Show that the $p$-series $\sum_{n=1}^\infty \frac{1}{n^p}$ is convergent iff $p \gt 1$. Show that the $p$-series $\sum_{n=1}^\infty \frac{1}{n^p}$ is convergent iff $p \gt 1$.
Attempt:
For the right direction:
Let $\sum_{n=1}^\infty \frac{1}{n^p}$ is convergent. Then, $\lim\limits_{n\to \infty} \frac{1}{n^p} = 0$.
In particular, $\lim\limits_{n\to \infty} \frac{1}{n^p} = 0$ for $p \gt 0$. But, for $0 \lt p \le 1$, the series $\sum_{n=1}^\infty \frac{1}{n^p}$ is divergent. Hence, we must have $p \gt 1$. $\Box$
Proof that series $\sum_{n=1}^\infty \frac{1}{n^p}$ is divergent:
We know that $n^p \le n$ for all positive integers $n$ and $0 \lt p \le 1$.
Then,
\begin{equation*}
\frac{1}{n} \le \frac{1}{n^p}.
\end{equation*}
Now, since the partial sums of the harmonic series are not bounded, this
inequality shows that the partial sums of the $p-$series are not bounded when $0 \lt p \le 1$. Hence, the $p-$series diverges for theses values of $p. \Box$
For the left direction:
Let $p \gt 1$. Let $f(x) = \frac{1}{x^p}$ for all $x \in [1.\infty)$. Then, $f$ is a positive, continous, and decreasing sequence. Hence, we can apply the Integral Test here. Notice that
\begin{align*}
\int_1^\infty \frac{1}{x^p} &= \lim\limits_{b \to \infty} \int_1^b \frac{1}{x^p} \\
&= \lim\limits_{b\to \infty} \left[\frac{1}{1-p}x^{1-p}\right]_1^b \\
&= \lim\limits_{b \to \infty} \left( \frac{1}{1-p}b^{1-p} - \frac{1}{1-p} \right) \\
&= \frac{1}{p-1} \lim\limits_{b \to \infty} \left(1 - \frac{1}{b^{p-1}} \right) \\
&= \frac{1}{p-1}.
\end{align*}
Thus,
\begin{equation*}
\int_1^\infty \frac{1}{x^p}dx
\end{equation*}
is convergent. Consequently, $\int_1^\infty \frac{1}{n^p}dn$ is convergent. By the Integral Test,
we have that
\begin{equation*}
\sum_{n=1}^\infty \frac{1}{n^p}
\end{equation*}
is convergent. $\Box$
Am I correct, especially for the right direction ?
 A: The case where $p \leqslant 0$ is quite simple to establish, so I will look only at the case where $p > 0$.  One way you can establish the convergence/divergence result is by using upper and lower bounds on the summation.  For all values $n \geqslant 1$ and $p>0$ we can easily establish the bounds:
$$\int \limits_{n}^{n+1} \frac{1}{r^p} \ dr \leqslant \frac{1}{n^p} \leqslant \int \limits_{n-1}^n \frac{1}{r^p} \ dr.$$
Applying these bounds to each term in the sum (except for the first term for the upper bound) gives:
$$\int \limits_1^\infty \frac{1}{r^p} \ dr \leqslant \sum_{n=1}^\infty \frac{1}{n^p} \leqslant 1 + \int \limits_1^\infty \frac{1}{r^p} \ dr.$$
Solving the definite integral we get:
$$R(p) \equiv \int \limits_1^\infty \frac{1}{r^p} \ dr 
= \begin{cases} 
\infty & & \text{if } 0 < p \leqslant 1, \\[8pt]
\frac{1}{p-1} & & \text{if } p > 1. \\[6pt]
\end{cases}$$
Substituting this form for the integral we therefore obtain:
$$R(p) \leqslant \sum_{n=1}^\infty \frac{1}{n^p} \leqslant 1+R(p).$$
We can now look at the conditions on $p$.  If we have $0 < p \leqslant 1$ then we get:
$$\infty \leqslant \sum_{n=1}^\infty \frac{1}{n^p} \leqslant \infty.$$
If we have $p>1$ then we get:
$$\frac{1}{p-1} \leqslant \sum_{n=1}^\infty \frac{1}{n^p} \leqslant \frac{p}{p-1}.$$
This establishes that the sum is divergent when $0 < p \leqslant 1$ and convergent when $p>1$.
A: The left direction is correct. An alternative way to prove this is using the Cauchy condensation test.
The right direction is not correct. You say: for $0 < p \le 1$ the series $\sum_n 1/n^p$ is divergent but you did not prove this!
Alternatively, you can also use the integral test or the Cauchy condensation test for this direction.
A: Here's something you may use:
Consider $p \leq 0$ first. In this case, $\frac{1}{n} = n^{-1} \geq 0$. The limit of this implies that it doesn't converge to $0$ as $n \rightarrow \infty$ and so $\sum_{n = 1}^{\infty} \frac{1}{n^p}$ does not converge.
Now assume $p > 0$. Then $\frac{1}{n^p}$ is a decreasing sequence of positive real numbers. By applying the Dyadic criterion, we have:
$\sum_{n = 1}^{\infty} \frac{1}{n^p}$ converges iff $\sum_{n = 0}^{\infty} 2^n \frac{1}{(2^n)^p}$ converges.
And so $\sum_{n = 0}^{\infty} (2^{1-p})^n$.
Let $q = 2^{1 - p}$ then the sum converges iff $q = 2^{1 - p} < 1$, which is iff $p > 1$.
