Ratio Test: sum{n=0,infty}((3^n)/(n+1)^n), sum{n=1,infty}(n/

[Oneiromancy]

I don't know how to find the limit on #27.

 

[Oneiromancy]

Re: Ratio Test

On 45 and 49 I encountered the limit n^(1/n) as n goes to infinite. I assumed it went to 1 but I just wanted to check on that so that'd I'd have an exact answer.

 

[tkhunny]

Re: Ratio Test

On 45 and 49 I encountered the limit n^(1/n) as n goes to infinite. I assumed it went to 1 but I just wanted to check on that so that'd I'd have an exact answer.

What does it's logarithm do? Investigate ln(n)/n.

 

[stapel]

For those who can't "see" the images, the poster's questions are as follows:

\(\displaystyle \mbox{27) }\, \, \sum_{n=0}^{\infty}\, \frac{3^n}{(n\, +\, 1)^n}\)

\(\displaystyle \mbox{45) }\, \, \sum_{n=1}^{\infty}\, \frac{n}{4^n}\)

\(\displaystyle \mbox{49) }\, \, \sum_{n=2}^{\infty}\, \frac{n}{\left(\ln{(n)}\right)^n}\)

 

[tkhunny]

Re: Ratio Test

I don't know how to find the limit on #27.

\(\displaystyle \frac{3^{n+1}}{(n+2)^{n+1}}*\frac{(n+1)^{n}}{3^{n}}\;=\;\frac{3}{n+2}*\frac{3^{n}}{(n+2)^{n}}*\frac{(n+1)^{n}}{3^{n}}\;=\;\frac{3}{n+2}*\left(\frac{n+1}{n+2}\right)^{n}\)

And we all know \(\displaystyle \left(\frac{n+1}{n+2}\right)^{n}\) heads for one (1), right? Wrong! It approaches 1/e.

Are we getting anywhere?

 

[skeeter]

Re: Ratio Test

for #27, rather than the ratio test, try the nth root test ... much easier.

 

[tkhunny]

Re: Ratio Test

for #27, rather than the ratio test, try the nth root test ... much easier.

...except that may not fly very well on a Ratio Test examination.

 

[skeeter]

Re: Ratio Test

...except that may not fly very well on a Ratio Test examination.

A big part of learning about series convergence/divergence tests is being able to choose the one or possibly two that will reveal convergence/divergence in an efficient manner ... I've yet to see an exam that only covers only a single test.

 

[Oneiromancy]

Re: Ratio Test

I got the help I needed thanks.

Also I'm a homeschooler so no tests to worry about.