Why Does the Divergence Test Yield Different Results for These Series?

[MillerGenuine]

Homework Statement

[tex]
\sum_{k=2}^\infty \frac{k^2}{k^2-1}
[/tex]


[tex]
\sum_{n=1}^\infty \frac{1+2^n}{3^n}
[/tex]

Homework Equations

I know that for the first problem i can apply the Divergence test by finding my limit as K goes to infinity. By doing this i get 1 which does not equal zero so i know it diverges.

Now my question is why can't i apply this same test to the 2nd problem? it seems as n approaches infinity i would get 1 as well. but that's not the case the correct solution for the 2nd problem is as follows..

[tex]
\sum_{n=1}^\infty \frac{1}{3^n} + \frac{2^n}{3^n}
[/tex]

Then by using a little algebra and sum of two convergent geometric series we get 5/2 to be the answer. which is convergent.

So why can't i find my limit as n approaches infinity in the 2nd problem? and why is 5/2 convergent? i thought if r>1 the series diverges?
Any help is appreciated.

 

[JThompson]

You can apply the Divergence Test to the 2nd problem, but it doesn't tell you anything because

[tex]\lim_{n\to\infty}\frac{1+2^{n}}{3^{n}}=0[/tex]

so the series may or may not diverge (obviously you already know that it converges, but in the future..).

 

[Mark44]

Homework Statement

[tex]
\sum_{k=2}^\infty \frac{k^2}{k^2-1}
[/tex]


[tex]
\sum_{n=1}^\infty \frac{1+2^n}{3^n}
[/tex]

Homework Equations

I know that for the first problem i can apply the Divergence test by finding my limit as K goes to infinity. By doing this i get 1 which does not equal zero so i know it diverges.

Now my question is why can't i apply this same test to the 2nd problem? it seems as n approaches infinity i would get 1 as well. but that's not the case the correct solution for the 2nd problem is as follows..

[tex]
\sum_{n=1}^\infty \frac{1}{3^n} + \frac{2^n}{3^n}
[/tex]

Then by using a little algebra and sum of two convergent geometric series we get 5/2 to be the answer. which is convergent.

The series is convergent because it is the sum of two convergent geometric series.

So why can't i find my limit as n approaches infinity in the 2nd problem? and why is 5/2 convergent? i thought if r>1 the series diverges?
Any help is appreciated.

It makes no sense to say that "5/2 is convergent." For the first series, r = 1/3, making it a convergent geometric series. For the second series, r = 2/3, making it also a convergent geometric series.

 

[MillerGenuine]

Awesome. Thank you both for the help. I missed out on a this lecture so i was trying to teach it to myself by using the text and was clearly having a hard time. Everything seems to make much better sense now