Question #96654
Show that the following series converges: (4)
(2n 3)(2n 5)
1
n 1 + + ∑

=
1
Expert's answer
2019-10-16T09:25:43-0400

Proof:

We need to show that the series converges.

Let i=1n\displaystyle\sum_{i=1}^n1(2n+3)(2n+5)=\frac {1} {(2n+3) (2n+5)} = i=1n\displaystyle\sum_{i=1}^n ui Here un =1(2n+3)(2n+5)\frac {1} {(2n+3) (2n+5)}



Let us consider the series i=1n\displaystyle\sum_{i=1}^n vi here vn =1n2\frac {1} {n^2}



Now, limn>\lim\limits_{n->\infin}\frac {} { }unvn\frac {u_n} {v_n} = limn>\lim\limits_{n->\infin} 1[(2n+3)(2n+5)]1(n2)\frac {\frac {1} {[(2n+3) (2n+5)]}} {\frac {1} {(n^2)}} = limn>\lim\limits_{n->\infin} n2(2n+3)(2n+5)\frac {n^2} {(2n+3) (2n+5)}



= limn>\lim\limits_{n->\infin}1(2+3n)(2+5n)\frac {1} {(2+\frac {3} {n} ) (2+\frac {5} {n})} = 1(2+0)(2+0)\frac {1} {(2+0) (2+ 0) } ( Here 3=0\frac {3} {\infin } = 0 and 5=0\frac {5} {\infin } = 0 )


= 14<1\frac {1} {4} < 1


limn>\lim\limits_{n->\infin}unvn<1\frac {u_n} {v_n} < 1 , then the given series converges.

Answer: the series converges.

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