Infinite Series and l'hopital's rule

[markraz]

Curious..... if you have a series as below and I am doing the "divergence test"....

\(\displaystyle \Large\lim_{n \to \infty} (\frac{ln(n)}{\sqrt{n}})\)

When I plug in \(\displaystyle \infty\) I would get an indeterminate form. I realize I technically have to use use l'hopital's rule at this point.
But my question (strictly in the context of Infinite Series) is, even though the numerator and the denominator are approaching
infinity at different rates, can I assume they collectively will still go to infinity and thus diverge? I realize when calculating limits
of continuous functions you can't assume this, but with series can assumptions be made?

thanks

 

[Ishuda]

Curious..... if you have a series as below and I am doing the "divergence test"....

\(\displaystyle \Large\lim_{n \to \infty} (\frac{ln(n)}{\sqrt{n}})\)

When I plug in \(\displaystyle \infty\) I would get an indeterminate form. I realize I technically have to use use l'hopital's rule at this point.
But my question (strictly in the context of Infinite Series) is, even though the numerator and the denominator are approaching
infinity at different rates, can I assume they collectively will still go to infinity and thus diverge? I realize when calculating limits
of continuous functions you can't assume this, but with series can assumptions be made?

thanks

Assuming I'm understanding your question, you can't assume that for series either. For example look at the example
\(\displaystyle lim_{n\to\infty} \frac{n}{n^2}\)
That limit converges to zero but both the numerator and denominator go to infinity.

 

[markraz]

Assuming I'm understanding your question, you can't assume that for series either. For example look at the example
\(\displaystyle lim_{n\to\infty} \frac{n}{n^2}\)
That limit converges to zero but both the numerator and denominator go to infinity.

doesn't the series of \(\displaystyle \frac{n}{n^2}\) Diverge though? if |m-n| = 1 then it will diverge, correct? Also would this be "inconclusive"? with the divergence test?

 

[Deleted member 4993]

doesn't the series of \(\displaystyle \frac{n}{n^2}\) Diverge though? if |m-n| = 1 then it will diverge, correct? Also would this be "inconclusive"? with the divergence test?

The series S_n = Σ(1/n) diverges - but the sequence a_n = 1/n converges.

 

[Ishuda]

doesn't the series of \(\displaystyle \frac{n}{n^2}\) Diverge though? if |m-n| = 1 then it will diverge, correct? Also would this be "inconclusive"? with the divergence test?

Sorry,

As Subhotosh Khan pointed out - I should have said sequence not series. For a series make that \(\displaystyle \frac{n}{n^3}\)