# Thread: Another friggin problem with series...

1. ## Another problem with series...

This series section is really bugging me. Am I allowed to use the word friggin? I can't get an answer that makes sense for n!/100^n with the Riemann symbol in front. I try to use the Ratio Test and then realize that the terms are positive so I don't need an absolute value.

I do some algebraic manipulation and come up with ((n+1)/(101)^n)) I don't see how I can take a limit/simplify the denominator here. Do I divide everything by ((101)^n))? Or did I do something wrong along the way and I should be getting a different result where I can see if it is <1 (converges) >1 (diverges) or = 1 (inconclusive).

2. $\displaystyle \sum \frac{n!}{100^n}$

Ratio test takes the infinite limit of: $\displaystyle \frac{\frac{(n+1)!}{100^{n+1}}}{\frac{n!}{100^n}}$ which simplifies to: $\displaystyle \frac{(n+1)}{100}$. The series diverges because the infinite limit is greater than one.

3. As an aside, and this doesn't necessarily constitute another problem (hence no new thread). When I'm given a series n/(ln n)^n, can I use the rules of ln in the denominator and get n/((n)(ln n)) so the n's cancel out and the rest is easy?

I don't see how my method would be breaking any rules in math. I just want to check if I'm not mistaken.

4. Originally Posted by Undefdisfigure
As an aside, and this doesn't necessarily constitute another problem (hence no new thread). When I'm given a series n/(ln n)^n, can I use the rules of ln in the denominator and get n/((n)(ln n)) so the n's cancel out and the rest is easy?

I don't see how my method would be breaking any rules in math. I just want to check if I'm not mistaken.
The n's would not cancel out as you would have $\displaystyle \frac{(n+1)}{n*ln|n|}$

5. I haven't gotten to logarithms yet as my teacher requested that I review algebra. I don't see how you got the n+1 in the numerator. There needs to be a 1 added to the numerator when the n comes down from being an exponent to multiply the denominator?

6. And ironically, by using your method colby I get the same answer as with my method. After an extra step is needed in your method colby, I get that
lim n--> infinity 1/(ln n) goes to zero and so the series converges.

7. Originally Posted by Undefdisfigure
I haven't gotten to logarithms yet as my teacher requested that I review algebra. I don't see how you got the n+1 in the numerator. There needs to be a 1 added to the numerator when the n comes down from being an exponent to multiply the denominator?
You are given $\displaystyle a_n=\frac{n}{(ln n)^n}$

Using the ratio test, you need to find $\displaystyle \frac{a_{n+1}}{a_n}$. This is equal to: $\displaystyle \frac{\frac{n+1}{(ln(n+1))^{n+1}}}{\frac{n}{(ln n)^n}} = \frac{(n+1)(ln|n|)^n}{(n)(ln|n+1|)^{n+1}}$. Note that the natural logs will not cancel each other out, but we can do the following..

$\displaystyle \frac{(n+1)(ln|n|)}{(n)(ln|n+1|)}\left(\frac{ln|n| }{ln|n+1|)}\right)^n$

The limit of that obviously goes to zero because the "bottom" grows faster than the "top".

8. Originally Posted by Undefdisfigure
This series section is really bugging me. Am I allowed to use the word friggin? I can't get an answer that makes sense for n!/100^n with the Riemann symbol in front. I try to use the Ratio Test and then realize that the terms are positive so I don't need an absolute value.

I do some algebraic manipulation and come up with ((n+1)/(101)^n)) I don't see how I can take a limit/simplify the denominator here. Do I divide everything by ((101)^n))? Or did I do something wrong along the way and I should be getting a different result where I can see if it is <1 (converges) >1 (diverges) or = 1 (inconclusive).
Stirling's formula tells us that (it does tell us what $\displaystyle k$ is but that is unimportant here):

$\displaystyle n! \sim k \frac{n^{n+1/2}}{e^n}$,

so:

$\displaystyle \frac{n!}{100^n} \sim k \frac{n^{n+1/2}}{e^{n(1+\ln(100))}}$,

and eventually $\displaystyle n^n$ dominates $\displaystyle e^{\alpha n}$ for any $\displaystyle \alpha$ so the terms diverge, and so the series itself diverges.

(that is for large enough $\displaystyle n: n>e^{\alpha}$ and so the terms are bounded below by a multiple of a geometric series with common ratio greater than 1)

RonL