an crude estimation of the sum $$S$$:

Since $$\ln(n)\leq n-1$$ for $$n\geq 1$$ a natural number we have $$n\leq S_n$$

We can do better since $$\ln(n)\leq q(n^\frac{1}{q}-1)$$ for $$n\geq 1$$ and $$q\geq 1$$

An obvious upper bound is :

$$S_n\leq \frac{n(n+1)}{2}$$ for $$n\geq 3$$

Finally I make the following conjecture :

There is always a prime number between $$S_n$$ and $$S_{n+2}$$

## My questions

Can someone improve the bound or give a theoretic representation ?

Have you an counter-example ?

Any helps is greatly appreciated..

## Update :

I propose The following conjecture for $$n\geq 100$$

There is always a prime number between $$S_n$$ and $$S_{n+1}$$

Maybe it's stronger than the Firoozbakht's conjecture

• We have $S_n \sim \frac{n^2}{2\log n}$. Thus your conjectures are (modulo verification for small numbers) stronger than Legendre's conjecture. Hence don't hold your breath waiting for a proof. – Daniel Fischer Apr 28 '20 at 15:07
• – Martin Sleziak May 3 '20 at 13:55

An upper bound can be given since $$S_n <\int_1^n {x \over \ln(x+1)} dx=li((n+1)^2)-2li(n+1)+li(2)-li(4)$$ In a very similar way you can obtain a lower bound, which is: $$\int _1^n {{x-1} \over ln(x) } = li(n^2)-li(n)-\ln(2)$$