Question

1/(3x5)+√3/(5x8)+ √5/(7x11) +... test the convergence

Answer 1

ANSWER: the series converges.

Explanation. $\frac { 1 }{ 3\cdot 5 } +\frac { \sqrt { 3 } }{ 5\cdot 8 } +\frac { \sqrt { 5 } }{ 7\cdot 11 } +\cdots +\frac { \sqrt { 2n-1 } }{ (2n+1)\cdot (3n+2) } +\cdots$

Let ${ a }_{ n }=\frac { \sqrt { 2n-1 } }{ (2n+1)\cdot (3n+2) } =\frac { \sqrt { n } }{ { n }^{ 2 } } \cdot \frac { \sqrt { 2-\frac { 1 }{ n } } }{ \left( 2+\frac { 1 }{ n } \right) \left( 3+\frac { 2 }{ n } \right) }$ , ${ b }_{ n }=\frac { 1 }{ { n }^{ \frac { 3 }{ 2 } } }$ . Since $\lim _{ n\rightarrow \infty }{ \frac { { a }_{ n } }{ { b }_{ n } } =\lim _{ n\rightarrow \infty }{ \frac { \sqrt { 2-\frac { 1 }{ n } } }{ \left( 2+\frac { 1 }{ n } \right) \left( 3+\frac { 2 }{ n } \right) } } } =\frac { \sqrt { 2 } }{ 2\cdot 3 } =\frac { 1 }{ 3\sqrt { 2 } } >0\quad$ and $\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { n }^{ \frac { 3 }{ 2 } } } }$ converges, hence (by Limit Comparison Test) $\sum _{ n=1 }^{ \infty }{ \frac { \sqrt { 2n-1 } }{ (2n+1)\cdot (3n+2) } }$ converges.