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Question Number 182875 by HeferH last updated on 15/Dec/22

$$$$$$FindthevalueofS:$$$$S=1+\frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}...$$$$$$

Answered by TheSupreme last updated on 16/Dec/22

$$S=1+\mathrm{\Sigma }\frac{1}{i(i+1)}=1+\mathrm{\Sigma }\frac{A}{i}+\mathrm{\Sigma }\frac{B}{i+1}$$$$1+\mathrm{\Sigma }\frac{Ai+A+Bi}{i(i+1)}$$$$A+B=0$$$$A=1$$$$S=1+\mathrm{\Sigma }\frac{1}{i}-\mathrm{\Sigma }\frac{1}{i+1}$$$$S_n=1+\frac{1}{2}-\frac{1}{n+1}$$$$S=\frac{3}{2}$$

Commented by Frix last updated on 16/Dec/22

$$S_n=1+\underset{i=1}{\sum ^n}\frac{1}{i}-\underset{i=1}{\sum ^n}\frac{1}{i+1}=1+1-\frac{1}{n+1}\Rightarrow$$$$S=2$$

Answered by manxsol last updated on 16/Dec/22

$$S=1+(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})...$$$$$$$$...(\frac{1}{n-2}-\frac{1}{n-1})+(\frac{1}{n-1}-\frac{1}{n})$$$$S=1+1-\frac{1}{n}$$$$n\Rightarrow \mathrm{\infty }\Rightarrow \frac{1}{n}=0$$$$S=2$$

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