Question

Examine the convergence of the series

a) 3×4/5^2+5×6/7^2+7×8/9^2+.........

b) 1+4x+4^2x^2+4^3x^3+......(x>0)

Answer 1

a)

$$\dfrac{3\times4}{5^2}+\dfrac{5\times6}{7^2}+\dfrac{7\times8}{9^2}+...$$

$$=\displaystyle\sum_{n=1}^{\infin}\dfrac{(2n+1)(2n+2)}{(2n+3)^2}$$

Use Test for Divergence

$$a_n=\dfrac{(2n+1)(2n+2)}{(2n+3)^2}$$

$$\lim\limits_{n\to \infin}a_n=\lim\limits_{n\to \infin}\dfrac{(2n+1)(2n+2)}{(2n+3)^2}$$

$$=\lim\limits_{n\to \infin}\dfrac{(\dfrac{2n}{n}+\dfrac{1}{n})(\dfrac{2n}{2}+\dfrac{2}{n})}{(\dfrac{2n}{n}+\dfrac{3}{n})^2}$$

$$=\lim\limits_{n\to \infin}\dfrac{(2+\dfrac{1}{n})(2+\dfrac{2}{n})}{(2+\dfrac{3}{n})^2}=\dfrac{(2+0)(2+0)}{(2+0)^2}=1\not=0$$

The series $\dfrac{3\times4}{5^2}+\dfrac{5\times6}{7^2}+\dfrac{7\times8}{9^2}+...$ diverges by the Test for Divergence.

b)

$$1+4x+4^2x^2+4^3x^3+......=\displaystyle\sum_{n=0}^{\infin}(4x)^n, x>0$$

This is a geometric series with $a=1$ and $r=4x.$

If $0<4x<1,$ the geometric series converges and its sum is

$$S=\dfrac{1}{1-4x}.$$

If $4x\geq1,$ the geometric series diverges.

Then the series

$$1+4x+4^2x^2+4^3x^3+......=\displaystyle\sum_{n=0}^{\infin}(4x)^n$$

converges for $0<x<\dfrac{1}{4}$ and diverges for $x\geq \dfrac{1}{4}.$