P-Series Test

Theorem

given a P-Series

• If $p\le 1$, then ${\sum }_{n=1}^{\infty }\frac{1}{n^p}$ diverges
• If $p>1$, then ${\sum }_{n=1}^{\infty }\frac{1}{n^p}$ converges

Proof

1. Let $p\in {\mathbb{R}}^{+}$
2. Given ${\sum }_{n=1}^{\infty }\frac{1}{n^p}$
3. For $\forall n\in \mathbb{N}$, Let $a_n=n^{-p}=f(n)$
4. Then, so $f(x)=x^{-p}$ on $[1,\infty )$
5. Note that $x>0,1>0⟹\frac{1}{x^p}>0$
6. Note that $\frac{1}{x^p}$ is continuous by Continuity Theorem
7. Note that $\frac{1}{x^p}$ is decreasing when $p>1$ as $\frac{d}{dx}\frac{1}{x^p}=-p\frac{1}{x^{p-1}}$. When $p\le 1$, the derivative is negative
8. Then, by Integral Test $f$ Converges