Before we get into this lesson, you should already know the formula for the sum of an arithmetic series and how to prove it. If not, there are many good resources online about this formula that you can find to learn about it.
Sample Problem 1. Find the sum of \(3+3^2+3^3+...+3^n\) in terms of n (give a closed form for the expression with a constant number of terms).
There is no clear solution at first, so we start by thinking about what aspects of this problem make it hard. The main challenge in this problem is that we don't know the number of terms. The other challenge in this problem is that the terms change by multiplication, but we are trying to find the sum of the terms. In order to reduce the number of terms to a constant number, we think about subtracting the sequence from itself somehow. We let \(S=3+3^2+3^3+...+3^n\). Evidently, just subtracting the sequence from itself won't help, so we think about how we could do something slightly different. Thinking about how each term is just 3 times the previous term, we think that maybe we should multiply the sequence by something and then subtract the product from the sequence. It seems natural to multiply the sequence by 3 because it essentially just makes the sequence start at \(3^2\) and end at \(3^{n+1}\), while still containing the unknown number of terms in the middle. Thus, we write \(3S=3^2+3^3+3^4+ . . . +3^{n+1}\). Because this sequence is so similar to \(S=3+3^2+3^3+...+3^n\), we see that subtracting them would get a lot of terms to cancel out. Thus, we have \(2S=3S-S=(3^2+3^3+3^4+ . . . + 3^{n+1})-(3+3^2+3^3+...+3^n)=3^{n+1}-3\). Finally, we see that we have \(S=\frac{3^{n+1}-3}{2}\).
We can generalize this example very easily by using the above process to see that \(a+ar+ar^2+ar^3+ . . . +ar^n=\frac{a(r^{n+1}-1)}{r-1}\). Note that this does work with the above example because \(3+3^2+3^3+...+3^n=3+3*3+3*3^2+3*3^3+ . . . + 3*3^{n-1}\). The process we used of subtract the sequence from a multiple of itself may seem a little bit odd, but it does indeed work perfectly fine. Note that the infinite geometric series \(a+ar+ar^2+ar^3+ar^4+ . . .\), going on forever, is equal to \(\frac{-a}{r-1}=\frac{a}{1-r}\) for r<1, because, for r<1, \(r^n\) goes to zero as n goes to infinity (this is far from a rigorous proof, but it should give an intuitive reason because any fully rigorous proof would be far above the level of this curriculum).
We can generalize this example very easily by using the above process to see that \(a+ar+ar^2+ar^3+ . . . +ar^n=\frac{a(r^{n+1}-1)}{r-1}\). Note that this does work with the above example because \(3+3^2+3^3+...+3^n=3+3*3+3*3^2+3*3^3+ . . . + 3*3^{n-1}\). The process we used of subtract the sequence from a multiple of itself may seem a little bit odd, but it does indeed work perfectly fine. Note that the infinite geometric series \(a+ar+ar^2+ar^3+ar^4+ . . .\), going on forever, is equal to \(\frac{-a}{r-1}=\frac{a}{1-r}\) for r<1, because, for r<1, \(r^n\) goes to zero as n goes to infinity (this is far from a rigorous proof, but it should give an intuitive reason because any fully rigorous proof would be far above the level of this curriculum).
Sample Problem 2. Find \(\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+ . . .}}}}\), with infinite square roots.
For this problem, we won't be fully rigorous with the idea of infinity, but you should be able to get an intuitive understand of why this solution works. Similar to what we did for the last problem, we will let \(S=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+ . . .}}}}\). However, because there are infinitely many square roots, there are still infinitely many square roots inside the first one. In other words, we see that \(S=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+ . . .}}}}=\sqrt{2+S}\) (again, we are not being fully rigorous with the idea of infinity here). Thus, we simply square both sides of \(S=\sqrt{S+2}\) to get \(S^2=S+2\). We then simply solve the quadratic to get S=2 or S=-1. However, the value of S is clearly positive, so -1 is an extraneous solution, so we have that S=2, and we are done.
Sample Problem 3. Find the value of \(\frac{1}{1*2}+\frac{1}{2*3}+\frac{1}{3*4}+. . . +\frac{1}{n*(n+1)}\).
This problem, again, is hard because the number of terms isn't constant. Further, in this problem, there doesn't appear to be an easy way to subtract it from itself as we did in the first problem. However, we recall that the key to the first problem was a ton of terms cancelling. Thus, we think of trying to rewrite this sequence to make a bunch of terms cancel. Because each pair of consecutive terms has a number in the denominator in common with the denominator of the second term, we think of using that. Thus, we guess that \(\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}\) because because it would make all of the terms cancel out nicely. As you can easily verify, this equality does indeed hold true. Therefore, we can rewrite this sum as \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+ . . . +\frac{1}{n}-\frac{1}{n+1}=1-\frac{1}{n+1}=\frac{n}{n+1}\). This method of rewriting the terms of the sequence in such a way to make almost all of the terms cancel out and leave us with a simple to evaluate expression is called telescoping because the series collapses like a folding telescope does. Telescoping can be done without rewriting the terms through cleverly multiplying by a term or other methods.
Exercises
- Prove that \(a+ar+ar^2+ar^3+ . . . +ar^n=\frac{a(r^{n+1}-1)}{r-1}\) using telescoping.
Solution 1
We can simply observe that \(a+ar+ar^2+ar^3+ . . . +ar^n=\frac{a(r^{n+1}-1)}{r-1}\leftrightarrow (a+ar+ar^2+ar^3+ . . . +ar^n)(r-1)=a(r^{n+1}-1)\). However, when we expand out \((a+ar+ar^2+ar^3+ . . . +ar^n)(r-1)\), we see that it telescopes nicely because \((a+ar+ar^2+ar^3+ . . . +ar^n)(r-1)=(ar+ar^2+ar^3+ . . . + ar^{n+1})-(a+ar+ar^2+ar^3+ . . . +ar^n)=ar^{n+1}-a=a(r^{n+1}-1)\), as desired.