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Numerade Educator



Problem 46 Hard Difficulty

If $ \sum a_n $ and $ \sum b_n $ are both convergent series with positive terms, is it true that $ \sum a_n b_n $ is also convergent?


Yes, it is true.


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Video Transcript

if the's sums are both conversion and we have dealing with positive terms, so this means and and be enter both positive. Is it true that this some also convergence? So let's try to see if this is true here. So fence, there's some converges by the diversions test. This is from section eleven point two. So let me write that on the side eleven point two. We have that be and must approach zero in the limit. Since B ends eventually are getting very close to zero. There exist and inside your end. So let's say a natural number and such that if we let the index little and be larger than that, then the fiend must be less than or equal to, Let's say, one. This is just a consequence of the limit being zero sense it zero. That means there's a point at which B n plus one B and plus two and so on. All these numbers will be less than or equal to one. So all I'm saying here, therefore, because of this fact, I can rewrite this some let's say there are using a starting point here unless you say the starting point is one. It doesn't really matter what the starting point is. You could replace that one with any other number, and this will still be true. So first, let me instead of writing is infinite. Some break this into two parts. So look at the sun for all the way up to Capitol. And then I'll look at the remaining part now in this some over here, our little and is bigger than capital and therefore, for these values and this sum over here in the green, I can go ahead and use the fact that being is less than or equal to one due to this fact. So that means that our Siri's over here is less than or equal to the sum from one to capitol and a NBN. But then over here I'm just replacing being with one, and I have the inequality here. So this inequality is what made me write this inequality here. And I just have the sum from capital and plus one to infinity of a end. All I'm doing here is a NBN less than or equal to a M times one equals and and finally I know that this some here let me use a different color this red some over here, which I'm now using blue. This some converges because we were given that there's some over here convergence and the sum from N plus one to infinity of a end is less than or equal to, actually strictly less than the sum from one toe infinity of AM so sense we're given at this one emerges then by Red, a comparison test. This also converges. So let's say, by comparison with the entire series of the A M so sense it converges and rolling, adding positive numbers. That just means that we just showed that the sum from N plus one to infinity of a M is less than infinity. However, this original sum over here in the green, which I'm not circling this is automatically less than infinity because it's only a finite sum. Therefore, we're adding two numbers together that are less than infinity, So the result will also be lesson infinity. So therefore, we have shown that the sum of the A m. Bien first of all, we know it has to be larger than zero because a end and beyond are positive. On the other hand, We just showed that it's less than infinity, so it has to converge to a real number because again we're only dealing with positive terms, and that's our final answer.