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Use the Integral Test to determine whether the series is convergent or divergent.

$ \displaystyle \sum_{n = 1}^{\infty} \frac {n}{n^2 + 1} $

divergent

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Campbell University

Oregon State University

Baylor University

Boston College

Hello. So here we have these somewhere we have and going from one to infinity of and over and squared plus one. So to use the integral test when we look at the function F of X is equal to X over X squared plus one. And we have to have the F of X is going to be continuous, positive and decreasing. Um on the interval from one to infinity. Well um f of X is certainly continuous and positive um on one to infinity. So then if we look at to determine if um we are decreasing, we look at the derivative. So look at fx again and take the derivative F prime of X. So by the quotient rule we have the F prime of X is equal to one minus X squared divided by X squared plus one quantity squared. And we see here that well, for all x in one to infinity. Um Yes, F prime of X is going to be less than zero because one minus X squared is less than zero for all X. Um in the interval from one to infinity. So therefore the function F of X. The function F of X which is going to be equal to well X over X squared plus one um is decreasing on the interval from one to infinity. So therefore by the integral test we then look at the integral from one to infinity of F of X dx. So that's the integral here from one to infinity of X over X squared plus one dx. Which we can then right as well, one half times the integral from one to infinity of two. X over X squared plus one D X. Um And then well this is gonna then going to be equal to is ready like this. This is then going to be equal to well one half um times the limit as um let's say T goes to infinity of well the natural log of the absolute value of t squared plus one evaluated from one to T. Uh And here we are. This is going to be equal to well infinity. So therefore we have that the integral from one to infinity from one to infinity of F of X. D X um is divergent. And therefore by the integral test we can say that the somewhere end goes from one to infinity of N over N squared plus one must also be divergent. All right, take care.

University of Wisconsin - Milwaukee