**Lemma** $\sum n^{-p}$ [[Convergent Real Series|converges]] if $p>1$ and does not if $p \leq 1$. **Proof** **(using [[Integral test for convergence of series of non-negative decreasing function]])** Note that the function $x \mapsto x^{-\alpha}$ is non-negative and decreasing. We have $\int_{1}^{n} x^{-\alpha} \, dx = \left[ \frac{1}{1-\alpha} x^{1-\alpha}\right] = \frac{1}{\alpha-1} (1-n^{-(\alpha-1)}). $This is bounded by $1/(\alpha-1)$ if $\alpha>1$, but tends to $\infty$ as $n \to \infty$ if $\alpha <1$. **Partial Proof (using [[Comparison Test for Series With Non-negative Terms (Corollary 1)]])** If $p \geq 2$, since $\frac{1}{n^{p}} \leq \frac{1}{n^{2}}$ and $\sum \frac{1}{n^{2}}< \infty$, $\sum \frac{1}{n^{p}}<\infty$ using [[Basel Problem]]. If $p\leq 1$, since $\frac{1}{n^{p}}\geq \frac{1}{n}$ and $\sum \frac{1}{n} = \infty$, $\sum \frac{1}{n^{p}} = \infty$ using [[Harmonic Numbers]]. We can't use comparison test for $1<p<2$.