Limit/Direct Comparison for Series Question

[Mad Season]

Homework Statement

$$\sum_{n=1}^{\infty} \frac{7n}{6n^2 ln(n)+2}$$

Determine whether the series converges or diverges.

Homework Equations

Denominator is growing faster, so the limit as n --> to infinity should equal zerio

The Attempt at a Solution

I tried isolating the highest power of the both the numerator and denominator. Which is:
$$\frac{7n}{6n^2 ln(n)}$$ = $$\frac{7}{6n ln(n)}$$

What would I do next? Would I compare the simplified bn to an for a limit comparison test?

I also tried a direct comparison through: $$\frac{1}{6n^2+2}$$
But I can't tell if that would work. Would the an be less than bn?Any feedback and help appreciated.

 

[jbunniii]

Does

$$\sum \frac{1}{n \ln n}$$

converge or diverge?

 

[206PiruBlood]

This might not be the only way but its pretty cool. Since the series is positive and decreasing you can test the series $$\sum 2^n a_{2^n}$$ where $$a_{n}$$ is your sequence after dividing the first factor of n. Your series converges if and only if the above series converges.

http://en.wikipedia.org/wiki/Cauchy_condensation_test

 

[206PiruBlood]

Does

$$\sum \frac{1}{n \ln n}$$

converge or diverge?

This can be easily calculated with the condensation test :D

 

[Mad Season]

Does

$$\sum \frac{1}{n \ln n}$$

converge or diverge?

I just did the integral test for this series, and it diverges. So basically, I can uses the limit comparison test with the bn:

$$\frac{7}{6n ln(n)}$$

So ultimately, the series diverges then. Right?

 

[206PiruBlood]

Yes it diverges.

 

[Mad Season]

Thanks for the help, appreciate it.