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Question Number 106465 by mohammad17 last updated on 05/Aug/20

Answered by Dwaipayan Shikari last updated on 05/Aug/20

$$$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{(n+2)(n+3)}$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n+2}-\frac{1}{n+3}$$$$\underset{n\to \mathrm{\infty }}{\lim }(\frac{1}{n+2}-\frac{1}{n+3})=\frac{1}{1+2}=\frac{1}{3}$$

Commented by mohammad17 last updated on 05/Aug/20

$$sirconvergeordiverg$$

Commented by Dwaipayan Shikari last updated on 05/Aug/20

$$Converges$$

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