Ratio and Root Test

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March 23rd 2009, 04:57 PM
TAG16

Ratio and Root Test

How would you work out if these series converge or diverge?

Problem 1

n=1 to infinity n!/10^n

Problem 2

n=1 to infinity ((n-2)/n)^n

Problem 3

n=1 to infinity (-2)^n/3^n
March 23rd 2009, 05:11 PM
o_O

What exactly are you having troubles with? You are already told which tests to use.

For example, the first one. Let $a_n = \frac{n!}{10^n}$ .

Then: $\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n}\right| = \lim_{n \to \infty} \left| \frac{(n+1)!}{10^{n+1}} \cdot \frac{10^n}{n!}\right| = \lim_{n \to \infty} \left|\frac{(n+1)n!}{10^n \cdot 10} \cdot \frac{10^n}{n!}\right| = \cdots$
March 24th 2009, 03:35 PM
TAG16

still not sure how to do the third problem though....can't use root test because of the negative term. There too many tests....
March 24th 2009, 05:04 PM
skeeter

Quote:

Originally Posted by TAG16

still not sure how to do the third problem though....can't use root test because of the negative term. There too many tests....

Quote:

Problem 3

n=1 to infinity (-2)^n/3^n

$\sum_{n=1}^{\infty} \frac{(-2)^n}{3^n} = \sum_{n=1}^{\infty} \left(\frac{-2}{3}\right)^n$

geometric series with $|r| < 1$ ???

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