Question

Find whether the following series are convergent or divergent:
$1+\frac{1}{2}.\frac{x^2}{4}+\frac{\mathrm{1.3.5}}{\mathrm{2.4.6}}.\frac{x^4}{8}+\frac{\mathrm{1.3.5.7.9}}{\mathrm{2.4.6.8.10}}.\frac{x^6}{12}+....$

Answer 1

Let the general term of the series as

$T_n=\frac{1\cdot 3\cdot 5\cdot 7\cdot 9\cdot ....\cdot (4n-3)x^{2n}}{2\cdot 4\cdot 6\cdot 8\cdot 10\cdot ....\cdot \left(4n\right)}{T}_{n+1}=\frac{1\cdot 3\cdot 5\cdot 7\cdot 9\cdot ....\cdot (4n-3)(4n+1)x^{2n+2}}{2\cdot 4\cdot 6\cdot 8\cdot 10\cdot ....\cdot \left(4n\right)(4n+4)}$

Applying ratio test

$\underset{n\to \infty }{lim}\frac{T_{n+1}}{T_n}=\underset{n\to \infty }{lim}\frac{(4n+1)x^2}{(4n+4)}=\underset{n\to \infty }{lim}\frac{(4+1/n)x^2}{(4+4/n)}=x^2$

if $x^2>1$, the series is divergent

$x\in (-\infty ,-1)\cup (1,\infty )$ The series is divergent

If $x^2<1$, the series is convergent

$x\in (-1,1)$ the series is convergent

for $x=1$, the ratio test fails.

We then apply Raabel's test

$\underset{n\to \infty }{lim}[\frac{T_n}{T_{n+1}}-1]n=\underset{n\to \infty }{lim}[\frac{4n+4}{4n+1}-1]n=\underset{n\to \infty }{lim}\frac{(4n+4-4n-1)n}{(4n+1)}=\underset{n\to \infty }{lim}\frac{3n}{4n+1}=\frac{3}{4}$

SInce, $\underset{n\to \infty }{lim}[\frac{T_n}{T_{n+1}}-1]n=\frac{3}{4}<1$

The series is divergent for $x=1$