Question

Find whether the following series are convergent or divergent:
$x^2+\frac{2^2}{3.4}{x}^4+\frac{2^2.4^2}{\mathrm{3.4.5.6}}{x}^6+\frac{2^2.4^2.6^2}{\mathrm{3.4.5.6.7.8}}{x}^8+...$

Answer 1

Let

$S_n=x^2[1+\frac{2^2}{3\cdot 4}{x}^2+\frac{2^2{4}^2}{3\cdot 4\cdot 5\cdot 6}{x}^4+\frac{2^2{4}^2{6}^2}{3\cdot 4\cdot 5\cdot 6\cdot 7\cdot 8}{x}^6+....]T_n=\frac{2^2{4}^2{6}^2....{\left(2n\right)}^2{x}^{2n+2}}{3\cdot 4\cdot 5\cdot 6\cdot 7\cdot 8\cdot ....(2n+1)(2n+2)}{T}_{n+1}=\frac{2^2{4}^2{6}^2....{\left(2n\right)}^2{(2n+2)}^2{x}^{2n+4}}{3\cdot 4\cdot 5\cdot 6\cdot 7\cdot 8\cdot ....(2n+1)(2n+2)(2n+3)(2n+4)}\underset{n\to \infty }{lim}\frac{T_{n+1}}{T_n}=\underset{n\to \infty }{lim}\frac{{(2n+2)}^2{x}^2}{(2n+3)(2n+4)}=\underset{n\to \infty }{lim}\frac{{(2+2/n)}^2{x}^2}{(2+3/n)(2+4/n)}=x^2$

For $x>1$ $S_n$ diverges

For $-1<x<1$ $S_n$ converges

For $x=1$, ratio test fails

Then we apply Raabe's Test

$\underset{n\to \infty }{lim}[\frac{T_n}{T_{n+1}}-1]n=\underset{n\to \infty }{lim}[\frac{(2n+3)(2n+4)}{{(2n+2)}^2}-1]n=\underset{n\to \infty }{lim}n\left[\frac{4n^2-14n+12-4n^2-4-9n}{{(2n+2)}^2}\right]=\underset{n\to \infty }{lim}n\left[\frac{6n+8}{{(2n+2)}^2}\right]=\underset{n\to \infty }{lim}\left[\frac{6n^2+8n}{4n^2+4+4n}\right]=\underset{n\to \infty }{lim}\left[\frac{6+\frac{8}{n^2}}{4+\frac{4}{n^2}+\frac{4}{n}}\right]=\frac{3}{2}$

Since, $\underset{n\to \infty }{lim}[\frac{T_n}{T_{n+1}}-1]n=\frac{3}{2}>1$

The series is convergent for $x=1$