Series Convergence/Divergence and Comparison Test

[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?

Thank you in advance.

 

[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...

 

[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.

 

[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.

 

[Bohrok]

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?

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

 

[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

 

[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 let's 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.

 

[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.

 

[rbpl]

Ok, so if <br /> \lim_{n\rightarrow\infty} \frac{(n+1)^{n-1}}{n^n} then <br /> \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.

 

[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) a_n clearly does not tend to zero which should be enough

 

[lanedance]

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

 

[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

 

[Bohrok]

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

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

The limit is not e³

 

[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

 

[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?

 

[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...

 

[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. :)