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

Question

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

Answer 1

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.