Answer to Question #95324 in Real Analysis for Rajni

Question #95324
Show that the following series converges :
summation (infinity , n= 1) [ 1/(2n+3)(2n+5)]?
1
Expert's answer
2019-09-27T09:45:10-0400

To prove that the series


"\\sum\\limits_{n = 1}^{ + \\infty } {\\frac{1}{{(2n + 3)(2n + 5)}}}"

converges, let's compare it with another series

"\\sum\\limits_{n = 1}^{ + \\infty } {\\frac{1}{{{n^2}}}}"


We can see that for any "n \\ge 1"

"\\frac{1}{{(2n + 3)(2n + 5)}} = \\frac{1}{{4{n^2} + 16n + 15}} < \\frac{1}{{{n^2}}}"

because "4{n^2} + 16n + 15 > {n^2}" for any "n \\ge 1" .

It means that if the series "\\sum\\limits_{n = 1}^{ + \\infty } {\\frac{1}{{{n^2}}}}" converges, then the series "\\sum\\limits_{n = 1}^{ + \\infty } {\\frac{1}{{(2n + 3)(2n + 5)}}}" converges too.

To prove that "\\sum\\limits_{n = 1}^{ + \\infty } {\\frac{1}{{{n^2}}}}" converges, we can use the Cauchy integral test: in short, the series "\\sum\\limits_{n = 1}^{ + \\infty } {f(n)}" converges or diverges simultaneously with the integral "\\int\\limits_1^{ + \\infty } {f(x)dx}". Thus, for "\\sum\\limits_{n = 1}^{ + \\infty } {\\frac{1}{{{n^2}}}}" we get


"\\int\\limits_1^{ + \\infty } {\\frac{1}{{{x^2}}}dx} = - \\left. {\\frac{1}{x}} \\right|_1^{ + \\infty } = 1"

So the series "\\sum\\limits_{n = 1}^{ + \\infty } {\\frac{1}{{{n^2}}}}" converges and then the series

"\\sum\\limits_{n = 1}^{ + \\infty } {\\frac{1}{{(2n + 3)(2n + 5)}}}"

converges too by the comparison theorem.


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