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Question Number 126952 by Study last updated on 25/Dec/20

$$\underset{n=1}{\sum ^{10000}}\frac{1}{n(n+1)}=???$$

Commented by Study last updated on 25/Dec/20

$$helpme$$

Answered by Dwaipayan Shikari last updated on 25/Dec/20

$$\underset{n=1}{\sum ^{10000}}\frac{1}{n}-\frac{1}{n+1}=(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...-\frac{1}{10001})=\frac{10000}{10001}$$

Answered by Olaf last updated on 25/Dec/20

$$\mathrm{S}=\underset{n=1}{\sum ^{10000}}\frac{1}{n(n+1)}$$$$\mathrm{S}=\underset{n=1}{\sum ^{10000}}[\frac{1}{n}-\frac{1}{n+1}]$$$$\mathrm{S}=\frac{1}{1}-\frac{1}{10001}=\frac{10000}{10001}$$

Answered by bramlexs22 last updated on 25/Dec/20

$$\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$$$$\underset{n=1}{\sum ^{10,000}}(\frac{1}{n}-\frac{1}{n+1})=1-\frac{1}{10,001}$$$$$$

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