Does the sequence converge or diverge?

[monomoco]

Does the sequence converge or diverge?

An= ln(n)/(n+1)^(1/2)

I tried using L'Hospital's rule, and got lim n ->infinity (1/x) / ( 1 / 2(n+1)^(1/2). Is this OK? And does this mean the sequence diverges because this limit is infinite?

 

[mmm4444bot]

... Is this OK?

Yes, your first application of L'Hospital's Rule is good.

-

And does this mean the sequence diverges because this limit is infinite?

No, it does not because the limit in your result does not equal infinity ... it's still infinity divided by infinity, just like the original limit! In other words, a single application of L'Hospital's Rule did not result in a limit that can be evaluated.

Apply L'Hospital's Rule a second time to your result, and evaluate the new limit to see whether or not it's still indeterminant.

Cheers,

~ Mark

 

[monomoco]

OK, so using the rule again, I get lim n-> inf of 1/(n+1)^(1/2). This evaluates to zero, so I can say A_n -> 0 also, right?

 

[mmm4444bot]

... I get lim n-> inf of 1/(n+1)^(1/2). This evaluates to zero, so I can say A_n -> 0 also, right?

Yup.

I would amend the statement to read A_n --> 0 as n --> infinity.

Good work.

~ Mark

 

[monomoco]

Thanks for your help!

 

[galactus]

Try the integral test. Then, the integral diverges and, therefore, so does the series.

 

[Deleted member 4993]

Try the integral test. Then, the integral diverges and, therefore, so does the series.

True... just to avoid confusion - the question was about the convergence of the sequence.

 

[galactus]

Oh, I'm sorry.Better pay closer attention.