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Question Number 144323 by qaz last updated on 24/Jun/21

$$\underset{\mathrm{i}=1}{\sum ^{\mathrm{n}}}\frac{{(-1)}^{\mathrm{n}+1}}{\mathrm{n}}=?$$

Answered by mathmax by abdo last updated on 25/Jun/21

$${\mathrm{A}}_{\mathrm{n}}=\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}-1}}{\mathrm{k}}\Rightarrow {\mathrm{A}}_{\mathrm{n}}=\mathrm{\Sigma }{(....)}_{\mathrm{k}=2\mathrm{p}}+\mathrm{\Sigma }{(....)}_{\mathrm{k}=2\mathrm{p}+1}$$$$=-\sum _{\mathrm{p}=1}^{\left[\frac{\mathrm{n}}{2}\right]}\frac{1}{2\mathrm{p}}+\sum _{\mathrm{p}=0}^{\left[\frac{\mathrm{n}-1}{2}\right]}\frac{1}{2\mathrm{p}+1}$$$$\mathrm{we}\mathrm{have}\sum _{\mathrm{p}=1}^{\left[\frac{\mathrm{n}}{2}\right]}\frac{1}{\mathrm{p}}={\mathrm{H}}_{\left[\frac{\mathrm{n}}{2}\right]}$$$$\sum _{\mathrm{p}=0}^{\mathrm{N}}\frac{1}{2\mathrm{p}+1}=1+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{2\mathrm{N}+1}=1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2\mathrm{N}}+\frac{1}{2\mathrm{N}+1}$$$$-\frac{1}{2}-\frac{1}{4}-...-\frac{1}{2\mathrm{N}}={\mathrm{H}}_{2\mathrm{N}+1}-\frac{1}{2}{\mathrm{H}}_{\mathrm{N}}\Rightarrow$$$$\sum _{\mathrm{p}=0}^{\left[\frac{\mathrm{n}-1}{2}\right]}\frac{1}{2\mathrm{p}+1}={\mathrm{H}}_{2\left[\frac{\mathrm{n}-1}{2}\right]+1}-\frac{1}{2}{\mathrm{H}}_{\left[\frac{\mathrm{n}-1}{2}\right]}\Rightarrow$$$${\mathrm{A}}_{\mathrm{n}}=-\frac{1}{2}{\mathrm{H}}_{\left[\frac{\mathrm{n}}{2}\right]}+{\mathrm{H}}_{2\left[\frac{\mathrm{n}-1}{2}\right]+1}-\frac{1}{2}{\mathrm{H}}_{\left[\frac{\mathrm{n}-1}{2}\right]}$$

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