Convergence of generalized harmonic series I'm trying to prove by comparison the convergence of generalized harmonic series, of the form $$\sum_{n=1}^{\infty} \frac{1}{n^{a+1}}$$ where $a>0$. I proved this for $a=1$, showing that $$\sum_{k=1}^{n} \frac{1}{k^2}<1+\sum_{k=2}^{n} \frac{1}{k(k-1)}$$
but I'm not succeeding in generalize this for any $a>0. $
Is it possible to prove it along the way of the first?

For $a>1$ you have $$n^{a+1}\ge n^2$$ 
Thus $$\frac {1}{n^{1+a}} \le\frac {1}{n^2}$$
By comparison we have $$\sum _1^\infty \frac {1}{n^{1+a}}\le \sum _1^\infty \frac {1}{n^2} =\pi^2/6$$ 
For $a>0$ the integral test will do.
