[proofplan] We write the series in the form $\sum_{n=1}^{\infty}(-1)^{n+1}a_n$ with $a_n=n^{-p}$. First we verify directly that $(a_n)$ is positive, decreasing, and tends to $0$, then prove convergence by showing that the even and odd partial sums are monotone and squeeze the same limit. Finally we identify the absolute series as $\sum n^{-p}$ and prove its convergence classification by dyadic grouping, obtaining convergence exactly for $p>1$. [/proofplan] [step:Verify that $n^{-p}$ is positive, decreasing, and tends to zero] Define the sequence $a:\mathbb{N}\to(0,\infty)$ by \begin{align*} a_n = \frac{1}{n^p}. \end{align*} Since $p>0$ and $n>0$, each $a_n$ is positive. The map $x\mapsto x^p$ is strictly increasing on $(0,\infty)$, so $(n+1)^p>n^p$ for every $n\in\mathbb{N}$. Taking reciprocals of positive numbers gives \begin{align*} a_{n+1}=\frac{1}{(n+1)^p}<\frac{1}{n^p}=a_n. \end{align*} Thus $(a_n)$ is decreasing. To prove $a_n\to0$, let $\varepsilon>0$. Choose an integer $N\in\mathbb{N}$ such that $N>\varepsilon^{-1/p}$. Then for every $n\geq N$, \begin{align*} 0<a_n=\frac{1}{n^p}\leq \frac{1}{N^p}<\varepsilon. \end{align*} Hence $\lim_{n\to\infty}a_n=0$. [/step] [step:Show that the alternating partial sums converge] For each $N\in\mathbb{N}$, define the $N$th partial sum $s_N\in\mathbb{R}$ by \begin{align*} s_N=\sum_{n=1}^{N}(-1)^{n+1}a_n. \end{align*} For $k\in\mathbb{N}$, define the even and odd subsequences of partial sums by \begin{align*} e_k=s_{2k}, \qquad o_k=s_{2k-1}. \end{align*} For every $k\in\mathbb{N}$, \begin{align*} e_{k+1}-e_k=a_{2k+1}-a_{2k+2}\geq0, \end{align*} because $(a_n)$ is decreasing. Thus $(e_k)$ is increasing. Also, \begin{align*} o_{k+1}-o_k=-a_{2k}+a_{2k+1}\leq0, \end{align*} so $(o_k)$ is decreasing. Moreover, \begin{align*} e_k=s_{2k}=s_{2k-1}-a_{2k}=o_k-a_{2k}\leq o_k. \end{align*} Since $(e_k)$ is increasing and bounded above by $o_1=s_1=a_1$, it converges to some $L\in\mathbb{R}$. Since $(o_k)$ is decreasing and bounded below by $e_1=s_2$, it converges to some $M\in\mathbb{R}$. The difference of the two subsequences is \begin{align*} o_k-e_k=a_{2k}. \end{align*} Because $a_{2k}\to0$, taking limits gives $M-L=0$, hence $L=M$. Every partial sum $s_N$ is either some $e_k$ or some $o_k$. Since both subsequences converge to the same limit $L$, the full sequence $(s_N)$ converges to $L$. Therefore \begin{align*} \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^p} \end{align*} converges. [guided] The point of separating even and odd partial sums is that alternating signs produce two monotone subsequences. Define $s_N=\sum_{n=1}^{N}(-1)^{n+1}a_n$, where $a_n=n^{-p}$. Then the even partial sums are \begin{align*} e_k=s_{2k}, \end{align*} and the odd partial sums are \begin{align*} o_k=s_{2k-1}. \end{align*} We compare consecutive even partial sums. The only new terms from $e_k=s_{2k}$ to $e_{k+1}=s_{2k+2}$ are $a_{2k+1}$ and $-a_{2k+2}$, so \begin{align*} e_{k+1}-e_k=a_{2k+1}-a_{2k+2}. \end{align*} Since $(a_n)$ is decreasing, $a_{2k+1}\geq a_{2k+2}$, and therefore $e_{k+1}\geq e_k$. Thus the even partial sums increase. Now compare consecutive odd partial sums. The new terms from $o_k=s_{2k-1}$ to $o_{k+1}=s_{2k+1}$ are $-a_{2k}$ and $a_{2k+1}$, hence \begin{align*} o_{k+1}-o_k=-a_{2k}+a_{2k+1}. \end{align*} Again using monotonicity, $a_{2k}\geq a_{2k+1}$, so $o_{k+1}\leq o_k$. Thus the odd partial sums decrease. The two subsequences also enclose each other. For every $k\in\mathbb{N}$, \begin{align*} e_k=s_{2k}=s_{2k-1}-a_{2k}=o_k-a_{2k}. \end{align*} Since $a_{2k}>0$, this gives $e_k\leq o_k$. In particular, $(e_k)$ is increasing and bounded above by $o_1$, so it converges to some real number $L$. Also, $(o_k)$ is decreasing and bounded below by $e_1$, so it converges to some real number $M$. It remains to see that these two limits are the same. Their gap is exactly the next positive term: \begin{align*} o_k-e_k=a_{2k}. \end{align*} From the previous step, $a_{2k}\to0$. Taking limits gives $M-L=0$, so $M=L$. Since every partial sum is either an even partial sum or an odd partial sum, the whole sequence $(s_N)$ converges to this common limit. Therefore the [alternating series](/page/Alternating%20Series) converges. [/guided] [/step] [step:Identify and classify the absolute series] For each $n\in\mathbb{N}$, \begin{align*} \left|\frac{(-1)^{n+1}}{n^p}\right|=\frac{1}{n^p}. \end{align*} Thus the absolute series is \begin{align*} \sum_{n=1}^{\infty}\frac{1}{n^p}. \end{align*} First suppose $p>1$. For each integer $k\geq0$, the dyadic block $\{2^k,2^k+1,\dots,2^{k+1}-1\}$ has $2^k$ terms, and every term in that block is at most $(2^k)^{-p}$. Hence \begin{align*} \sum_{n=2^k}^{2^{k+1}-1}\frac{1}{n^p}\leq 2^k\frac{1}{(2^k)^p}=2^{-k(p-1)}. \end{align*} Therefore, for every $K\in\mathbb{N}$, \begin{align*} \sum_{n=1}^{2^{K+1}-1}\frac{1}{n^p}\leq \sum_{k=0}^{K}2^{-k(p-1)}. \end{align*} Since $p-1>0$, the geometric series $\sum_{k=0}^{\infty}2^{-k(p-1)}$ converges. The partial sums of $\sum n^{-p}$ are increasing and bounded above, so $\sum_{n=1}^{\infty}n^{-p}$ converges. Now suppose $0<p\leq1$. For every $n\in\mathbb{N}$, $n^p\leq n$, and hence \begin{align*} \frac{1}{n^p}\geq\frac{1}{n}. \end{align*} For each integer $k\geq0$, \begin{align*} \sum_{n=2^k}^{2^{k+1}-1}\frac{1}{n}\geq 2^k\frac{1}{2^{k+1}}=\frac{1}{2}. \end{align*} Thus the dyadic block sums of $\sum 1/n$ are bounded below by $1/2$, so the harmonic series diverges. By comparison from below, $\sum_{n=1}^{\infty}n^{-p}$ diverges. [/step] [step:Conclude absolute and conditional convergence] The preceding step shows that \begin{align*} \sum_{n=1}^{\infty}\left|\frac{(-1)^{n+1}}{n^p}\right| \end{align*} converges exactly when $p>1$. Therefore the alternating $p$-series converges absolutely if and only if $p>1$. When $0<p\leq1$, the alternating series itself converges by the partial-sum argument above, while its absolute series diverges. By the definition of conditional convergence, the alternating $p$-series converges conditionally exactly for $0<p\leq1$. This proves all asserted convergence classifications. [/step]