# What constant $c$ will make $\sum_{k=2}^{N}c^{\frac{1}{k\log k}}=N ?$

What constant $$c$$ will make this equality valid for any $$N$$ chosen?

$$\sum_{k=2}^{N}c^{\frac{1}{k\log k}}=N.$$

I tried getting a rough idea of what $$c$$ should be and got about $$1.46$$ when $$N=1000$$ but I don't have much perspicacity on how to proceed.

I do not think that there is a single value of $$c$$. So, let us consider that you look for the zero of function $$f(c_n)=\sum_{k=2}^{n}c_n^{\frac{1}{k\log k}}-n$$ which does not make much problems to solve numerically.
For illustration purposes, let $$n=10^k$$; computing, we should obtain the following values $$\left( \begin{array}{cc} k & c_{10^k} \\ 1 & 1.7140673 \\ 2 & 1.4978998 \\ 3 & 1.4189990 \\ 4 & 1.3760092\\ 5 & 1.3480802 \end{array} \right)$$ I gave up for $$k=6$$ (my computer too !).