Problem with limit/direct comparison for series

[Edder]

\(\displaystyle \sum_{n=2}^{\infty} \frac{2}{7n\ln(n)+4n}\)

Can someone just help push me in the right direction? Would I use the limit or direct comparison test? If I used the limit comparison test, \(\displaystyle \frac{ln(n)}{n}\) would be reasonable I think. Which would then make the series divergent??

Appreciate any feedback and help.

 

[pka]

\(\displaystyle \sum_{n=2}^{\infty} \frac{2}{7n\ln(n)+4n}\)

Can someone just help push me in the right direction? Would I use the limit or direct comparison test? If I used the limit comparison test, \(\displaystyle \frac{ln(n)}{n}\) would be reasonable I think. Which would then make the series divergent??

I would let\(\displaystyle a_n=\dfrac{2}{7n\ln(n)+4n}\;\;\)\(\displaystyle \&~b_n=\dfrac{1}{n}~.\)

 

[Edder]

I would let\(\displaystyle a_n=\dfrac{2}{7n\ln(n)+4n}\;\;\)\(\displaystyle \&~b_n=\dfrac{1}{n}~.\)

Thanks for the tip. However, I am curious as to why you picked \(\displaystyle \frac{1}{n}\).
Is it because the 4n is the more dominant power compared to 7n ln(n)?

 

[pka]

Thanks for the tip. However, I am curious as to why you picked \(\displaystyle \frac{1}{n}\).
Is it because the 4n is the more dominant power compared to 7n ln(n)?

Because n is a factor of the denominator.

 

[Edder]

I see, thanks.