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I need to prove the following Identity $$\frac1{n+1}+\frac1{n+2}+\cdots+\frac1{2n-1}+\frac1{2n}=1-\frac12+\frac13-\cdots+\frac1{2n-1}-\frac1{2n}$$ Which in compact form is: $$\sum_{j=1}^n \frac1{n+j}=\sum_{k=1}^{2n} \frac{(-1)^{k+1}}{k}$$

I can also express the previous equality as: $$\sum_{j=1}^n \frac1{n+j}=\sum_{j=1}^{n} \frac1{2j-1}-\sum_{j=1}^{n} \frac1{2j}$$ However I can't seem to go any further. Any hint is appreciated!

DMH16
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  • Use mathematical induction. – Ted Shifrin Apr 08 '17 at 16:34
  • @TedShifrin I thought about that, however, I don't think my book wants mathematical induction, it wants me to work with properties of sums (operations of summations). Also, math induction is explained chapters after the one I found the exercise on. – DMH16 Apr 08 '17 at 16:36

1 Answers1

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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \sum_{j = 1}^{n}{1 \over n + j} & = \sum_{j = 1 + n}^{2n}{1 \over j} = \sum_{j = 1}^{2n}{1 \over j} - \sum_{j = 1}^{n}{1 \over j} = \bracks{\sum_{j = 1}^{n}{1 \over 2j} + \sum_{j = 1}^{n}{1 \over 2j - 1}} - 2\sum_{j = 1}^{n}{1 \over 2j} \\[5mm] & = \sum_{j = 1}^{n}{1 \over 2j - 1} - \sum_{j = 1}^{n}{1 \over 2j} = \bbx{\ds{\sum_{j = 1}^{2n}{\pars{-1}^{n + 1} \over j}}} \end{align}

Felix Marin
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  • Can you explain what you did on your third step? I can see that is is true but I don't see how you went from one to the other. Thank you! Specifically: $\sum_{j=1}^{2n}\frac1j=\sum_{j = 1}^{n}{1 \over 2j} + \sum_{j = 1}^{n}{1 \over 2j - 1}$ – DMH16 Apr 09 '17 at 04:59
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    @DMH16 You can partition the integers between $1$ and $2n$ into the set of odd ones and the set of even ones – Astyx Apr 09 '17 at 08:13