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JH
Numerade Educator

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Problem 41 Hard Difficulty

(a) Suppose that $ \sum a_n $ and $ \sum b_n $ are series with positive terms and $ \sum b_n $ is divergent. Prove that if

$ \displaystyle \lim_{n \to \infty} \frac {a_n}{b_n} = \infty $

then $ \sum{a_n} $ is also divergent.

(b) Use part (a) to show that the series diverges.

(i) $ \displaystyle \sum_{n = 2}^{\infty} \frac {1}{\ln n } $

(ii) $ \displaystyle \sum_{n = 1}^{\infty} \frac {\ln n}{n} $

Answer

a. $\lim _{n \rightarrow \infty} \frac{a_{n}}{b_{n}}=\infty,$ there is an integer $N$ such that $\frac{a_{n}}{b_{n}}>1$ whenever $n>N$ Then $a_{n}>b_{n}$ whenever $n>N$ and since $\sum b_{n}$ is divergent, $\sum a_{n}$ is also divergent by the Comparison Test.
b. i.(a), $\sum_{n=2}^{\infty} \frac{1}{\ln n}$ is divergent.
ii $\sum_{n=1}^{\infty} a_{n}$ diverges by part (a).

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Video Transcript

suppose we have to Siri's with positive terms. This just means and and being a positive and were given that the second Siri's diverges. Now we'd like to use this fact over here at the limit is infinity to show that this Siri's also diverges. So here, let me say it this way. Since the limit is infinity, there exists an end. This is just by definition of the limit, such that if we take a little end to be bigger than big, then me write it this way. Zen, we have a N over BN is bigger than one. Eventually, it has to be bigger than one since it goes to infinity. And this implies a M is bigger than the end, if little and it's bigger than big. So here in part A. We can rewrite this Siri's. It's not giving us a starting point. It doesn't matter what the starting value for any solicitous stay from one to infinity. Now I can rewrite this by splitting it up into two sons. So some from one to capital in and then from n plus one to infinity. And now we use the fact that in the second some that little and is bigger than Biggins. So we have this condition. Therefore, we know that a N is bigger than d n a. So here I'll just leave the first some as it is. So that was from the previous part and that here we have. So I did was I kept this some of the same and sense an is bigger than be in. I just replaced ends would be ends. But because of the inequality here, that's why I have inequality. And now we use the fact that this infinite some diverges. This is just given info that was given in the very beginning. The first sentence, therefore, by the comparison test our Siri's and also diverges. So that's the verification for part A. Now let's go ahead and use part A on the next page to show that these two thumbs both diverge. So let's go on to the next page here. Part B. And then we had Parwan. If party, we're looking at the sum from two to infinity of one over the natural law argument. Now let's call this R A N. That's one over natural log event. Let's take the end to be won over end. We know that bien diverges. This is just a harmonic series. Now what's also before we use part? A. We have tio get a limit of infinity when we look at and over Bien. So we have limit and goes to infinity of end over natural log of n, and this will be infinity. And if you want hear, you could use Low Patel's rule If it's unclear why we're getting infinity so we can go ahead and use part, eh? The sum of the a end, which in our case is just the sum from two to infinity of one over natural. Other men also diverges. Now let's go ahead and use part B. Our let's go ahead and use party to do Part two for this part of the problem here. This time we're looking at a different some from n equals one to infinity natural lot of end over end. So that'LL be my and this time in my being and I'll still keep this as one over end. So as before, we know this being will diverge because it's harmonic. Now we go ahead and take the limit and over Bian. So here's my a n and then I dividing by bn is the same thing is multiplying by an over one. Those ends cancel. And as we take the limit of this log that goes to infinity so we can use part, eh? So my part, eh? This some of the a end this sum over here that in part two also beverages and that's our final answer.