Math for fun, find the value using calculus!

Math for fun, Series of n/3^n = ?
Value of the sum of n/3^n, from n = 1 to inf.

[DrQuatsch]

"Let me do this in blue, so people will comment down below." You, sir, are a legend XD

[alkankondo]

I found another way to approach this problem. As one of my favorite math professors once told me, if you don't know what to do with problems like these, just start writing terms down and see if you can spot a pattern. It worked in this case!

To start, notice that we can rewrite the expanded sum "1/(3^1) + 2/(3^2) + 3/(3^3) + 4/(3^4) + ...." as

[1/(3^1) + 1/(3^2) + 1/(3^3) + ] + ...
[1/(3^2) + 1/(3^3) + 1/(3^4) + ] + ...
[1/(3^3) + 1/(3^4) + 1/(3^5) + ] + ...

This is really the key insight. Note that each term in the "[ ]" is of the form [ (1/(3^n)) / (1 - (1/3)) ]. From there, you can see that you basically have a geometric series OF geometric series. Simplifying gets (3/2)^2 * (1/3), which simplifies to 3/4. Such a simple answer for such a complicated-looking problem! Math is amazing!

[An-htso]

"We'll fix that later, maybe" XD

[hypercent]

I like that solution, it's quite interesting. I also like the solution in other comments, calculating the sum S by seeing that 3S-S is a geometric series. My solution is similar to that, maybe a bit more complicated, but I thought I'd share nonetheless:

S = 1/3 + 2/9 + 3/27 + 4/81 + 5/243 + ...
= 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + ... (S1)
+ 1/9 + 1/27 + 1/81 + 1/243 + ... (S2)
+ 1/27 + 1/81 + 1/243 + ... (S3)
+ 1/81 + 1/243 + ... (S4)
+ 1/243 + ... (S5)
+ ...

So S can be seen as an infinite sum of geometric series, S1, S2, and so on.

S1 = 1/3 * (1 + 1/3 + 1/9 + 1/27 + 1/81 + ...)
S2 = 1/9 * (1 + 1/3 + 1/9 + 1/27 + 1/81 + ...)
S3 = 1/27 * (1 + 1/3 + 1/9 + 1/27 + 1/81 + ...)
S4 = 1/81 * (1 + 1/3 + 1/9 + 1/27 + 1/81 + ...)
S5 = 1/243 * (1 + 1/3 + 1/9 + 1/27 + 1/81 + ...)
...

Summing all Sn, factoring out the common geometric series, gives you S as a product of two geometric series:

S = S1 + S2 + S3 + S4 + S5 + ...
= (1/3 + 1/9 + 1/27 + 1/81 + 1/243...) * (1 + 1/3 + 1/9 + 1/27 + 1/81 + ...)
= 1/3 * (1 + 1/3 + 1/9 + 1/27 + 1/81 + ...)^2
= 1/3 * (1/(1-1/3))^2
= 1/3 * (3/2)^2
= 3/4

[IoT_]

As a university teacher, I really appreciate your way of teaching. Your scrupulous explanations are off the charts.

[RasperHelpdesk]

Start video
Hear doorbell
Pause video
Check the door, nobody there
Restart video... oh, I'm an idiot.

[KrzaqDBP]

People there are like 10 possible ways to solve it, why everyone thinks that the way you just mind out is the right one...
This genius dude just show us the beautiful way to use Calculus(which is summing all real number from A to B) to sum only NATURAL numbers ;)! Aprreciate it guys.

I love this method, thank you for showing it :)

[lackdejuranez]

There was this simple technique my professor once told me, I probably told this earlier on a video of yours, but here it is again.

S = 1/3 + 2/3² + 3/3³ +....
S/3 = 0 + 1/3² + 2/3³ +

S-S/3 = 1/3 + 1/3² + 1/3³ + ....
2S/3 = 1/3/(1-1/3)
S = 3/4

[jonashammerich]

Math actually is so freaking interesting! I just love it!

[basedblueboy]

For any base x, the solution can be generalized as x/(x^2-2x+1).

For base 3, the solution is 3/(3^2-3*2+1) = 3/4

[DerToasti]

you can also do sum=K then 3K-K=2K=1+1/3+1/9+...=3/2 so sum=3/4

[KwongBaby]

you are a talent!!
d/dx makes summation to be n=1
let x be 1/3
finally multiply both sides by 1/3

I never thought these 3 steps. thank you

[wellbangok]

Do you plan on doing any videos on Complex Analysis? Like complex integration.

[donmoore]

Having just watched Prof Leondard's videos, I actually saw the derivative coming. Very cool!

[TheReligiousAtheists]

There're these things called 'Arithmetico-Geometric Progressions', which are basically series with their 'n'th term of the form (a+(n-1)d)r^(n-1)... it's basically the product of a geometric progression whose 'n'th term is of the form r^(n-1) and an arithmetic progression whose 'n'th term is of the form a+(n-1)d. This is referred to as an 'AGP' and you can derive the formula for the sum to n terms is a similar way as you do for geometric progressions. At first I thought that you'd do something similar to the derivation of that formula, but still watched on and you seriously took me by surprise with that differentiation step. Damn elegant.

[ffggddss]

Nicely done & explained! And you can generalize somewhat, replacing the "3" with "a" :
x = 1/a, with |a| > 1. And then, running through all the same steps with 1/a in place of x, you get

∑₀ºº(n/aⁿ) = a/(a–1)²

In particular, if you take a = 10, you can easily verify the result on a calculator ... by taking 10/81 ... and be careful to note where n goes from 9 to 10, causing a carry ...

[TheProloe]

I decided to solve this one in a pretty convoluted and not at all efficient way, but it was fun to do so I'm gonna share it.
First I noted that 1/3 + 2/9 + 3/27 + 4/81 + ... (the sum we're setting out to solve), can be written as a sum of sums:
(1/3 + 1/9 + 1/27 + 1/81 +...) + (1/9 + 1/27 + 1/81 + ...) + (1/27 + 1/81 + ...) + ...
In more direct notation, Sum(n/3^n)[1, inf] = Sum(Sum(3^-m)[n, inf])[1, inf]

Next step is to find a general solution for Sum(3^-m)[n, inf]. Fortunately this is pretty simple:
S = 1/3^n + 1/3^(n+1) + 1/3^(n+2) + ...
S/3 = 1/3^(n+1) + 1/3^(n+2) + 1/3^(n+3) +...
And S-1/3^n = 1/3^(n+1) + 1/3^(n+2) + 1/3^(n+3) + ...
So, S/3 = S-1/3^n, giving S = 1/(2*3^(n-1))

Plugging that back into the sum of sums about, we get the solution to our desired sum is: Sum(n/3^n)[1, inf] = Sum(1/(2*3^(n-1)))[1, inf]
With only one instance of n in the equation, this becomes a lot easier to reason about, and we can use exactly the same trick:
S = 1/(2*3^0) + 1/(2*3^1) + 1/(2*3^2) + 1/(2*3^3) + ...
S/3 = 1/(2*3^1) + 1/(2*3^2) + 1/(2*3^3) + 1/(2*3^4) + ...
And S - 1/(2*3^0) = 1/(2*3^1) + 1/(2*3^2) + 1/(2*3^3) + 1/(2*3^4) + ...
So, S/3 = S - 1/(2*3^0) = S - 1/2
And rearranging this we get:
S(1/3 - 1) = -1/2
2S/3 = 1/2
S = 3/4

Convoluted? Yes. Fun? Absolutely.

[fabiogenduso]

Thank you for your kind math divulgation. It should be useful to remind that term to term differentiation of a power series takes sense only if the same one is uniformly convergent on a given interval and certainly this is true for the geometric series, but not for many other series being simply convergent only.
Cheers from Italy.

[twwc]

You mention at the start that the ratio test shows you that the series converges, but that is not enough. If you are going to differentiate both sides of the expression for ∑x^n you need to show it is uniformly convergent (which it is, at least in a neighborhood of x=1/3).

[arbitrarilyarbitrary]

You know what’s up when he asks “wouldn’t it be nice?”