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

repeat question Shiw that : Σ_(i=1) ^n ((1/(2i−1))−(1/(2i)))=Σ_(i=1) ^n (1/(n+i)) ?

$${repeat}\:{question} \\ $$$${Shiw}\:{that}\:: \\ $$$$\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\:\left(\frac{\mathrm{1}}{\mathrm{2}{i}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}{i}}\right)=\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\:\frac{\mathrm{1}}{{n}+{i}}\:\:? \\ $$

Answered by witcher3 last updated on 10/Jul/23

n=1⇒1−(1/2)=(1/2)=(1/(1+1)) ∀n∈Z_+ suppose TrueΣ_(i=1) ^n (1/(2i−1))−(1/(2i))=Σ(1/(n+i)) Σ_(i=1) ^(n+1) (1/(2i−1))−(1/(2i))=^? Σ_(i=1) ^n (1/(n+1+i))=Σ_(i=1) ^(n−1) (1/(n+1+i))+(1/(2n+1)) ⇔Σ_(i=1) ^n ((1/(2i−1))−(1/(2i)))+(1/(2n+1))−(1/(2n+2))= Σ_(i=1) ^n (1/(n+i))+(1/(2n+1))−(1/(2n+2))=Σ_(n=2) ^n (1/(n+i))+(1/(2n+1))+(1/(n+1))−(1/(2(n+1))) =Σ_(i=1) ^(n−1) (1/(n+1+i))+(1/(n+1+n))+(1/(n+1+n+1)) =Σ_(i=1) ^(n+1) (1/((n+1)+i))....True

$$\mathrm{n}=\mathrm{1}\Rightarrow\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{1}} \\ $$$$\forall\mathrm{n}\in\mathbb{Z}_{+} \mathrm{suppose}\:\mathrm{True}\underset{\mathrm{i}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{2i}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}\boldsymbol{\mathrm{i}}}=\Sigma\frac{\mathrm{1}}{\mathrm{n}+\mathrm{i}} \\ $$$$\underset{\mathrm{i}=\mathrm{1}} {\overset{\mathrm{n}+\mathrm{1}} {\sum}}\frac{\mathrm{1}}{\mathrm{2i}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2i}}\overset{?} {=}\underset{\mathrm{i}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}+\mathrm{i}}=\underset{\mathrm{i}=\mathrm{1}} {\overset{\mathrm{n}−\mathrm{1}} {\sum}}\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}+\mathrm{i}}+\frac{\mathrm{1}}{\mathrm{2n}+\mathrm{1}} \\ $$$$\Leftrightarrow\underset{\mathrm{i}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{2i}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2i}}\right)+\frac{\mathrm{1}}{\mathrm{2n}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2n}+\mathrm{2}}= \\ $$$$\underset{\mathrm{i}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{n}+\mathrm{i}}+\frac{\mathrm{1}}{\mathrm{2n}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2n}+\mathrm{2}}=\underset{\mathrm{n}=\mathrm{2}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{n}+\mathrm{i}}+\frac{\mathrm{1}}{\mathrm{2n}+\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{n}+\mathrm{1}\right)} \\ $$$$=\underset{\mathrm{i}=\mathrm{1}} {\overset{\mathrm{n}−\mathrm{1}} {\sum}}\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}+\mathrm{i}}+\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}+\mathrm{n}}+\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}+\mathrm{n}+\mathrm{1}} \\ $$$$=\underset{\mathrm{i}=\mathrm{1}} {\overset{\mathrm{n}+\mathrm{1}} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}\right)+\mathrm{i}}....\mathrm{True} \\ $$$$ \\ $$

Commented by MM42 last updated on 10/Jul/23

excellent ★

$${excellent}\:\:\bigstar \\ $$

Commented by witcher3 last updated on 11/Jul/23

thank You sir

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

Answered by MM42 last updated on 10/Jul/23

(1−(1/2))+((1/3)−(1/4))+...+((1/(2n−1))−(1/(2n))) =(1+(1/3)+...+(1/(2n−1)))−((1/2)+(1/4)+...+(1/(2n))) +2((1/2)+(1/4)+...+(1/(2n)))−2((1/2)+(1/4)+...+(1/(2n))) =(1+(1/2)+(1/3)+...+(1/n)+(1/(n+1))+...+(1/(2n)))−(1+(1/2)+(1/3)+...+(1/n)) =(1/(n+1))+(1/(n+2))+...+(1/(2n)) ✓

$$\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right)+\left(\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{4}}\right)+...+\left(\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}{n}}\right) \\ $$$$=\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}+...+\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}\right)−\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}+...+\frac{\mathrm{1}}{\mathrm{2}{n}}\right) \\ $$$$+\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}+...+\frac{\mathrm{1}}{\mathrm{2}{n}}\right)−\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}+...+\frac{\mathrm{1}}{\mathrm{2}{n}}\right) \\ $$$$=\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+...+\frac{\mathrm{1}}{{n}}+\frac{\mathrm{1}}{{n}+\mathrm{1}}+...+\frac{\mathrm{1}}{\mathrm{2}{n}}\right)−\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+...+\frac{\mathrm{1}}{{n}}\right) \\ $$$$=\frac{\mathrm{1}}{{n}+\mathrm{1}}+\frac{\mathrm{1}}{{n}+\mathrm{2}}+...+\frac{\mathrm{1}}{\mathrm{2}{n}}\:\checkmark \\ $$$$ \\ $$

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