[proofplan]
We split the proof into the cases $p > 1$ and $p \leq 1$. For $p > 1$, we group the positive terms into dyadic blocks and bound each block by a geometric sequence with ratio $2^{1-p} < 1$. For $p \leq 1$, we compare from below with the harmonic series, and prove the harmonic series diverges by the same dyadic blocking method.
[/proofplan]
[step:Define the partial sums and reduce convergence to boundedness]
For the fixed real number $p \in \mathbb{R}$, define the sequence of terms
\begin{align*}
a_p: \mathbb{N} &\to (0,\infty) \\
n &\mapsto n^{-p},
\end{align*}
and define the partial sums
\begin{align*}
S_N(p) := \sum_{n=1}^{N} a_p(n) = \sum_{n=1}^{N} n^{-p}
\end{align*}
for each $N \in \mathbb{N}$.
Since $a_p(n) > 0$ for every $n \in \mathbb{N}$, the sequence $(S_N(p))_{N=1}^{\infty}$ is increasing. Therefore the series $\sum_{n=1}^{\infty} n^{-p}$ converges in $\mathbb{R}$ exactly when the increasing sequence $(S_N(p))_{N=1}^{\infty}$ is bounded above.
[/step]
[step:Bound the dyadic blocks when $p > 1$]
Assume $p > 1$. For each integer $k \geq 0$, define the dyadic block
\begin{align*}
B_k := \{n \in \mathbb{N} : 2^k \leq n \leq 2^{k+1}-1\}.
\end{align*}
The set $B_k$ has exactly $2^k$ elements. For every $n \in B_k$, we have $n \geq 2^k$, and since the map $t \mapsto t^{-p}$ is decreasing on $(0,\infty)$ for $p > 0$,
\begin{align*}
n^{-p} \leq (2^k)^{-p}.
\end{align*}
Hence
\begin{align*}
\sum_{n=2^k}^{2^{k+1}-1} n^{-p}
\leq \sum_{n=2^k}^{2^{k+1}-1} (2^k)^{-p}
= 2^k (2^k)^{-p}
= 2^{k(1-p)}.
\end{align*}
Define the real number
\begin{align*}
r := 2^{1-p}.
\end{align*}
Since $p > 1$, we have $0 < r < 1$, and the preceding estimate becomes
\begin{align*}
\sum_{n=2^k}^{2^{k+1}-1} n^{-p} \leq r^k.
\end{align*}
[guided]
We want to prove convergence by showing that all partial sums are bounded above. Because the terms are positive, it is natural to group them into blocks whose sizes grow geometrically. For each integer $k \geq 0$, define
\begin{align*}
B_k := \{n \in \mathbb{N} : 2^k \leq n \leq 2^{k+1}-1\}.
\end{align*}
This block contains exactly $2^{k+1}-2^k = 2^k$ integers.
Now fix $n \in B_k$. Then $n \geq 2^k$. Since $p > 1$, in particular $p > 0$, and the function $t \mapsto t^{-p}$ decreases on $(0,\infty)$. Therefore
\begin{align*}
n^{-p} \leq (2^k)^{-p}.
\end{align*}
So every term in the $k$-th block is bounded above by the first term of that block. Summing this uniform bound over the $2^k$ terms in the block gives
\begin{align*}
\sum_{n=2^k}^{2^{k+1}-1} n^{-p}
\leq \sum_{n=2^k}^{2^{k+1}-1} (2^k)^{-p}
= 2^k (2^k)^{-p}
= 2^{k(1-p)}.
\end{align*}
The exponent $1-p$ is negative, so these block bounds form a decaying geometric sequence. Define
\begin{align*}
r := 2^{1-p}.
\end{align*}
Because $p > 1$, we have $0 < r < 1$, and hence
\begin{align*}
2^{k(1-p)} = r^k.
\end{align*}
Thus the entire $k$-th block is bounded by $r^k$.
[/guided]
[/step]
[step:Sum the block estimates to prove convergence for $p > 1$]
Let $N \in \mathbb{N}$. Choose an integer $m \geq 0$ such that
\begin{align*}
N \leq 2^{m+1}-1.
\end{align*}
Since all terms are positive,
\begin{align*}
S_N(p)
\leq \sum_{n=1}^{2^{m+1}-1} n^{-p}
= \sum_{k=0}^{m} \sum_{n=2^k}^{2^{k+1}-1} n^{-p}
\leq \sum_{k=0}^{m} r^k.
\end{align*}
Using the finite geometric sum identity,
\begin{align*}
\sum_{k=0}^{m} r^k
= \frac{1-r^{m+1}}{1-r}
\leq \frac{1}{1-r}.
\end{align*}
Thus $S_N(p) \leq (1-r)^{-1}$ for every $N \in \mathbb{N}$. Hence $(S_N(p))_{N=1}^{\infty}$ is bounded above, so $\sum_{n=1}^{\infty} n^{-p}$ converges.
[guided]
Let $N \in \mathbb{N}$ be arbitrary. We choose an integer $m \geq 0$ such that
\begin{align*}
N \leq 2^{m+1}-1.
\end{align*}
This means the first $N$ terms are contained in the union of the dyadic blocks $B_0,\dots,B_m$. Since every term $n^{-p}$ is positive, enlarging the finite sum can only increase its value:
\begin{align*}
S_N(p)
= \sum_{n=1}^{N} n^{-p}
\leq \sum_{n=1}^{2^{m+1}-1} n^{-p}.
\end{align*}
The integers from $1$ to $2^{m+1}-1$ are exactly partitioned by the blocks $B_0,\dots,B_m$, so
\begin{align*}
\sum_{n=1}^{2^{m+1}-1} n^{-p}
= \sum_{k=0}^{m} \sum_{n=2^k}^{2^{k+1}-1} n^{-p}.
\end{align*}
From the previous step, each block is at most $r^k$, where $r = 2^{1-p}$ and $0 < r < 1$. Therefore
\begin{align*}
S_N(p)
\leq \sum_{k=0}^{m} r^k.
\end{align*}
The finite geometric sum identity gives
\begin{align*}
\sum_{k=0}^{m} r^k
= \frac{1-r^{m+1}}{1-r}.
\end{align*}
Since $0 < r < 1$, we have $0 < r^{m+1} < 1$, and hence
\begin{align*}
\frac{1-r^{m+1}}{1-r} \leq \frac{1}{1-r}.
\end{align*}
Thus
\begin{align*}
S_N(p) \leq \frac{1}{1-r}
\end{align*}
for every $N \in \mathbb{N}$. This upper bound is independent of $N$, so the increasing sequence of partial sums is bounded above. Therefore the series converges.
[/guided]
[/step]
[step:Prove the harmonic series diverges by dyadic lower bounds]
Define the harmonic partial sums
\begin{align*}
H_N := \sum_{n=1}^{N} \frac{1}{n}
\end{align*}
for $N \in \mathbb{N}$. For each integer $k \geq 0$ and every $n \in B_k$, we have $n \leq 2^{k+1}-1 < 2^{k+1}$, so
\begin{align*}
\frac{1}{n} > \frac{1}{2^{k+1}}.
\end{align*}
Therefore
\begin{align*}
\sum_{n=2^k}^{2^{k+1}-1} \frac{1}{n}
> 2^k \cdot \frac{1}{2^{k+1}}
= \frac{1}{2}.
\end{align*}
Consequently, for every integer $m \geq 0$,
\begin{align*}
H_{2^{m+1}-1}
= \sum_{k=0}^{m} \sum_{n=2^k}^{2^{k+1}-1} \frac{1}{n}
> \sum_{k=0}^{m} \frac{1}{2}
= \frac{m+1}{2}.
\end{align*}
Since $(m+1)/2 \to \infty$ as $m \to \infty$, the sequence $(H_N)_{N=1}^{\infty}$ is unbounded, and hence the harmonic series diverges.
[guided]
We now need a lower-bound argument. Define
\begin{align*}
H_N := \sum_{n=1}^{N} \frac{1}{n}
\end{align*}
for each $N \in \mathbb{N}$. We use the same dyadic blocks
\begin{align*}
B_k := \{n \in \mathbb{N} : 2^k \leq n \leq 2^{k+1}-1\}.
\end{align*}
For $n \in B_k$, the largest possible value of $n$ is $2^{k+1}-1$, so in particular
\begin{align*}
n < 2^{k+1}.
\end{align*}
Taking reciprocals reverses the inequality because both sides are positive:
\begin{align*}
\frac{1}{n} > \frac{1}{2^{k+1}}.
\end{align*}
The block $B_k$ has $2^k$ terms, and each term is larger than $2^{-(k+1)}$. Hence
\begin{align*}
\sum_{n=2^k}^{2^{k+1}-1} \frac{1}{n}
> 2^k \cdot \frac{1}{2^{k+1}}
= \frac{1}{2}.
\end{align*}
Now sum these lower bounds from $k=0$ through $k=m$. The blocks $B_0,\dots,B_m$ partition the integers from $1$ to $2^{m+1}-1$, so
\begin{align*}
H_{2^{m+1}-1}
= \sum_{k=0}^{m} \sum_{n=2^k}^{2^{k+1}-1} \frac{1}{n}
> \sum_{k=0}^{m} \frac{1}{2}
= \frac{m+1}{2}.
\end{align*}
The right-hand side tends to infinity as $m \to \infty$. Therefore the harmonic partial sums are not bounded above. Since the harmonic partial sums are increasing, unboundedness implies that the harmonic series diverges.
[/guided]
[/step]
[step:Compare with the harmonic series to prove divergence when $p \leq 1$]
Assume $p \leq 1$. For every $n \in \mathbb{N}$, since $n \geq 1$ and $p \leq 1$,
\begin{align*}
n^p \leq n.
\end{align*}
Taking reciprocals of positive numbers gives
\begin{align*}
n^{-p} = \frac{1}{n^p} \geq \frac{1}{n}.
\end{align*}
Hence, for every $N \in \mathbb{N}$,
\begin{align*}
S_N(p)
= \sum_{n=1}^{N} n^{-p}
\geq \sum_{n=1}^{N} \frac{1}{n}
= H_N.
\end{align*}
The sequence $(H_N)_{N=1}^{\infty}$ is unbounded by the previous step, so $(S_N(p))_{N=1}^{\infty}$ is also unbounded. Therefore $\sum_{n=1}^{\infty} n^{-p}$ diverges when $p \leq 1$.
Combining convergence for $p > 1$ with divergence for $p \leq 1$, the series $\sum_{n=1}^{\infty} n^{-p}$ converges if and only if $p > 1$.
[/step]
