How to prove statement about sum We have such sum: $$\sum_{n=1}^{\infty}\frac{(-1)^{(n+1)}}{n^p}, 0<p<1$$
How to prove that sum is between 1/2 and 1?
 A: Let $ \quad f(x) = 1/x^{p} \quad \colon \{ x \ge 1 \,, \space 0 \lt p \lt 1 \} \space\Rightarrow\space 1/x \lt 1/x^{p} \lt 1  $
 $ \Rightarrow \quad f^{\prime}(x) = -p/x^{p+1} \lt 0 \space\Rightarrow\space f^{\prime\prime}(x) = p(p+1)/x^{p+2} \gt 0 \space\Rightarrow\space f(x) \space $ is convex
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\begin{align}
& \\
& \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^{p}} = 1 + \sum_{n=2}^{\infty} \frac{(-1)^{n+1}}{n^{p}} = 1 + \sum_{n=1}^{\infty} \frac{(-1)^{n+2}}{(n+1)^{p}} = 1 - \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{(n+1)^{p}} \Rightarrow \\
& \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{(n+1)^{p}} = 1 - \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^{p}} \Rightarrow \small \sum_{n=1}^{\infty} \left[ \frac{1}{(2n)^{p}} - \frac{1}{(2n+1)^{p}} \right] = 1 - \sum_{n=1}^{\infty} \left[ \frac{1}{(2n-1)^{p}} - \frac{1}{(2n)^{p}} \right] \\
\end{align}
$$
$$
\begin{align}
& \\
& \because \space 2n \lt 2n+1 \Rightarrow \frac{1}{2n} \gt \frac{1}{2n+1} \Rightarrow \frac{1}{(2n)^{p}} \gt \frac{1}{(2n+1)^{p}} \Rightarrow \frac{1}{(2n)^{p}} - \frac{1}{(2n+1)^{p}} \gt 0 \Rightarrow \\
& \sum_{n=1}^{\infty} \left[ \frac{1}{(2n)^{p}} - \frac{1}{(2n+1)^{p}} \right] = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{(n+1)^{p}} = 1 - \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^{p}} \gt 0 \Rightarrow \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^{p}} \color{red}{\lt 1} \\
\end{align}
$$
$$
\begin{align}
& \\
& \because \space f(x) \space \text{convex} \space\Rightarrow\space f(x) \space \text{midpoint convex} \space\Rightarrow\space \small f \left( \frac{x_{1} + x_{2}}{2} \right) \le \frac{f(x_{1}) + f(x_{2})}{2} \\
& \text{Let} \space \small x_{1} = 2n - 1 \,, \space x_{2} = 2n + 1 \normalsize \Rightarrow \small f \left( \frac{2n - 1 + 2n + 1}{2} \right) = f(2n) \le \frac{f(2n - 1) + f(2n + 1)}{2} \normalsize \Rightarrow \\
& \frac{1}{(2n-1)^{p}} + \frac{1}{(2n+1)^{p}} \ge \frac{2}{(2n)^{p}} \quad\Rightarrow\quad \frac{1}{(2n-1)^{p}} - \frac{1}{(2n)^{p}} \ge \frac{1}{(2n)^{p}} - \frac{1}{(2n+1)^{p}} \space\space\Rightarrow \\
& \sum_{n=1}^{\infty} \left[ \small \frac{1}{(2n-1)^{p}} - \frac{1}{(2n)^{p}} \normalsize \right] \ge \sum_{n=1}^{\infty} \left[ \small \frac{1}{(2n)^{p}} - \frac{1}{(2n+1)^{p}} \normalsize \right] \Rightarrow \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^{p}} \ge 1 - \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^{p}} \Rightarrow \\
& 2 \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^{p}} \ge 1 \Rightarrow \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^{p}} \color{red}{\ge 1/2} \\
\end{align}
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