The sum of the series $\sum_{n=9}^{\infty}\frac{1}{\sqrt(N)}$

Question

Proving that the sum of $\sum_{n=9}^{\infty}\frac{1}{\sqrt(N)}= -\frac{1}{\sqrt(3)}$

Hi, I am trying to proving the sum above where $N$ is all the odd composites , any hint please ?

Answer 1

Since $\frac{1}{\sqrt{n}} > \frac1n$, and $$\sum_{n=9}^\infty \frac1n$$ diverges, it follows that the series $$\sum_{n=9}^\infty\frac{1}{\sqrt N}$$ also diverges.