Question :
Test the convergence of the series
$\displaystyle
\frac{x^2}{3 \sqrt{2}} + \frac{x^4}{4 \sqrt{3}} + \frac{x^6}{5 \sqrt{4}} + ...
$ for all x
I am using the ratio test to test the series ...
$\displaystyle \lim_{n \to \infty} \frac{a_n + 1 }{a_n} = \lim_{n \to \infty} \frac{x^{2(n+1)}}{(n + 3) \sqrt{n + 2}} . \frac{(n + 2) \sqrt{n + 1}}{x^{2n}}$
= $\displaystyle \lim_{n \to \infty} \frac{x^{2n} . x^2}{(n + 3) \sqrt{n + 2}} . \frac{(n + 2) \sqrt{n + 1}}{x^{2n}}$
= $\displaystyle \lim_{n \to \infty} \frac{x^2}{(n + 3) \sqrt{n + 2}} . \frac{(n + 2) \sqrt{n + 1}}{1}$
= $\displaystyle \lim_{n \to \infty} \frac{x^2 \left[(n + 2) \sqrt{n + 1} \right]}{(n + 3) \sqrt{n + 2}} $
I am stuck here!!!!!!
From your last step, you can pull out the $\displaystyle x^2$ (because it has no $\displaystyle n$ in it), so
$\displaystyle \lim_{n \to \infty} \frac{x^2(n + 2) \sqrt{n + 1}}{(n + 3) \sqrt{n + 2}} = x^2\lim_{n\to\infty}\frac{\sqrt{(n+2)^2(n+1)}}{\sq rt{(n+3)^2(n+2)}}=x^2\lim_{n\to\infty}\frac{\sqrt{ n^3+5n^2+8n+4}}{\sqrt{n^3+8n^2+21n+18}}$
Spoiler:
Substitute x = 1 in the original series we get
$\displaystyle
\frac{(-1)^2}{3 \sqrt{2}} + \frac{(-1)^4}{4 \sqrt{3}} + \frac{(-1)^6}{5 \sqrt{4}}
$
$\displaystyle
u_n = \frac{(-1)^2n}{(n + 2) \sqrt{n + 1}}
$ and $\displaystyle
v_n = \frac{1}{n^{\frac{3}{2}}} $
$\displaystyle
\lim_{n \to \infty} \frac{u_n}{v_n} = \lim_{n \to \infty} \frac{\frac{(-1)^2n}{(n + 2) \sqrt{n + 1}} }{\frac{1}{n^{\frac{3}{2}}}}$ = $\displaystyle \lim_{n \to \infty} \frac{(-1)^2n}{(n + 2) \sqrt{n + 1}} . n^{\frac{3}{2}}$
Is this correct????????