I'm supposed to prove the following theorem:
Theorem: The p-series $\sum_{n=1}^{\infty}\frac{1}{n^p}$,
i) converges if $p>1$
ii) diverges if $p \le 1$
That's how I did it:
First I proved the following property
Property (E): The improper integral $\int_{a}^{\infty}\frac{1}{x^p}\,dx$, $\,$ $0<a<\infty$ $\,$ , converges if $p>1$ and diverges if $p \le 1$.
Case $p=1$.
$\int_{a}^{\infty}\frac{1}{x^p}\,dx=\int_{a}^{\infty}\frac{1}{x}\,dx=\lim_{t \to \infty}(\int_{a}^{t}\frac{1}{x}\,dx)=\lim_{t \to \infty}(\left.\ln(x)\right|_a^t)=\lim_{t \to \infty}(\ln(t)-\ln(a))=\lim_{t \to \infty}\ln(t)-\lim_{t \to \infty}\ln(a)=\infty-\ln(a)=\infty$. Therefore, the improper integral diverges in this case.
Case $p>1$.
$\int_{a}^{\infty}\frac{1}{x^p}\,dx=\int_{a}^{\infty}x^{-p}\,dx=\lim_{t \to \infty}(\int_{a}^{t}x^{-p}\,dx)=\lim_{t \to \infty}(\left.(\frac{1}{1-p}x^{1-p})\right|_a^t)=\lim_{t \to \infty}(\frac{1}{1-p}t^{1-p}-\frac{1}{1-p}a^{1-p})=(\lim_{t \to \infty}(\frac{1}{1-p}t^{1-p}))-(\lim_{t \to \infty}(\frac{1}{1-p}a^{1-p}))=(\lim_{t \to \infty}(\frac{1}{1-p}t^{1-p}))-\frac{1}{1-p}a^{1-p}$.
Since $1-p<0$, we have that $\lim_{t \to \infty}t^{1-p}=0$, and, therefore, $\lim_{t \to \infty}(\frac{1}{1-p}t^{1-p})=0$. This way, $\int_{a}^{\infty}\frac{1}{x^p}\,dx=(\lim_{t \to \infty}(\frac{1}{1-p}t^{1-p}))-\frac{1}{1-p}a^{1-p}=-\frac{1}{1-p}a^{1-p}$. From this, the improper integral converges in this case.
Case p<1.
$\int_{a}^{\infty}\frac{1}{x^p}\,dx=\int_{a}^{\infty}x^{-p}\,dx=\lim_{t \to \infty}(\int_{a}^{t}x^{-p}\,dx)=\lim_{t \to \infty}(\left.(\frac{1}{1-p}x^{1-p})\right|_a^t)=\lim_{t \to \infty}(\frac{1}{1-p}t^{1-p}-\frac{1}{1-p}a^{1-p})=(\lim_{t \to \infty}(\frac{1}{1-p}t^{1-p}))-(\lim_{t \to \infty}(\frac{1}{1-p}a^{1-p}))=(\lim_{t \to \infty}(\frac{1}{1-p}t^{1-p}))-\frac{1}{1-p}a^{1-p}$.
Since $1-p>0$, we have that $\lim_{t \to \infty}t^{1-p}=\infty$, and, therefore, $\lim_{t \to \infty}(\frac{1}{1-p}t^{1-p})=\infty$. This way, $\int_{a}^{\infty}\frac{1}{x^p}\,dx=(\lim_{t \to \infty}(\frac{1}{1-p}t^{1-p}))-\frac{1}{1-p}a^{1-p}=\infty-\frac{1}{1-p}a^{1-p}=\infty$. From this, the improper integral diverges in this case.
Also, use the following definition for the integral test:
If f is continuous in $[1,\infty)$, is decreasing in this interval and $f(x) \ge 0$ in this interval, then the series $\sum_{n=1}^{\infty}f(n)$ converges if and only if the improper integral $\int_{1}^{\infty}f(x)$ converges. If the improper integral diverges, then the series diverges.
Now, with these two tools we can prove the theorem as follows:
Let be the p-series $\sum_{n=1}^{\infty}\frac{1}{n^p}$, and let $f(x)=\frac{1}{x^p}$ be the function associated to the sequence $a_n=\frac{1}{n^p}$, for $x \ge 1$.
i) If $p>1$, we have that, for $x \in [1,\infty)$, $f(x)=\frac{1}{x^p}$ is decreasing and also $f(x) \ge 0$. Therefore, the integral test applies.
Then, $\int_{1}^{\infty}\frac{1}{x^p}\,dx=-\frac{1}{1-p}1^{1-p}=-\frac{1}{1-p}$, by property (E).
Therefore, since this integral converges, we have that the series $\sum_{n=1}^{\infty}\frac{1}{n^p}$ converges in this case.
ii) If $0<p \le 1$, we have that, for $x \in [1,\infty)$, $f(x)=\frac{1}{x^p}$ is decreasing and also $f(x) \ge 0$. Therefore, the integral test applies.
This way, $\int_{1}^{\infty}\frac{1}{x^p}\,dx=\infty$, by property (E).
Therefore, since this integral diverges, we have that the series $\sum_{n=1}^{\infty}\frac{1}{n^p}$ diverges in this case.
If $p=0$, we have that $\sum_{n=1}^{\infty}\frac{1}{n^p}=\sum_{n=1}^{\infty}\frac{1}{n^0}=\sum_{n=1}^{\infty}(1)$, and, therefore, the series diverges in this case.
If $p<0$, we have that $\lim_{x \to \infty}f(x)=\lim_{x \to \infty}\frac{1}{x^p}=\lim_{x \to \infty}x^{-p}$. Since, $-p>0$, we have that $\lim_{x \to \infty}x^{-p}=\infty$. Therefore, $\lim_{n \to \infty}\frac{1}{n^p}=\infty$ and, by the nth-term test, the series $\sum_{n=1}^{\infty}\frac{1}{n^p}$ diverges in this case.
I hope this is correct. The way I found to solve it was to break it into many steps.