Answer to Question #199718 in Calculus for Moe

Question

Consider the series:

$∞$

$Σ$ [ $(x^2 + 9)$ / 25 $]^a$

$a = 1$

The values of $y$ of which this series converges form an internal $(-k, k)$ .

Give the value for $k$

Answer 1

It is given that,

$\displaystyle\sum_{a=1}^\infin\lbrack\frac{(x^2+9)}{25}\rbrack^a=\displaystyle\sum_{a=1}^\infin\lbrack M_a\rbrack$

$\implies M_a=\lbrack\frac{(x^2+9)}{25}\rbrack^a$

By the ratio test,

$L=\lim_{a\rightarrow \infty} \vert\frac{M_{a+1}}{M_a}\vert$

$L=\lim_{a\rightarrow \infty }\vert\frac{\lbrack\frac{ (x^2+9)}{25}\rbrack ^{a+1}}{\lbrack \frac{(x^2+9)}{25}\rbrack^a}\vert$

$L=lim_{ a\rightarrow \infty}\space \vert\lbrack\frac{(x^2+9)}{25}\rbrack\vert$

$L=\vert \lbrack\frac{(x^2+9)}{25}\rbrack\vert$

For interval of convergence,

L<1

$\vert\frac{(x^2+9)}{25}\vert<1$

$\frac{x^2+9}{25}<1$

$x^2+9<25\\x^2+9-25<0\\x^2-16<0\\(x-4)(x+4)<0\\-4<x<4$

Hence the interval of convergence is $x\in(-4,4)$

and $k=4$