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Question Number 91837 by zainal tanjung last updated on 03/May/20

$$1).\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}(\frac{5}{\mathrm{n}+2}-\frac{5}{\mathrm{n}+3})=...$$$$$$$$2).\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\left(\frac{1}{{4\mathrm{n}}^2-1}\right)=...$$$$$$$$3).\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\left(\frac{3\mathrm{n}}{5\mathrm{n}-1}\right)=...$$$$$$$$\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\left(\frac{\mathrm{n}}{\mathrm{n}+1}\right)=...$$

Commented by Tony Lin last updated on 03/May/20

$$1)=5(\frac{1}{3}-\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{n})=\frac{5}{3}$$$$2)=\frac{1}{2}\underset{n=1}{\sum ^{\mathrm{\infty }}}(\frac{1}{2n-1}-\frac{1}{2n+1})$$$$=\frac{1}{2}(1-\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{2n+1})=\frac{1}{2}$$$$3)\underset{n\to \mathrm{\infty }}{\lim }\frac{3n}{5n-1}=\frac{3}{5}\ne 0\to divergent$$$$4)\underset{n\to \mathrm{\infty }}{\lim }\frac{n}{n+1}=1\ne 0\to divergent$$

Commented by zainal tanjung last updated on 03/May/20

$$$$$$$$$$\mathrm{corection}\mathrm{and}\mathrm{revition}$$$$1).\mathrm{false}\to \mathrm{true}\mathrm{answered}:\frac{5}{3}$$$$2).\mathrm{true}.$$$$3).\mathrm{true}$$$$4).\mathrm{true}$$$$$$

Commented by Tony Lin last updated on 03/May/20

$$Ihavecorrected$$

Commented by zainal tanjung last updated on 03/May/20

$$\mathrm{okey}.\mathrm{thanks}$$

Commented by john santu last updated on 03/May/20

You seem like you're testing

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