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### Course: Calculus, all content (2017 edition)>Unit 7

Lesson 11: Estimating infinite series

# Worked example: Series estimation with integrals

See how we can use improper integrals to approximate the infinite sum of 1/n².

## Want to join the conversation?

• , why did Sal have to introduce a new variable `b → ∞` to take the limit?
• Sal made a mistake right there. He was correct to introduce a new variable, but what he did next was incorrect, and made it confusing as to why he introduced the variable.

The problem we're dealing with at that point in the video is evaluating an integral from 6 to infinity. That's an improper integral: the fundamental theorem of calculus tells us how to evaluate the integral from 6 to some other finite number (assuming there are no "blow-up" problems in between), but it doesn't tell us how to evaluate an integral that goes to infinity. We solve the problem by dividing it into two steps. First, we find the integral from 6 to some unspecified finite upper bound, which we'll call "b" in this case. And second, once we see what the integral would be with an upper bound of b, we find what the limit would be as b goes to infinity.

Sal's mistake here was that immediately after introducing the variable b, he again writes the integral with an upper bound of infinity, when he intended to write it with an upper bound of b for the reason explained in the previous paragraph. If this still doesn't make sense, you may want to go back and review the videos on improper integrals, which begin here:

• I know this is not too relevant, but didn't it bother anyone else that he forgot to change "n" to "x" when doing the integrals?
the integral of 1/n^2 dx is equal to x/n^2.

I know it's a very minor mistake, but it could be misleading.
• This is not just a mistake, this is a BIG ( and in some costly ) mistake. I'm reporting this mistake.
• Is there a series of videos about integrals with bounds at infinity as used in this video? I'm having trouble finding any
• As k increases your estimate becomes more accurate. Could one find the exact sum by taking the limit as k→∞?
• Well, Euler proved that the sum from n=1 -> infinity of 1/n^2 is actually equal to (pi^2)/6
I'm not sure why that is, but you can definitely check out his proof of it on the internet :)
• Why do we have to split the sum into Sk and Sr in order to take the estimate? Why cant we start from n=1?
• We want to break the unknowable sum into a piece that is clear and known and a piece that can be estimated. Then we can use the estimated piece to set a bounds on all the values that the sum can possibly be. You usually cannot just start from n= 1 and keep adding terms to infinity.
` known sum of first 4 terms` + `estimate of the terms from 5 to ∞` fair estimate
` known sum of first 20 terms` + `estimate of the terms from 21 to ∞` better estimate
` known sum of first 100 terms` + `estimate of the terms from 101 to ∞` even better
• At Sal sets up an integral 1/n^2 dx. Shouldn't he be using f(x) not f(n)? His integral is with respect to x yet he integrates with respect to n.
• Yes, I actually reported this error many moons ago, but still have seen no correction made to the video. If you click "Report a Mistake" on the right side of the comment section you can view and submit errors you find in videos.
• At , near the bottom, Sal writes that the summation is approximately __, but isn't it exactly equal to __ instead?
• This is because Sal wrote an approximate value from the calculator (he used 4 significant figures), and not an exact value.
(1 vote)
• At about Sal writes that the integrand is -n^-1. How'd he get the negative out in front of the n? I'm sure this is a simple question, I'm just blanking on this subject. I understand why there's a negative on the exponent and why it went from -2 to -1.
• So when integrating the format will be x^(n + 1)/(n + 1) which in this leads to n^-2 becoming n^(-2 + 1)/(-2 + 1) or n^-1/-1 which is simply -n^-1. So basically make sure when you're integrating this problem you don't forget about the denominator part of the problem!
(1 vote)
• I'm confused because if you chose a different value of k, say 125 for example, when you apply the same integral and inequality don't you get a different number for the convergence?