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# (a) Suppose that $\sum a_n$ and $\sum b_n$ are series with positive terms and $\sum b_n$ is convergent. Prove that if$\displaystyle \lim_{n \to \infty} \frac {a_n}{b_n} = 0$then $\sum a_n$ is also convergent. (b) Use part (a) to show that the series converges.(i) $\displaystyle \sum_{n = 1}^{\infty} \frac {\ln n }{n^3}$(ii) $\displaystyle \sum_{n = 1}^{\infty} \frac { \ln n}{\sqrt n e^n}$

## a. Click to seeb. SEE SOLUTION

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suppose we have to sums here, but positive terms so and and be enter strictly a larger than zero and there's some converges now for part A. Let's suppose that this limit zero and then use this to show that this sum and also convergence. So since this limit is zero, eventually a Zen gets really, really big. This fraction is very close to zero. So this house is there exist and natural number end such that if little and his data and begin then and over bien is less than one. So if you're close to zero, then you're eventually less than one. That's all I'm saying here. So this also implies that a M is less than the end, so as long as little and it's bigger than big. So this tells me that for part A, I could take the sum and and then re write it, as is some upto capital in. And then from n plus one all the way to infinity. Now, for the second some we have little and is bigger than begin. So therefore we can replace a end with bien. So here I am replacing and with being and then I'm using an inequality due to this fact over here and now I could say that this some converges due to this fact up here. Oh, so that's given information. However, this first song, this's only a finite sum. So in automatically commercialise so converges. And if we add two converging sums together, the result the entire right hand side converges. Therefore, the left hand side. The smaller side also converges by the comparison test, so that takes care of the party, eh? Now let's go ahead and use part A to show that these two Siri's in part B converged. So I'm running on a room here. Let me go on to the next page for part one. We had to some natural log over and cute. So it's treatises are hoops as are and two. Then let's steak being to be won over and squared. We know this some commercials here. The reason you can use the pee test and then you have P equals two, which is bigger than one that applies convergence. Now let's look at the limit of a N over bian and hopefully this is zero so that we can use part, eh? So we have Ellen of end over end cubed times and square over one. And then we have natural log of end over end, and that's equals zero. You can use locals house rule here if you need to, so we can use part, eh? This Siri's also Kam urges, So that takes care of the first theories in part and the second part of this problem, Part B. Now let's go to the remaining Siri's and party, the last part of this problem, this one natural log of end over the square root of n times either. Then, once again, let's call this our and and then let's take B In this time, let's think that to be won over either then or if you want one over E to the end, then there's some convergence. And why is that converges here? It's geometric with our equals one over e, and this are satisfies the inequality, absolute value of our less than one so converges. Now let's look at the living to see if we can use part of this problem. The limit should be zero, and then we're dividing by beating the end. So that's the same thing. It's multiplying by either the end over one and then we have natural Aga n over the squared of end and you can use low. Patel's rule here is well, and this limit will be zero so we can use part. Eh? The sum of the A n which in this part of the problem is natural on Damen over this food of end. Either the end converges and that's our final answer.

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