Sum of n/2^n

In this video, I calculate an interesting sum, namely the series of n/2^n. For this we'll use an incredibly clever trick of splitting up and using a telescoping sum. Enjoy this beautiful adventure through calculus and analysis.

[skwbusaidi]

We start with sum of (x^n) = 1/(1-x)
The sum is from n=0 to inf
Take the drivative of both side we get
Sum of (n x^(n-1)) = 1/(1-x)^2
The sum now is from n=1 to inf
Sub x=1/2
Sum of( n/2^(n-1) = 4
Multply both side by 1/2
Sum of (n/2^n) = 2

[ПетроЗарицький-жф]

The coolest thing is that I've actually calculated this sum in more general case a year ago.
Let's consider the following sum where k is just an arbitrary positive number. Then we can rewrite it in this weird looking way (every row is a part of our sum):
e^-1k+e^-2k+e^-3k+...
+e^-2k+e^-3k+...
+e^-3k+...
...
Then we just use geometric progression formula for each row to get the following:

=(e^-1+e^-2k+...)/(1-e^-k)
And now we just use geometric progression formula again:


I find the answer 1/4sinh²(k/2) really elegant and satisfying especially compared to such a messy way to derive it with this double geometric progression. And yes, we definitely can get the answer to the case considered in the video by taking k=ln(2).

[DrWeselcouch]

My thoughts at 1:28 "No way this is gonna work"
1 minute later.... **MIND BLOWN**

Nice video as always Dr. P!

[pierreabbat]

The way I do it is squaring a geometric series.

[ahmadkalaoun]

It is rare to see solutions that do not involve calculus..very cool ;)

[racheltdm]

This guy's passion scares me... also gives me smiles when im struggling with calculas :D

[yoav]

i did not check but i think this is the only math channel that thanks us for watching!

[tgx]

One more idea for this problem. We can take the matrix (there is upper part, diagonal, bottom part). In first column all elements are 1/2, second column 1/(2^2) ....). Then the series summa(k/(2^k)) is the sumation upper part of matrix+ diagonal of matrix. One row is 1/2+1/(2^2 )+....+1/(2^N)=1-(1/2)^N. All matrix has summation N*(1-(1/2)^N). Bottom part under diagonal has - 1/2). So the origin series is limit from matrix- bottom part, it's limit from N*(1-(1/2)^N) - (N-1) -2*((1/2)^N - 1/2) it's go to 2

[Nick-kgsk]

I worked out the partial sums of the goal series. Get a common denominator of 2^n and the numerator becomes the sum of k2^(n-k) for k =1, n. Which equals 2^(n+1) - n -2. There’s a neat way to compute this finite sum by drawing a “triangle” of powers of 2 and summing the diagonals.

[eliyasne]

I wouldn't have expected that to work, thats a pleasent surprise!

[Adventurin_hobbit]

Evaluate the following:
\begin{center}
$\displaystyle \dfrac{\sum\limits _{k=1}^{\infty }\frac{k}{2^{k}}}{\sum\limits _{k=1}^{\infty }\frac{k^{2}}{2^{k}}} .$
\end{center}

Let us assume that
\begin{center}
${\displaystyle S_{1} =\sum _{k=1}^{\infty }\frac{k^{2}}{2^{k}} =\underbrace{\sum\limits _{k=1}^{\infty -\frac{1}{4}\left( 2+\underbrace{\sum\limits _{k=2}^{\infty +\underbrace{\sum\limits _{k=1}^{\infty 2}}$
\end{center}


Hence we have that
\begin{center}
${\displaystyle \frac{S_{1}}{4} =\frac{3}{2} \Longrightarrow \boxed{S_{1} =6.}}$
\end{center}
And we have that
\begin{center}
$\displaystyle S_{2} =\sum _{k=1}^{\infty }\frac{k}{2^{k}} \underbrace{\sum\limits _{k=1}^{\infty 1}} +\underbrace{\sum\limits _{k=1}^{\infty }\frac{k^{2}}{2^{k}}}_{=S_{1} =6} _{k=1}^{\infty }\frac{( k-1)^{2}}{2^{k-1}}}_{=S_{1} =6}\right) =\frac{1+6-3}{2} =\boxed{2}$

$\displaystyle \Longrightarrow \frac{S_{2}}{S_{1}} =\frac{2}{6} =\boxed{\frac{1}{3}}$
\end{center}

[emanuellandeholm]

Really cool solution! I was thinking of solving it using a generating function.

[vibhupandya]

Ok I can finally say I found a question in this channel that I can finally solve😂😂
I'm an 11th grader from India, currently preparing for JEE. This question is a classic example of the Arithmetico-Geometric series, a sequence in which each term is a product of two terms, one of which is from an arithmetic progression and the other from a geometric progression.
We were taught the following method to solve this:

S = 1/2 + 2/4 + 3/8 + 4/16 + • • • (1)
Now if this was not an infinite series, we'd have to write the general term, but in this case it's not needed.
In the next step you multiply both sides by the common ratio of the geometric series which in this case is 1/2 to obtain:
S/2 = 1/4 + 2/8 + 3/16 + 4/32 + • • •
Now we use this weird trick, we write this as:
S/2= 0 + 1/4 + 2/8 + 3/16 + • • • (2)
Now we subtract the terms of (2) from the corresponding terms of (1).

S - S/2 = (1/2 - 0) +(2/4 - 1/4) + (3/8 - 2/8) + • • •
So
S/2 = 1/2 + 1/4 + 1/8 + 1/16 + • • •
Now it's easy,
S = 1 + 1/2 + 1/4 + • • • = 2
(By the BEST FRIEND)

P.S. : I've always wanted to ask, how does it feel to be the best friend of a guy who says his best friend is the sum of the infinite geometric series? xD xD

[noahtaul]

Dr Peyam at 4:30: split up our sum into something looking like what we just calculated, plus a geometric series

Me, an intellectual, at 4:30: multiply by 4

[tgx]

Nice trick to circumvent the derivation of the circle of convergence.

[Kdd]

This was superb!! 😍😍But u can also do it by the way in which your buddy max did it on bprp's channel 😆

[RalphDratman]

That is almost funny where everything in the middle cancels.
You keep blocking the whiteboard

[lucaswilkins]

This is a very fine video of your back.

[lecinquiemeroimage]

Hi Dr Peyam; for your viewers, this one is also very interesting : E = Σ1/[n (n + 1)(n + 2)] = 1/(1.2.3) + 1/(2.3.4) + 1/(3.4.5) + = ? (the sum is from n = 1 to infinity).
Greetings from Paris. (January 9, 2021).

[tgx]

We can also use suma (1/2 * x)^N=x/(2-x), this series is convergent for x=1, then d/dx(x/(2-x))=2/(2-x)^2, if I take x=1, I will get  the result 2