Question Number 116057 by Study last updated on 30/Sep/20
$$\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\mathrm{3}}{\mathrm{3}{n}+\mathrm{1}}=? \\ $$
Answered by mindispower last updated on 30/Sep/20
$$\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{3}{n}+\mathrm{1}\right)}? \\ $$
Commented by mnjuly1970 last updated on 01/Oct/20
$${answer}::=\:\frac{\pi}{\mathrm{3}\sqrt{\mathrm{3}}}+\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{3}}\:\checkmark \\ $$
Commented by mnjuly1970 last updated on 01/Oct/20
$$\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=−\infty} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{3}{n}+\mathrm{1}\right)}\:\neq\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{3}{n}+\mathrm{1}} \\ $$$${l}.{h}.{s}=\frac{\mathrm{1}}{\mathrm{2}}\left[.....+\frac{\mathrm{1}}{\mathrm{14}}−\frac{\mathrm{1}}{\mathrm{11}}\:+\frac{\mathrm{1}}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{5}}\:\:\:+\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{4}}++\frac{\mathrm{1}}{\mathrm{7}}−\frac{\mathrm{1}}{\mathrm{10}}+...\right. \\ $$$$ \\ $$$$\:\: \\ $$
Commented by mnjuly1970 last updated on 01/Oct/20
$$\:{because} \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{3}{n}+\mathrm{1}}\:=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{\mathrm{3}{n}} {dx}\:\: \\ $$$$=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{3}} }\:{dx}=\frac{\pi}{\mathrm{3}\sqrt{\mathrm{3}}}+\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{3}}\:\:\checkmark \\ $$$$\:\:\:\:\:.{m}.{n}.{july}.\mathrm{70} \\ $$$$\: \\ $$
Answered by mathmax by abdo last updated on 30/Sep/20
$$\mathrm{this}\:\mathrm{serie}\:\mathrm{is}\:\mathrm{divergent}\:\mathrm{due}\:\mathrm{to}\:\frac{\mathrm{3}}{\mathrm{3n}+\mathrm{1}}\sim\frac{\mathrm{1}}{\mathrm{n}} \\ $$