Answer
The series converges when $c=1$
Work Step by Step
$\Sigma_{n=1}^{\infty}\frac{c}{n}-\frac{1}{n+1}\lt \int_{1}^{\infty}\frac{c}{x}-\frac{1}{x+1}dx$
$=[clnx-ln(x+1)]_{1}^{\infty}$
$=ln2+\lim\limits_{x \to \infty} ln(\frac{x^{c}}{x+1})$
when $c\ne 1$ then $\lim\limits_{x \to \infty} ln(\frac{x^{c}}{x+1})= \pm \infty $
when $c =1$ then $\lim\limits_{x \to \infty} ln(\frac{x}{x+1})= 0 $
which is convergent
The series converges when $c=1$
