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..
...Thanks in advance for all your contributions.
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
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: $$ \int _1^n {{x-1} \over ln(x) } = li(n^2)-li(n)-\ln(2)$$
I don't know if the sum has a closed form.