Answer to Question #123127 in Real Analysis for Wachira Ann Wangari

Question #123127
Prove that the series below converges and has a sum<1
1) The summation of 1/(4n-1)(4n+3)as n tends to infinity
2) The summation of1/4n^2-1
1
Expert's answer
2020-06-22T18:05:21-0400

1). Given,

"\\sum_{n=1}^{\\infty}\\dfrac{1}{(4n-1)(4n+3)}"

Thus,

"\\sum_{n=1}^{\\infty}\\dfrac{1}{(4n-1)(4n+3)}=\\frac{1}{4}\\sum_{n=1}^{\\infty}\\bigg(\\dfrac{1}{4n-1}-\\frac{1}{4n+3}\\bigg)"

Now, by Telescoping method, we get

"\\sum_{n=1}^{\\infty}\\dfrac{1}{(4n-1)(4n+3)}=\\frac{1}{4}\\lim_{n\\rightarrow \\infty}(\\frac{1}{3}-\\frac{1}{4n+3})=\\frac{1}{12}<1"

2).

Given,

"\\sum_{n=1}^{\\infty}\\dfrac{1}{4n^2-1}=\\sum_{n=1}^{\\infty}\\dfrac{1}{(2n-1)(2n+1)}=\\frac{1}{2}\\sum_{n=1}^{\\infty}\\bigg(\\dfrac{1}{2n-1}-\\frac{1}{2n+1}\\bigg)"

Again, by Telescoping sum we get

"\\sum_{n=1}^{\\infty}\\dfrac{1}{4n^2-1}=\\frac{1}{2}\\lim_{n\\rightarrow \\infty}(1-\\frac{1}{2n+1})=1\/2<1"

Hence, we are done.


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