series and convergence

**nngktr** · May 18th 2010, 09:41 AM

This might seem super easy, but could anyone give me some hint how to show the following An is convergent?
1.n!/n^n
2.(1-1/n)^(n^2)
3. 1-cos (1/n)
I've tried the limit test/ratio test, but not working. Maybe comparison test?
Also, could anyone help calculating the series using taylor expansion of cos?
the sum of all (cos n)/(2^n) where n goes from 0 to infinity?
Any help is greatly appreciated!

**seams192** · May 18th 2010, 11:29 AM

Use the root test for these

**nngktr** · May 18th 2010, 12:08 PM

thanks, I've figured out how to do the convergence tests. Could anyone please give me a hint how to calculate the sum of cos n/2^n?

**Drexel28** · May 18th 2010, 08:47 PM

Originally Posted by nngktr

thanks, I've figured out how to do the convergence tests. Could anyone please give me a hint how to calculate the sum of cos n/2^n?

$\sum_{n=0}^{\infty}\frac{\cos(n)}{2^n}=\text{Re}\s um_{n=0}^{\infty}\frac{e^{in}}{2^n}$ and since $\left|\frac{e^i}{2}\right|<1$ we see that this is equal to $\text{Re}\frac{1}{1-\frac{e^i}{2}}=\text{Re}\frac{2}{2-e^i}=\cdots=\frac{2(\cos(1)-2)}{4\cos(1)-5}$

I've rigorously justified nothing. I leave that to you.

EDIT: P.S. that is not the Taylor expansion for $\cos(x)$ . Try the Maclaurin $\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{(2n)!}$

**chisigma** · May 19th 2010, 03:24 AM

Because is...

$\cos x = \sum_{k=0}^{\infty} (-1)^{k} \frac{x^{2k}}{(2k)!}$ (1)

... is also...

$1 - \cos \frac{1}{n} = \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{n^{2k} (2k)!}$ (2)

... so that is...

$\sum_{n=1}^{\infty} (1-\cos \frac{1}{n}) = \sum_{n=1}^{\infty} \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{n^{2k} (2k)!} = \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{(2k)!} \sum_{n=1}^{\infty} \frac{1}{n^{2k}} = \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{(2k)!} \zeta (2k)$ (3)

For $k\ge 2$ is $1 < \zeta (k) < 2$ so that the series in second term of (3) converges... and converges also very quickly!

...

Kind regards

$\chi$ $\sigma$

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