Does this series converge?

[Ainzley]

The series is n²ⁿ/3ⁿ(2n)!

I've tried using the ratio test, but I can't simplify the monster that is produced by doing so. As for the comparison test, I have no idea what to begin with. Most of the examples I've done haven't been this annoying.
Any help would be greatly appreciated
Cheers

 

[daon2]

Whats the problem with the ratio test? Do you know the limit of this?


\(\displaystyle \dfrac{(n+1)^{2n+2}}{3n^{2n}(2n+2)(2n+1)} = \dfrac{1}{3}\left(1+\dfrac{1}{n}\right)^{2n}\cdot \dfrac{(n+1)(n+1)}{(2n+2)(2n+1)}\)

 

[lookagain]

The series is n²ⁿ/3ⁿ(2n)!

The series isn't that expression, but a general term is intended by it. And you
need grouping symbols around the denominator, such as this:

n²ⁿ/[3ⁿ(2n)!]


Otherwise, what you have is equivalent to \(\displaystyle \ \dfrac{n^{2n}}{3^n}(2n)!\)

 

[Ainzley]

Thanks guys, really appreciate it.
I managed to get the answer (made a stupid mistake in my algebra).
The final result of the ratio test (that is a_n+1/a_n) is: e²/12 which is less than 1 so the series converges
I hadn't thought to use (1+(a/n))ⁿ=e^a which was pretty neat once I figured it out. Everything fell out nicely after that.

Thanks again guys

 

[soroban]

Hello, Ainzley!

Your answer correct.
I got the same result.

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

The Ratio Test works fine; it just takes some Olympic-level gymnastics.

\(\displaystyle \displaystyle R \;=\;\frac{(n+1)^{2(n+1)}}{3^{n+1}(2n+2)!} \cdot \frac{3^n(2n)!}{n^{2n}} \;=\; \frac{(n+1)^{2n+2}}{n^{2n}}\cdot\frac{3n}{3^{n+1}} \cdot\frac{(2n)!}{(2n+2)!} \)

\(\displaystyle \displaystyle \;\;\; =\;(n+1)^2\frac{(n+1)^{2n}}{n^{2n}}\cdot\frac{1}{3}\cdot\frac{1}{(2n+2)(2n+1)}\;=\;\left(\frac{n+1}{n}\right)^{2n}\frac{(n+1)^2}{3\cdot2(n+1)(2n+1)}\)

\(\displaystyle \displaystyle \;\;\;=\; \left(1 + \frac{1}{n}\right)^{2n}\frac{n+1}{6(2n+1)}\;=\; \left[\left(1+\frac{1}{n}\right)^n\right]^2\frac{1 + \frac{1}{n}}{6(2+\frac{1}{n})} \)


\(\displaystyle \displaystyle\lim_{n\to\infty}R \;=\; \left[\lim_{n\to\infty}\left(1 +\frac{1}{n}\right)^n\right]^2 \lim_{n\to\infty}\frac{1+\frac{1}{n}}{6(2+\frac{1}{n})} \;=\;e^2\frac{1+0}{6(2+0)} \;=\;\frac{e^2}{12} \;<\;1\)

The series converges.