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Check to see if the series is a geometric series Or p-series: If yes, then use the conclusion given for these series Example. Determine whether each of the followin...

Question

Check to see if the series is a geometric series Or p-series: If yes, then use the conclusion given for these series Example. Determine whether each of the following series diverges O1 converges. k-[2 0.38)*-1 Ex 22 6+2+3 9 27 Find lim an of thc scrics Ok: lim On + 0,, thcn by thc Test for Divcrgcncc, thc scrics K=T T =0 divcrgcs: Example: Determine whether each of the following series diverges or converges 9k3 + 5k2 2 k5/2 + 4 (b) 2(1+2) k2 + 5k 23+56 k= [ 2x(3)If the seriesak, has only positiv

Check to see if the series is a geometric series Or p-series: If yes, then use the conclusion given for these series Example. Determine whether each of the following series diverges O1 converges. k-[ 2 0.38)*-1 Ex 22 6+2+3 9 27 Find lim an of thc scrics Ok: lim On + 0,, thcn by thc Test for Divcrgcncc, thc scrics K=T T =0 divcrgcs: Example: Determine whether each of the following series diverges or converges 9k3 + 5k2 2 k5/2 + 4 (b) 2(1+2) k2 + 5k 23+56 k= [ 2x(3) If the series ak, has only positive terms and meets the conditions of the Integral Test, find the related function f, use the iutegral test if Jo f(r) dx is easy to find. Example: Dctcrminc whcthcr cach of thc following scrics divcrgcs or convcrgcs. 2, kVnk 2 ke (c) 2 2h



Answers

Determine whether each series converges (absolutely or conditionally) or diverges. Use any applicable test. $\sum_{k=1}^{\infty} \frac{3 k+2}{k^{3}+1}$

Hello. So given our series we take um going to use the ratio test and take the absolute value of a seven plus one over a sub N. So that gives us um M plus one factorial over to M plus two factorial. And then multiplying here by the reciprocal as we get to N factorial over N factorial. Which is then going to be equal to what we have an N plus one times N factorial. We have an N plus one times N factorial divided by two N plus two times two. N plus one times to end factorial. And then multiplying by two N. Factorial divided by N. Factorial. So this is just going to give us just a N plus one over or to M plus two times two. N plus one is equal to two times N plus one times two N plus one. So the M plus one factor cancels out here. Um And we're just this is going to be equal to one divided by two times two. N plus one. Okay, then the limit we take, they take them the limit as N goes to infinity of this here, which is then going to give us the limit as N tends to infinity of 1/4 N times one plus one over to N and his N goes to infinity. Um This is this going to go to zero. So this the limit here is 00 is less than one. So therefore by the ratio test are given series is going to be convergent and it's going to be absolutely convergent.

Series converges or diverges Go ahead and apply the diversions test. Take the limit as N approaches infinity of N plus one over, and mine is, too. That's equal to one. Just doesn't equal zero. Therefore, they're serious diverges by the diversions test.

Hello. So from our given series we get that a sub K is equal to K factorial divided by K plus two factorial. So therefore a sub K plus one is going to be equal to Well, K plus one factorial divided by well K plus one plus two is K plus three factorial. And then we do a sub K plus one divided by a sub K. So we get that a sub K plus one over a sub K. Well that is going to be equal to K plus one factorial over K plus two factorial times K plus two factorial over K. Factorial is going to just give us K plus one over K plus three. We then um this is then going to be equal to, we can write this as one plus one over K and then divided by one plus three over K is dividing everything through by K and K approaches infinity. This here is going to approach just 1, 1/1. So therefore um the limit as K approaches infinity of a So keep this one over. It's okay. The limit here is going to be equal to one as K approaches infinity. So therefore by using the root test here we get that are given series is going to be convergent


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