# Series Convergence/Divergence

1. Sep 28, 2009

### rbpl

I have two questions one is just about part of the problem and the other one I want to make sure I am going in the right direction.

The directions are "For each of the following series, determine whether it converges, converges absolutely, or diverges"

1) $$\sum$$(n^n)/(n!n!)
here I can use ratio test
[(n+1)^(n+1)/[(n+1)!(n+1)!]]/[n^n/(n!n!)]
[[(n+1)^(n+1)]*(n!n!)]/[n^n*[(n+1)!(n+1)!]]
[(n+1)(n+1)^n(n!)(n!)]/[(n+1)n!(n+1)n!(n^n)]
(n+1)^n/[(n+1)n^n]

and here is my problem I know that we can write (n+1)! as (n+1)n! and (n+1)^n as n but then what is (n+1) and n^n.

2) $$\sum$$(n^2+1)^(1/2)-1
in order to figure out if it converges/diverges could I use the comparison test where:
[(n^2+1)^(1/2)-1]/(n^(2/2)
[(1+(1/n))^(1/2)-1] so 1^(1/2) is 1 hence in converges
If this is not the way to do it am I at least using the correct test?

2. Sep 29, 2009

### lanedance

for the first one, cancel another (n+1), then you have

$$\frac{a_{n+1}}{a_n} = \frac{(n+1)^{n-1}}{n^n}$$
now have a think about the terms you would get when you expand the top... consider a binomial expansion if you need it explicitly

for the 2nd, not sure if i can read the term correctly, is it:
$$\sum_n (n^2+1)^{1/2}-1$$
if so, it looks pretty divergent...

Last edited: Sep 29, 2009
3. Sep 29, 2009

### Bohrok

For 1), you got rid of the factorials, so now you can use the natural log. Using the limit L in the ratio test as n→∞, now try finding

$$\lim_{n\rightarrow\infty} e^{\ln(L)}$$

Find the limit of ln(L) first and then e to that limit.

4. Sep 29, 2009

### rbpl

For the first one, we have not used binomial expansion in order to find limits of series, so I don't think that the teacher expects it. However, I wanted to make some thing clear:
if (n+1)^n/[(n+1)n^n] goes to [(n+1)^(n-1)]/(n^n) it seems that it jumps around a little but it is approaching 0. But that is just trying a few numbers, is there any other way to see it without using the bionmial expansion?

For the second my question is if I am using the right test, I'm guessing the point of this exercise was to see if we can apply the tests in the right way. If I did everything correctly then could you clarify how using ln(L) would be helpful?

Thank you guys for such a quick response.

5. Sep 29, 2009

### Bohrok

Yes; try what I suggested. Did you understand what I wrote?

6. Sep 29, 2009

### rbpl

I understand it...to some extent. I am not sure why would I should use natural log, am I forgetting something from calculus classes?

Also am I right about the (n^2+1)^(1/2)-1

7. Sep 29, 2009

### Bohrok

It's pretty hard to deal with this limit as it is
$$\lim_{n\rightarrow\infty} \frac{(n+1)^{n-1}}{n^n}$$

Rewriting it like this makes it easier to work with since ln lets you break up those exponents
$$e^{\ln\left(\lim_{n\rightarrow\infty} \frac{(n+1)^{n-1}}{n^n}\right)}$$

For 2), did lanedance write it correctly? We're not too sure how it's written.

Last edited: Sep 29, 2009
8. Sep 29, 2009

### rbpl

Yes lanedance got it right, right now I'm working on a problem for my abstract algebra class, after that I will try to use your method for the first problem.

I really appreciate you helping me.

9. Sep 29, 2009

### rbpl

Ok, so if $$\lim_{n\rightarrow\infty} \frac{(n+1)^{n-1}}{n^n}$$ then $$\lim_{n\rightarrow\infty} {(n+1)^{n-1}}{(1/(n^n))}$$=$${(n+1)^{n-1}}{((n^{-n}))}$$ then $${ln(n+1)^{n-1}}{(ln(n^{-n}))}$$=$${[(n-1)ln(n+1)]}{[(-n)ln(n)]}$$=$$\frac{ln(n+1)}{({n-1})^{-1}}}*\frac{ln(n)}{({-n})^{-1}}}$$=$$\frac{(n+1)^-1}{-({n-1})^{-2}}}*\frac{(n)^-1}{({n})^{-2}}}$$=$${{-(n+1)^{-1}}{(n-1)^2}{n}}$$ should I keep going and evaluate this, that way I will end up with $$\frac{{3n^2}-{2n^2}+{1}}{n+1}$$ or can I get the limit without the evaluation, and did I apply the l'Hopitals rule correctly, it has been a while, I tried to look at my old Calculus book and that is what I came up with.

10. Sep 29, 2009

### lanedance

the idea behind the binomial expansion was to show

$$\frac{(n+1)^{n-1}}{n^n} = \frac{an^{n-1}+bn^{n-2}+...+c}{n^n}$$
for contants a,b,...
you should now be able to seprate each numerator and apply the limit no worries

though you should be also able to do it as bohrok implies

for 2) an clearly does not tend to zero which should be enough

11. Sep 29, 2009

### lanedance

also on your last post, i think ln(a/b) = lna - lnb

12. Sep 29, 2009

### rbpl

Yes, you are right I can't believe I forgot about that, in that case what I get is 3. Since: ln(((n+1)^(n-1))/(n^n))
=ln(n+1)^(n-1)-ln(n^n)
=((n-1)ln(n+1))-((n)ln(n))
=[ln(n+1)]/(n-1)^-1-[ln(n)/(n^-1)]
=(n+1)^-1/(-(n-1)^-2)-(n^-1)/(-(n)^-2)
=(-(n+1)^-1)((n-1)^2)+n
=(-(n-1^2)/(n+1))+n
=(-n^2+2n+1+n^2+n)/(n+1)
=(3n+1)/(n+1)
and so it approaches 3??
then put it in a form lim e^ln(((n+1)^(n-1))/(n^n))=e^3??

for second I guess I was over analyzing it

Last edited: Sep 29, 2009
13. Sep 29, 2009

### Bohrok

What happened to the ln's between the bold lines?

The limit is not e3

14. Sep 29, 2009

### rbpl

I was trying to follow an example from my old calculus book the example was: lim x^x goes to lim(x^x)=lim(xlnx)=lim(lnx/x^-1)=lim(x^-1/-x^-2)=lim(-x)=0
So I tried to apply it there

15. Sep 29, 2009

### Bohrok

My way is tricky, but it can be done.

$$\ln\left(\frac{(n+1)^{n-1}}{n^n}\right) = \ln(n + 1)^{n-1} - \ln(n^n) = (n - 1)\ln(n + 1) - n\ln n$$ This was the last thing you did in the second line.
$$= n\ln(n + 1) - \ln(n + 1) - n\ln n = n(\ln(n + 1) - \ln n) - \ln(n + 1)$$

$$= n\ln\left(\frac{n + 1}{n}\right) - \ln(n + 1) = \ln\left(1 + \frac{1}{n}\right)^n - \ln(n + 1)$$

Now,
$$\lim_{n\rightarrow\infty}\ln\left(\frac{(n+1)^{n-1}}{n^n}\right) = \lim_{n\rightarrow\infty}\ln\left(1 + \frac{1}{n}\right)^n - \lim_{n\rightarrow\infty}\ln(n + 1)$$

Can you finish it? Do you recognize any special limits in the line above?

Edit: For the second series, try pulling out an x from the square root. Can you show if the terms in the sequence even go to 0?

Last edited: Sep 29, 2009
16. Sep 29, 2009

### lanedance

my computer doesn't display tex well, but that does look tricky

i think expansion or a repeated appliacation of L'Hopital's rule would be easier...

17. Sep 29, 2009

### Bohrok

Now that I look more at your binomial expansion (usually I can't really follow along with binomial expansions on this forum), I like how nicely you can take the limit of each term which go to 0. Yet I like the (1+1/n)n as n→∞ in mine. :)