Super Cool Problem! 🤯

Añadido: 2026-05-19

[00:00:00]Let us solve a beautiful double infinite series problem.

[00:00:04]The question is to evaluate a double summation where n goes from zero to infinity and for every fixed value of n, the variable m goes from zero to n.

[00:00:16]Inside the summation, we have 1 / 2 raised to n + m.

[00:00:23]Our goal is to find the value of this double summation. So, can you solve it?

[00:00:28]Okay, as a first step, what we can do is try to separate variables n and m so that the inner summation becomes easier to handle. And after that, we can solve for outer summation.

[00:00:41]Notice that 1 / 2 raised to n + m can be separated using exponent rules. We can rewrite the fraction as 1 / 2 raised to m multiplied by 1 / 2 raised to n.

[00:00:56]This is a very important step because now the term involving n can be separated outside the inner summation.

[00:01:04]This is because the inner summation only depends upon m and not on n. So, the double summation becomes summation over n from zero to infinity of 1 / 2 raised to n multiplied by summation over m from zero to n of 1 / 2 raised to m.

[00:01:26]Suddenly, the expression now looks much cleaner and more manageable.

[00:01:30]Now, let us focus only on the inner summation. For a fixed value of n, the inner terms become 1 + 1 / 2 + 1 / 4 + 1 / 8 and so on up to 1 / 2 raised to n.

[00:01:46]This is clearly a geometric progression because the first term is 1. The common ratio is 1 / 2. And since powers go from zero to n, the total number of terms is n plus one.

[00:02:00]So, now we can use this formula for the sum of a finite geometric progression.

[00:02:05]Substituting these values gives 1 - 1 / 2 raised to n + 1 all divided by 1 - 1 / 2.

[00:02:15]The denominator becomes half, and dividing by half is the same as multiplying by two.

[00:02:21]Since two is just a constant, we can take it outside the summation.

[00:02:26]We get this.

[00:02:28]The expression may still look large, but now double summation is converted into a single summation.

[00:02:35]Next, distribute the multiplication inside the bracket carefully.

[00:02:39]1 / 2 raised to n multiplied by 1 simply remains 1 / 2 raised to n.

[00:02:48]Then 1 / 2 raised to n multiplied by 1 / 2 raised to n + 1 becomes 1 / 2 raised to n + n + 1 or 2 n + 1.

[00:03:02]Now, separate this into two different infinite summations.

[00:03:06]This gives two times this summation minus two times this summation.

[00:03:12]Let us evaluate the first infinite series. It becomes 1 + 1 / 2 + 1 / 4 + 1 / 8 and so on forever.

[00:03:22]This is an infinite geometric progression with first term one and common ratio 1 / 2.

[00:03:29]For an infinite geometric progression, the sum is first term divided by 1 minus common ratio. So, this becomes 1 / 1 minus 1 / 2, which simplifies to two.

[00:03:42]Since we already had a two multiplied outside the summation, the first entire part becomes four.

[00:03:49]Now, move to the the infinite series.

[00:03:52]Put n = 0 to get 1 by 2.

[00:03:56]Then put n = 1 to get 1 over 2 raised to 2 + 1, or 1 over 2 cubed, or 1 over 8.

[00:04:04]So, its terms are 1 by 2 + 1 by 8 + 1 by 32 and so on forever.

[00:04:11]Again, this is a geometric progression.

[00:04:14]The first term is 1 by 2 and each next term is obtained by multiplying by 1 by 4. So, the common ratio is 1 by 4.

[00:04:24]Using the infinite geometric progression formula again, the sum becomes 1 by 2 divided by 1 minus 1 by 4.

[00:04:34]The denominator becomes 3 by 4. We get 4 by 3 * 1/2. This simplifies to 2 by 3.

[00:04:42]Since there was already a 2 multiplied outside, this entire second part becomes 4 by 3.

[00:04:48]Finally, subtract both results. We get 4 - 4 by 3. Therefore, the value of the entire double summation is 8 divided by 3 and that's it.

[00:05:00]This was a super-duper cool problem.

[00:05:03]Like, share, and subscribe. So good.

[00:05:12]>> [music]