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Question Number 194559 by MM42 last updated on 09/Jul/23

$$repeatquestion$$$$Shiwthat:$$$$\underset{i=1}{\sum ^n}(\frac{1}{2i-1}-\frac{1}{2i})=\underset{i=1}{\sum ^n}\frac{1}{n+i}?$$

Answered by witcher3 last updated on 10/Jul/23

$$\mathrm{n}=1\Rightarrow 1-\frac{1}{2}=\frac{1}{2}=\frac{1}{1+1}$$$$\mathrm{\forall }\mathrm{n}\in {\mathbb{Z}}_{+}\mathrm{suppose}\mathrm{True}\underset{\mathrm{i}=1}{\sum ^{\mathrm{n}}}\frac{1}{2\mathrm{i}-1}-\frac{1}{2\mathbf{i}}=\mathrm{\Sigma }\frac{1}{\mathrm{n}+\mathrm{i}}$$$$\underset{\mathrm{i}=1}{\sum ^{\mathrm{n}+1}}\frac{1}{2\mathrm{i}-1}-\frac{1}{2\mathrm{i}}\stackrel{?}{=}\underset{\mathrm{i}=1}{\sum ^{\mathrm{n}}}\frac{1}{\mathrm{n}+1+\mathrm{i}}=\underset{\mathrm{i}=1}{\sum ^{\mathrm{n}-1}}\frac{1}{\mathrm{n}+1+\mathrm{i}}+\frac{1}{2\mathrm{n}+1}$$$$\iff \underset{\mathrm{i}=1}{\sum ^{\mathrm{n}}}(\frac{1}{2\mathrm{i}-1}-\frac{1}{2\mathrm{i}})+\frac{1}{2\mathrm{n}+1}-\frac{1}{2\mathrm{n}+2}=$$$$\underset{\mathrm{i}=1}{\sum ^{\mathrm{n}}}\frac{1}{\mathrm{n}+\mathrm{i}}+\frac{1}{2\mathrm{n}+1}-\frac{1}{2\mathrm{n}+2}=\underset{\mathrm{n}=2}{\sum ^{\mathrm{n}}}\frac{1}{\mathrm{n}+\mathrm{i}}+\frac{1}{2\mathrm{n}+1}+\frac{1}{\mathrm{n}+1}-\frac{1}{2(\mathrm{n}+1)}$$$$=\underset{\mathrm{i}=1}{\sum ^{\mathrm{n}-1}}\frac{1}{\mathrm{n}+1+\mathrm{i}}+\frac{1}{\mathrm{n}+1+\mathrm{n}}+\frac{1}{\mathrm{n}+1+\mathrm{n}+1}$$$$=\underset{\mathrm{i}=1}{\sum ^{\mathrm{n}+1}}\frac{1}{(\mathrm{n}+1)+\mathrm{i}}....\mathrm{True}$$$$$$

Commented by MM42 last updated on 10/Jul/23

$$excellent★$$

Commented by witcher3 last updated on 11/Jul/23

$$\mathrm{thank}\mathrm{You}\mathrm{sir}$$

Answered by MM42 last updated on 10/Jul/23

$$(1-\frac{1}{2})+(\frac{1}{3}-\frac{1}{4})+...+(\frac{1}{2n-1}-\frac{1}{2n})$$$$=(1+\frac{1}{3}+...+\frac{1}{2n-1})-(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2n})$$$$+2(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2n})-2(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2n})$$$$=(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{2n})-(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n})$$$$=\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}✓$$$$$$

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