Problem 38 Hard Difficulty

Find the sum of the series $ \sum_{n = 1}^{\infty} ne^{-2n} $ correct to four decimal places.

Answer

Watch More Solved Questions in Chapter 11

Video Transcript

in this problem were to find this sum this infinite Siri's correct to four decimal places. So here we see that our end term a N is given by N E to the minus to end. So that implies we should define the continuous version f to be X e to the minus two x where X is any real number at least one. So now the reason for noting this is that we can talk about the ear if we want to be correct to four decimal places, that puts a restriction on the air. So this is given by our end. If you use in terms and your partial sung, this is given by the sum minus the and partial some and using the inner qualities in this section In the textbook, this is less than the integral from end to infinity f of X, the X, which in our case we can just plug in that f there and technically this is a improper integral So we should go ahead. And instead of writing infinity here, let's just go ahead and write. This's a limit. Lemonis Kay goes to infinity. So this is the inner improper integral we'LL have to deal with, but in a moment. But in the meantime, let's go ahead and know what we want. We want his hair to be less than zero point zero zero zero zero five Dude surrounding if the air is less than this number datil insurers that our approximation agrees with the actual value s up to four decimal places. So what we should be solving? Is this here for end? Or we just at least want some end value that makes this true. So in this case, this integral ignoring the limit right now, we would have to evaluate this integral here. And to do this you might recognise integration by parts. That would be the way to go here. And since that's not the focus of this chapter, I'll just leave it to you to carry out the integration by parts here. But the left hand side, after taking the limit, becomes too, and plus one over four e to the to end. And we want that to be less than zero point zero zero zero zero and then five. And now it's just a matter of finding some and value that works. I try and equals five, and that just does not lead to a true inequality here. However, when I plug in and equal six in this case, the inequality becomes so two times six twelve plus one that's thirteen over for each of the twelve and the calculator, this is about point zero zero zero zero two, which is definitely less than this number here. So in that case, the inequality is true. Oh, and that means that if we want to have a son that's correct to four decimal places, we'LL just go ahead and take any pool six. This will imply that the remainder is less than the desire quantity. Well, you showed this on the previous page, so now we could actually just go ahead and write this out. This is the exact value us. This is approximately equal to a partial son. So one could write all these terms out. I'll do it. I'LL write out a few here and go into the calculator. That's about points one a one zero. So that will be the first four digits of this entire sum over here. And that resolves the problem