# Series converge or diverge?

• May 21st 2008, 01:05 AM
Rui
Series converge or diverge?
Use what test?

a) ∑(from n=1 to infinity) (arctan n/ n^1.2)
b) ∑(from n=1 to infinity) ( (n^n) / (2n)! )

Not sure how to do these...when i do it using ratio test, it turns out really weird..so yea i need help.

Thanks.
• May 21st 2008, 01:23 AM
flyingsquirrel
Hi
Quote:

Originally Posted by Rui
Use what test?

a) ∑(from n=1 to infinity) (arctan n/ n^1.2)

$\displaystyle \forall x\in \mathbb{R},\,\,|\arctan x|<\frac{\pi}{2}$ : you may try the comparison test

Quote:

b) ∑(from n=1 to infinity) ( (n^n) / (2n)! )
Ratio test : $\displaystyle \left|\frac{a _{n+1}}{a_n}\right|=\frac{(n+1)^{n+1}}{(2(n+1))!}\ cdot \frac{(2n)!}{n^n}=\frac{(n+1)^{n+1}}{n^n(2n+1)(2n+ 2)}$ because $\displaystyle (2(n+1))!=(2n+2)!=(2n+2)\cdot(2n+1)\cdot(2n)!$.

Then, you can use $\displaystyle n+1<n$ to say that $\displaystyle (n+1)^{n+1}<n^{n+1}$ and see what it gives.
• May 21st 2008, 01:49 AM
angel.white
Quote:

Originally Posted by Rui
a) ∑(from n=1 to infinity) (arctan n/ n^1.2)

Comparison test:

let $\displaystyle a_n = \frac{arctan(n)}{n^{1.2}}$, and let $\displaystyle b_n = \frac{\pi}{n^{1.2}}$

Then $\displaystyle a_n \leq b_n$ (because $\displaystyle \lim_{n\to\infty} arctan(n) = \frac {\pi}2$)

And $\displaystyle b_n$ is a p-series with p=1.2, which is greater than 1, therefore $\displaystyle \sum b_n$ converges, and because $\displaystyle \sum a_n$ is monotonic and increasing, and bounded above by $\displaystyle \sum b_n$ we know that $\displaystyle \sum a_n$ converges by the comparison test.

Quote:

Originally Posted by Rui
b) ∑(from n=1 to infinity) ( (n^n) / (2n)! )

Ratio Test:
$\displaystyle a_n =\frac{n^n}{(2n)!}$, and $\displaystyle a_{n+1} = \frac{(n+1)^{n+1}}{(2(n+1))!} = \frac{(n+1)^{n+1}}{(2n+1)(2n+2)(2n)!}$

$\displaystyle \lim_{n\to\infty} \left| \frac {a_{n+1}}{a_n}\right| = \lim_{n\to\infty} \left| \frac{(n+1)^{n+1}}{(2n+1)(2n+2)(2n)!} * \frac{(2n)!}{n^n}\right|$

$\displaystyle =\lim_{n\to\infty} \left| \frac{(n+1)^{n+1}}{(2n+1)(2n+2)n^n}\right|$

$\displaystyle =\lim_{n\to\infty} \left| \frac{(n+1)(n+1)^n}{(2n+1)(2n+2)n^n}\right|$

Here multiply numerator and denominator by 1 over n^n to get

$\displaystyle =\lim_{n\to\infty} \left| \frac{(n+1)(1+1/n)^n}{(2n+1)(2n+2)}\right|$

Now multiply the numerator and denominator by 1/n to get

$\displaystyle =\lim_{n\to\infty} \left| \frac{(1+1/n)(1+1/n)^n}{(2+1/n)(2n+2)}\right|$

It should be relatively clear that this converges to zero, (because the numerator converges to e, and the denominator diverges) but if you need me to explain that more, I can.

So because $\displaystyle \lim_{n\to\infty} \left| \frac {a_{n+1}}{a_n}\right| = 0 < 1$ the series $\displaystyle \sum_n^\infty \frac{n^n}{(2n)!}$ is absolutely convergent and therefore convergent.
• May 21st 2008, 03:23 AM
Mathstud28
Quote:

Originally Posted by angel.white
Comparison test:

let $\displaystyle a_n = \frac{arctan(n)}{n^{1.2}}$, and let $\displaystyle b_n = \frac{\pi}{n^{1.2}}$

Then $\displaystyle a_n \leq b_n$ (because $\displaystyle \lim_{n\to\infty} arctan(n) = \frac {\pi}2$)

And $\displaystyle b_n$ is a p-series with p=1.2, which is greater than 1, therefore $\displaystyle \sum b_n$ converges, and because $\displaystyle \sum a_n$ is monotonic and increasing, and bounded above by $\displaystyle \sum b_n$ we know that $\displaystyle \sum a_n$ converges by the comparison test.

Ratio Test:
$\displaystyle a_n =\frac{n^n}{(2n)!}$, and $\displaystyle a_{n+1} = \frac{(n+1)^{n+1}}{(2(n+1))!} = \frac{(n+1)^{n+1}}{(2n+1)(2n+2)(2n)!}$

$\displaystyle \lim_{n\to\infty} \left| \frac {a_{n+1}}{a_n}\right| = \lim_{n\to\infty} \left| \frac{(n+1)^{n+1}}{(2n+1)(2n+2)(2n)!} * \frac{(2n)!}{n^n}\right|$

$\displaystyle =\lim_{n\to\infty} \left| \frac{(n+1)^{n+1}}{(2n+1)(2n+2)n^n}\right|$

$\displaystyle =\lim_{n\to\infty} \left| \frac{(n+1)(n+1)^n}{(2n+1)(2n+2)n^n}\right|$

Here multiply numerator and denominator by 1 over n^n to get

$\displaystyle =\lim_{n\to\infty} \left| \frac{(n+1)(1+1/n)^n}{(2n+1)(2n+2)}\right|$

Now multiply the numerator and denominator by 1/n to get

$\displaystyle =\lim_{n\to\infty} \left| \frac{(1+1/n)(1+1/n)^n}{(2+1/n)(2n+2)}\right|$

It should be relatively clear that this converges to zero, (because the numerator converges to e, and the denominator diverges) but if you need me to explain that more, I can.

So because $\displaystyle \lim_{n\to\infty} \left| \frac {a_{n+1}}{a_n}\right| = 0 < 1$ the series $\displaystyle \sum_n^\infty \frac{n^n}{(2n)!}$ is absolutely convergent and therefore convergent.

For the second one also consider

That there $\displaystyle \exists{N}\backepsilon\forall{n>N},\frac{n}{(n)!}> \frac{n^n}{(2n)!}$

Which implies convergence by the comparison test
• Jun 1st 2008, 08:12 AM
Mathstud28
Quote:

Originally Posted by Rui
Use what test?

a) ∑(from n=1 to infinity) (arctan n/ n^1.2)
b) ∑(from n=1 to infinity) ( (n^n) / (2n)! )

Not sure how to do these...when i do it using ratio test, it turns out really weird..so yea i need help.

Thanks.

This is an old one, but for b) consider using the Root Test

$\displaystyle \lim_{n\to\infty}\bigg|\frac{n^n}{(2n)!}\bigg|^{\f rac{1}{n}}=\lim_{n\to\infty}\bigg|\frac{n^n}{\sqrt {4\pi{n}}(n^n)^2(e^{-n})^24^n}\bigg|^{\frac{1}{n}}=\lim_{n\to\infty}\bi gg|\frac{n}{(4\pi{n})^{\frac{1}{2n}}4n^2e^{-2}}\bigg|=0<1$

therefore convergent by root test