Sum $\sum\limits_{n=1}^{\infty}\frac{(-1)^n}{\sqrt{n}}$

Question

I have the following infinite sum: $$ \sum\limits_{n=1}^{\infty}\frac{(-1)^n}{\sqrt{n}} $$ Because there is a $(-1)^n$ I deduce that it is a alternating series. Therefore I use the alternating series test: $$ \lim_{n\to\infty} \frac{1}{\sqrt{n}} $$ Because this limit is decreasing and approaching $0$ I thought it should therefore be convergent. However in the answer key it uses a different method to get a different answer. It instead takes the absolute value of the series:

$\left|\frac{(-1)^n}{\sqrt{n}}\right| = \frac{1}{\sqrt{n}}$ and says because this is a divergent p series that the series is divergent.

Why is the alternating series test not applied here?

Answer 1

You are correct that the series is convergent, for the reasons you indicated. The answer key is checking for absolute convergence. The series is convergent, but not absolutely convergent. In other words, it is conditionally convergent.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{{\displaystyle #1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\sr}[2]{\,\,\,\stackrel{{#1}}{{#2}}\,\,\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} & \color{#44f}{\sum_{n = 1}^{\infty}{\pars{-1}^{n} \over \root{n}}} = \lim_{N \to \infty}\,\sum_{n = 1}^{N}{\pars{-1}^{n} \over \root{n}} \\[5mm] = & \ \lim_{N \to \infty}\bracks{% \sum_{n = 1}^{\left\lfloor\,{N/2}\,\right\rfloor}{1 \over \root{2n}} - \pars{\sum_{n = 1}^{N}{1 \over \root{n}} - \sum_{n = 1}^{\left\lfloor\,{N/2}\,\right\rfloor}{1 \over \root{2n}}}} \\[5mm] = & \ \lim_{N \to \infty}\pars{% \root{2}\sum_{n = 1}^{\left\lfloor\,{N/2}\,\right\rfloor}{1 \over \root{n}} - \sum_{n = 1}^{N}{1 \over \root{n}}}\label{1}\tag{1} \end{align} With a Riemann Zeta Identity: \begin{align} & \sum_{n = 1}^{m}{1 \over \root{n}} \\ = & \ 2\root{m} + \zeta\pars{1 \over 2} + {1 \over 2}\int_{m}^{\infty}\!\!{\braces{x} \over x^{3/2}}\,\dd x \!\!\sr{{\rm as}\ m\ \to\ \infty}{\sim} \bbx{2\root{m} + \zeta\pars{1 \over 2}} \end{align} This identity and (\ref{1}) lead to: \begin{align} & \color{#44f}{\sum_{n = 1}^{\infty}{\pars{-1}^{n} \over \root{n}}} \\ = & \ \lim_{N \to \infty}\braces{\root{2}\bracks{2\root{\left\lfloor\,{N \over 2}\,\right\rfloor} + \zeta\pars{1 \over 2}} - \bracks{2\root{N} + \zeta\pars{1 \over 2}}} \\[5mm] = & \ \bbx{\color{#44f}{\pars{\root{2} - 1}\zeta\pars{1 \over 2}}} \approx -0.6049 \\ & \end{align} Note that $\ds{\lim_{N \to \infty}\pars{2\root{2}\root{\left\lfloor\,{N \over 2}\,\right\rfloor} - 2\root{N}} = \color{red}{\Large0}}$.