4

I have a proof but it does not seem elegant. Is there a more elegant solution? Thanks.

Consider $X = \sum_{i = 1}^{2k}\frac{(-1)^{i+1}}{i} = X_1 + X_2$ where

$X_1 = 1 - \frac{1}{2} + ... + \frac{1}{k-1} - \frac{1}{k}$

$X_2 = \frac{1}{k+1} - \frac{1}{k+2} + ... + \frac{1}{2k-1} - \frac{1}{2k}$

$k$ is even.

Every negative term $d$ in $X_2$ is of the form $\frac{1}{2^my}$ where $y$ is odd and could be $1$. Also $y \le k$

For every such $d$, we will have the following terms $\frac{1}{y} - \frac{1}{2y} - ... - \frac{1}{2^{m-1}y}$ in $X_1$. Adding $d$ to the above, we get

$\frac{1}{y} - \frac{1}{2y} - ... - \frac{1}{2^{m-1}y} - \frac{1}{2^my} = \frac{1}{2^my}$

So every negative term $d$ in $X_2$ is now replaced with a term of the same magnitude but with a positive sign.

Hence $X = \frac{1}{k+1} + \frac{1}{k+2} + ... + \frac{1}{2k-1} + \frac{1}{2k}$

Jyrki Lahtonen
  • 133,153
sku
  • 2,643

4 Answers4

2

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\sum_{i = 1}^{2k}{\pars{-1}^{i + 1} \over i} = \sum_{i = k + 1}^{2k}{1 \over i}:\ {\large ?}}$.

\begin{align} \sum_{i = 1}^{2k}{\pars{-1}^{i + 1} \over i} & = \sum_{i = 1}^{k}{1 \over 2i - 1} - \sum_{i = 1}^{k}{1 \over 2i} = \pars{\sum_{i = 1}^{2k}{1 \over i} - \sum_{i = 1}^{k}{1 \over 2i}} - \sum_{i = 1}^{k}{1 \over 2i} \\[5mm] & = \sum_{i = 1}^{2k}{1 \over i} - \sum_{i = 1}^{k}{1 \over i} = \bbx{\ds{\sum_{i = k + 1}^{2k}{1 \over i}}} \end{align}

Felix Marin
  • 89,464
2

Let $S_k=\sum_{k=1}^{2k}(-1)^{i+1}/i$ and $T_k=\sum_{i+1}^{2i}1/i$. Then $$T_k-T_{k-1}=\frac1{2k-1}+\frac1{2k}-\frac1k=\frac1{2k-1}-\frac1{2k}= S_k-S_{k-1}.$$ Obviously, $S_0=T_0$ so by induction....

Angina Seng
  • 158,341
1

You're right, there's a much simpler way:

$$S=\sum^{2k}_{i=1}\left[\frac{(-1)^{(i+1)}}{i}\right]=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6} \cdot\cdot\cdot +\frac{1}{2k-1}-\frac{1}{2k}$$

$$S=\color{blue}{\left[\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6} \cdot\cdot\cdot +\frac{1}{2k-1}+\frac{1}{2k}\right]}-2\color{red}{\left[\frac{1}{2}+\frac{1}{4}+\cdot\cdot\cdot\frac{1}{2k}\right]}$$ $$S=\sum^{2k}_{i=1}\frac{1}{i}-2\sum^k_{i=1}\frac{1}{2i}=\sum^{2k}_{i=1}\frac{1}{i}-\sum^k_{i=1}\frac{2}{2i}=\color{purple}{\sum^{2k}_{i=1}\frac{1}{i}-\sum^k_{i=1}\frac{1}{i}}=\sum^{2k}_{i=k+1}\frac{1}{i}$$

Hence proved.

Note: $$\sum^{2k}_{i=1}\frac{1}{i}=\sum^{k}_{i=1}\frac{1}{i}+\sum^{2k}_{i=k+1}\frac{1}{i}$$

Panglossian Oporopolist
  • 5,223
0

We can use harmonic numbers $H_k=\sum_{i=1}^k\frac{1}{i}$.

We obtain \begin{align*} \sum_{i=k+1}^{2k}\frac{1}{i}=\color{blue}{H_{2k}-H_k} \end{align*}

on the other hand we get \begin{align*} \sum_{i=1}^{2k}\frac{(-1)^{i+1}}{i}-H_{2k} &=-2\left(\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2k}\right)=-H_k\\ \end{align*}

which implies \begin{align*} \sum_{i=1}^{2k}\frac{(-1)^{i+1}}{i} &=\color{blue}{H_{2k}-H_k} \end{align*} and the claim follows.

Markus Scheuer
  • 108,315