# Math Help - Interval of convergence and ratio test

1. ## Interval of convergence and ratio test

I was wondering if I could get some help on these two problems...

Find the interval of convergence....

$\sum^{\infty}_{n=2} \frac{(x-7)^{n}}{(2 ^ {n} {n}^2 ln n)}$

And use the ratio test to find convergence or divergence....

$\sum^{\infty}_{n=1} \frac{(n!)(2n)!}{(3n)!}$

Thanks, I'd really appreciate it. Some of it is done, but I am unsure of what to do from what I've got.

2. Hello, Rosey!

Here's the second one . . .

Use the ratio test to find convergence or divergence: . $\sum^{\infty}_{n=1} \frac{(n!)(2n)!}{(3n)!}$

$R \;=\;\frac{a_{n+1}}{a_n} \;=\;\frac{(n+1)!(2n+2)!}{(3n+3)!} \cdot\frac{(3n)!}{(n!)(2n)!} \;=\;\frac{(n+1)!}{n!}\cdot\frac{(2n+2)!}{(2n)!}\c dot\frac{(3n)!}{(3n+3)!}$

. . $= \;\frac{n+1}{1}\cdot\frac{(2n+2)(2n+1)}{1}\cdot\fr ac{1}{(3n+3)(3n+2)(3n+1)} \;= \;\frac{(n+1)(2n+2)(2n+1)}{(3n+3)(3n+2)(3n+1)}$

Divide top and bottom by $n^3$

. . $\frac{\left(\dfrac{n+1}{n}\right)\left(\dfrac{2n+2 }{n}\right)\left(\dfrac{2n+1}{n}\right)} {\left(\dfrac{3n+3}{n}\right)\left(\dfrac{3n+2}{n} \right)\left(\dfrac{3n+1}{n}\right)} \;=$ . $\frac{\left(1 + \dfrac{1}{n}\right)\left(2 + \dfrac{2}{n}\right)\left(2 + \dfrac{1}{n}\right)} {\left(3 + \dfrac{3}{n}\right)\left(3 + \dfrac{2}{n}\right)\left(3 + \dfrac{1}{n}\right)}$

Then: . $\lim_{n\to\infty} \frac{\left(1 + \dfrac{1}{n}\right)\left(2 + \dfrac{2}{n}\right)\left(2 + \dfrac{1}{n}\right)} {\left(3 + \dfrac{3}{n}\right)\left(3 + \dfrac{2}{n}\right)\left(3 + \dfrac{1}{n}\right)} \;=\;\frac{(1+0)(2+0)(2+0)}{(3+0)(3+0)(3+0)} \;=\;\frac{4}{27}$

Since $\lim_{n\to\infty} R \;=\;\frac{4}{27}\;<\;1$, the series converges.

3. [quote=Rosey;299537]
Find the interval of convergence....

$\sum^{\infty}_{n=2} \frac{(x-7)^{n}}{(2 ^ {n} {n}^2 ln n)}$

Ratio test would break this guy

$lim_{n-->\infty}\frac{a_{n+1}}{a_n} \$
$lim_{n-->\infty}\frac{(x-7)^{n+1}}{(2^{n+1})(n+1)^2(ln(n+1)} \$ $/ \frac{(x-7)^{n}}{(2^{n})(n)^2(ln(n)} \$

Simplifying we get ...

$lim_{n-->\infty}\frac{(x-7)n^2 lnn}{(2)(n+1)^2(ln(n+1))} = \frac{x-7}{2}\$

5<x<9.

We test the end points by pluging 5 and 9 into the original question.

Doing that for 5 we obtain.

$\sum^{\infty}_{n=2} \frac{(-2)^{n}}{(2 ^ {n} {n}^2 ln n)}$

Which converges by alternating series test.

When we plug in 9 we get

$\sum^{\infty}_{n=2} \frac{1}{( {n}^2 ln n)}$

Which converges by comparison test

$\frac{1}{n^2lnn} < \frac{1}{n^2} \$

So the interval of convergence is

[5,9]