Answer to Question #203172 in Real Analysis for Rajkumar

Question

Examine the convergence of the following series:

i) (3×4)/5² + (5×6)/7² + (7×8)/9²....

ii) 1 + 4x + 4²x² + 4³x³ +....(x > 0)

Answer 1

$$\displaystyle\sum_{i=1}^{\infin}\dfrac{(2n+1)(2n+2)}{(2n+3)^2}=\dfrac{3\times4}{5^2}+\dfrac{5\times6}{7^2}+\dfrac{7\times8}{9^2}+...$$

Use the Test for Divergence

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

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

$$=\dfrac{2(2)}{(2)^2}=1\not=0$$

The given series diverges by the Test for Divergence.

(ii)

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

The geometric series

$$\displaystyle\sum_{i=0}^{\infin}ar^n$$

is convergent if $|r|<1.$

If $|r|\geq1,$ the geometric series diverges.

We have $r=4x, x>0$

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