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JH

# Find the values of $p$ for which the series is convergent.$\displaystyle \sum_{n = 1}^{\infty} n(1 + n^2)^P$

## $P<-1$

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##### Catherine R.

Missouri State University

##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

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### Video Transcript

let's use the Inderal test and we'LL see why this works out nicely in a second. So just replace and with the ex and first thing is a cz usual. You could just write this as a improper, integral and then if it helps, you can also do a use up here. Now the steppe is not mandatory, but it'LL require less work. So I'm just going to convert this entire integral into the new variable you So notice here, this was X equals one. So plugging that in for X over here, you see that you have one equals two. That's our new lower limit. And similarly, if you plug in the upper bomb, be in for X, you get our upper limit b squared plus one and there's the rest of our internal rule. So now we should consider some cases here, case one p equals negative one. And in that case, the integral becomes ln absolute value over too. But when you plug in the B squared plus one into the log and take the limit, this will go to infinity. So diverges so there. It means that we do not want to consider p equals minus one. Let's go to the next page here. So now we consider P not equal to minus one. And when we do that, we can go ahead and use the power rule to evaluate the integral. This is why we're considering two cases because the anti derivative is different, depending whether or not pee is equal to negative one. So now we're plugging in, and so at this point, we need to consider cause notice that one plus b squared goes to infinity. So we should consider whether or not this exponents is positive or negative. If P is less than negative one, then P plus one is less than zero. And this will ensure that now we have a negative exponents. This will ensure that the infinity goes in the denominator and that will end up being a zero. So if P is less the negative one, the integral just becomes this term here, which is a real number. You're not dividing by zero so converges in this case, and then consider the other case he's bigger than negative one. This implies P plus whoops. Sorry about that. This implies P plus one is bigger than zero. But that only mean that the infinity has a positive exponents and that will ensure that the entire in a girl is infinity. So, out of all the cases, the only one that we had conversions was a P was strictly less than negative one.

JH

#### Topics

Sequences

Series

##### Catherine R.

Missouri State University

##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

Lectures

Join Bootcamp