Pure Goosebumps! 🤯

These types of questions give me pure joy. We have a summation of n equal to 0 to infinity 1 over 3n plus 1 minus infinity 1 over 3n plus 2. At first glance, this problem feels like a normal sequence and series question. A student may start by writing a few terms and quickly notice that their difference looks strange and meaningless. Usual tricks from sequence and series do not help. And subtracting term by term gives no clear result. So algebra alone refuses to simplify the expression. At this point, it is important to look at the problem differently. Because, of course, it is a G, advanced problem, which focuses more on quality questions, like the ones which make you think harder, instead of a routine method. Therefore, we have to come up with something very clever. What else can we think of when we look at a summation sign and infinity in it? Yes, right! We can think of integration, because integration is nothing but adding infinitely many very small quantities. Here we have 1 over something, so the natural question is whether 1 over a number can appear as the result of an integration. If we think carefully, the integral of x to the power of some constant k from 0 to 1 gives the result as x raised to k plus 1 over k plus 1. When we evaluate it from 0 to 1, we get 1 over k plus 1 as the answer. Wow! It feels like we are going in the right direction. So, the integral of x raised to k minus 1 from 0 to 1 will be equal to 1 over k, right? Now, here we have 1 over 3n plus 1 and 1 over 3n plus 2. So, in place of k, put these values to get the integral of x raised to 3n from 0 to 1 as 1 over 3n plus 1. And the integral of x raised to 3n plus 1 from 0 to 1 as 1 over 3n plus 2. Oh my God! That was so refreshing! Substitute both these values here to get this. Now, separate both these terms like this. At this stage, we notice that we have a summation and an integral together. And it becomes natural to ask whether we can change their order. See, we are allowed to interchange summation and integration here because x goes from 0 to 1. And therefore, the powers of x stay between 0 and 1, which means the terms decrease smoothly and nothing blows up inside the interval. So, adding first and then finding area gives the same result as finding area first and then adding. Therefore, after changing the order to summation and integral, we get this minus this. Now, both these integrals are the same and therefore, we will combine them to get this. Next, we can rewrite this x raised to 3n plus 1 as x times x raised to 3n. Since this summation depends upon n, therefore, this x acts just as a constant inside this summation. And therefore, we can take this x outside the summation. We can't do the same for this x raised to 3n because it contains n. Take x raised to 3n as common to get this as 1 minus x times x raised to 3n. So, we are now left with this. Now, let us focus on this part. We can rewrite it as x raised to 0 or 1 plus x cubed plus x to the 6 and so on till infinity. So, this is a geometric series whose first term is 1 and the common ratio is x cubed. Since x goes from 0 to 1, therefore, this common ratio is also between 0 and 1. And thus, we can use the standard result for the sum of a geometric series which is a over 1 minus r, which is 1 over 1 minus x cubed for this case. Substitute it here. Wow! See how beautifully we have converted a series problem to an integral problem. Now, 1 can be written as 1 cubed and thus, we can use a cubed minus b cubed identity, which is this to factor out 1 minus x cubed as 1 minus x times 1 plus x plus x squared. Put it here in the denominator. The factor 1 minus x in the numerator cancels with the same factor inside 1 minus x cubed, leaving us with 1 over 1 plus x plus x squared. Now, this is super easy. Rewrite this denominator as the sum of perfect squares. We can write it as x squared plus 2 times x times 1 half plus 1 half squared, which is 1 over 4. To make it as 1, add the remaining 3 over 4. Now, using a plus b whole squared formula, this becomes x plus half whole squared and this can be written as root 3 over 2 whole squared. Put it here in the denominator. This integral is a standard 1 and evaluates to 2 over root 3 times tan inverse of 2 over root 3 times x plus half evaluated from 0 to 1. Put x as 1 here to get this. This becomes 3 over 2 and thus, this entire thing becomes root 3. Tan inverse root 3 is pi over 3. So, this becomes 2 pi over 3 root 3. Now, put x as 0 here to get this. This becomes 1 over 2 and thus, this entire thing becomes 1 over root 3. Tan inverse 1 over root 3 is pi over 6. So, this becomes pi over root 3. Therefore, the integration is equal to this minus this or pi over root 3. Hence, the difference of these two infinite summations is pi over 3 root 3. This method just blew away my mind. 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