💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here!



Numerade Educator



Problem 39 Hard Difficulty

Estimate $ \sum_{n = 1}^{\infty} (2n + 1)^{-6} $ correct to five decimal places.




You must be signed in to discuss.

Video Transcript

and this problem will estimate this sum, this infinite series correct to five decimal places. So to do that, we should go ahead and use the upper bound given in this section for the air when using in terms. So what we mean by that is is that you're looking at the end partial sum from one to capital and and then a n where n is given by this over here. Now, the upper bound for this expression is given in the section. And for our capital end, this will be the integral capital and to infinity and then f a b x d x where f of X is given by replacing n with X and A n and because ends at least one that implies X is at least one. So that tells us that over here we can rewrite this integral by just replacing F with the formula. Now we'd like to start thinking about what technique we should use to integrate this. But before we do that, since we want to be correct, if I decimal places, yeah, that tells us that we should take the air to be no larger than the following that way when we round off, this will be correct to five decimal places. And yeah, So this, uh, I guess before we go on, we could just write this this scientific notation, you'll see why that will be helpful in a few moments. Mhm. So now let's go ahead and solve this inequality given by this here, we would like to solve that for capital in, because if we can do that, that will tell us that will let us know how to compute this partial sum, and we'll have our answer. So I'll go on to the next page for some more room here. Mhm. And we have an improper integral on the left and this decimal on the right. So in this case, you can go ahead and use a U substitution for the honor roll, but that one half due to this and then it's up to you whether or not you want to change the limits or back substitute. But if you if you want to keep things in terms of you, then the lower limit becomes 21 plus one and the upper limit. If you plug infinity, multiply it by two and add one Well, that's still infinity, so that doesn't change. So we used the U substitution to change the limits of integration there. Now, this you recognize you can use the power rule here and then simplify that. Yeah. And now, finally bringing that inequality. So if we bring, um, in this case, let's just go ahead and solve this for, um and again. So simplifying this, we'll end up with two n plus one larger than five. So at this point, you can go ahead and multiply this tent city, other side. And when you do that, you'll get five times 10 to the minus five and then raise both sides to the 1/5 Power simplified, in any case, but this inequality we have and to be, um, at least two. So in this case, you can just use s to as an approximation. So this will be approximately equal to the infinite zone. Yeah, and then this in the calculator. And so that's our approximation, which is correct up to five decimal places