Question

Any cool tricks to find this sum
\(\sum \limits_{n=1}^{\infty} \frac{n^2}{10^n} \)

Answer 1

Yeah, I found it out in the last thread. But I'll leave it for someone else to figure out if you want the challenge. =P

Answer 2

I didn't post it there though, so don't bother looking. >=)

Answer 3

It doesn't look to simplify (by gauss approach)

x = 1^2/10 + 2^2/10^2 + 3^2/10^3 + ...

10x = 1^2 + 2^1/10 + 3^2/10^3 + ...
-----------------------------------------------------
9x = 1 + 3/10 + 5/10^2 + 7/10^3 + ....

Stuck i guess

Answer 4

i have an idea, don't know if it is right

Answer 5

I think a Taylor-Lagrange would do it.

Answer 6

write
\[f(x)=\sum\frac{n^2x^n}{10^n}\]

Answer 7

You're on the right track @ganeshie8 just notice that you might be able to do the trick again. =)

Answer 8

You can write that as
\[ \sum_{n=1}^\infty \frac{1}{\ln(10)^2} \frac{\partial^2}{\partial \lambda^2} 10^{-\lambda n} \]
Interchange the summation and differentiation, then set lambda = 1 at the very end.

Answer 9

\[\sum_{n=1}^\infty \frac{1}{\ln(10)^2} \frac{\partial^2}{\partial \lambda^2} 10^{-\lambda n} = \frac{1}{\ln(10)^2} \frac{\partial^2}{\partial \lambda^2} \sum_{n=1}^\infty10^{-\lambda n}\]

like this ?

Answer 10

and thats an infinite covnerging geometric series i guess xD

Answer 11

Yep

Answer 12

I'll go ahead and show how you could derive the most general version of it I can by another method, but it's slower. This method also allows you to easily see how to bootstrap your way up to higher powers of n, like n^3/10^n.

differentiate a couple times
\[\sum_{n=1}^\infty x^n = \frac{1}{1-x}\]\[\sum_{n=1}^\infty nx^{n-1} = \frac{1}{(1-x)^2}\]\[\sum_{n=1}^\infty n(n-1)x^{n-2} = \frac{2}{(1-x)^3}\]

notice that the last one looks like this \[\sum_{n=1}^\infty n^2x^{n-2}-\sum_{n=1}^\infty nx^{n-2} = \frac{2}{(1-x)^3}\]
Well that's convenient since our second one above can be changed by a simple division by x. \[\sum_{n=1}^\infty nx^{n-2} = \frac{1}{x(1-x)^2}\] so we get:\[\sum_{n=1}^\infty n^2x^{n-2}=\frac{1}{x(1-x)^2}+ \frac{2}{(1-x)^3}\]

Perhaps there's a very general formula suggested by this?

Answer 13

That is equivalent to using the parameter lambda n times:
\[ \frac{1}{\ln(10)^n} \frac{\partial^n}{\partial \lambda^n} \sum_{p=1}^\infty 10^{-\lambda p} = \frac{1}{10\ln(10)^n} \frac{\partial^n}{\partial \lambda ^n} \frac{1}{1-10^{-\lambda}}\]

Answer 14

Oops- there should be a (-1)^n in there as well.

Answer 15

But this is really the hard way, if you keep going with your first attempt @ganeshie8 you might find it. Just notice that all your numbers are really now turned into odd coefficients so if you move them down by 10 you will have all 2's.

So wait did I just derive that basically @Jemurray3 ? Good to know I guess lol.

Answer 16

woah woah fantastic !
It looks recursive ? we need to find the n/10^n series first right ?

Answer 17

Yeah, but by simply taking the derivative, we had found that as one of our earlier derivatives so it's already there and I plugged it in for us so we didn't have to look at it as a sum anymore.

But you're better off continuing:

9x=1+3/10+5/100+7/1000+...

90x=10+3+5/10+7/100+9/1000+...

90x-9x=12+2/10+2/100+2/1000+...

Should be straight forward now I think? =P

Answer 18

brilliant ! got it completely xD

Answer 19

There is a simpler way to do all these things

Answer 20

and yes lol looks you have just derived the lagrange-taylory thingy by taking nth derivatives :)

Answer 21

shall i tell my approach

Answer 22

Go for it, I'm interested haha.

Answer 23

I actually came up with a generalize summation formula that has a variable step size instead of just increasing by 1 each time it can increase by 1/2 or pi or whatever. And the great part is if you plug in 0 you get an integral. If anyone's interested in seeing that, I can show it. The fun part is, it's all algebraic and there's no limit taking.

Answer 24

sure @No.name im also very curious to see another way to do this !

Answer 25

If u see the pattern

sum((n^0)/(B^n)) is (1 /((B-1)^1))

sum((n^1)/(B^n)) is (B /((B-1)^2))

sum((n^2)/(B^n)) is (B(B+1))/((B-1)^3)

Answer 26

Take small numbers and evaluate the pattern i bet its right

So for this you need to take pattern in patterns

Answer 27

Wait, I don't really know what you're saying @No.name but it sounds pretty much like what I just said above.

Answer 28

you have got very good eyes @No.name yes it works perfectly !!

Answer 29

B= 10

So, for n^2
[10(10+1)]/(10-1)^3
110/729

Answer 30

No need to take partial derivatives

Answer 31

I disagree. What is the next term? For n = 3?

Answer 32

haha that ends the nice pattern :(

Answer 33

try it with any calculus method u would get the same answer as mine

Answer 34

What is the continuation of that pattern for n = 3?

Answer 35

or n^3, rather

Answer 36

It won't

Answer 37

What do you mean, it won't?

Answer 38

sum((n^A)/(B^n)) is vA/((B-1)^(A+1))

Answer 39

I actually came up with a generalize summation formula that has a variable step size instead of just increasing by 1 each time it can increase by 1/2 or pi or whatever. And the great part is if you plug in 0 you get an integral. If anyone's interested in seeing that, I can show it. The fun part is, it's all algebraic and there's no limit taking.

you mean n = real number @Kainui ? you're gona explode my head today lol ;=;

Answer 40

sum((n^A)/(B^n)) is vA/((B-1)^(A+1))

Answer 41

Here's an interesting and slightly useful identity I discovered. Haven't really used it but eh.\[\log \prod_{n=1}^{\infty}f(n)=\sum_{n=1}^\infty \log[f(n)]\]

Answer 42

So it dosen't end the nice pattern
sum((n^A)/(B^n)) is vA/((B-1)^(A+1))

Answer 43

That's not true.

Answer 44

what say @ganeshie8

Answer 45

Also, notice that you can change the step size by replacing n with h*n, where h is the step size. In that case, the sum becomes
\[ \frac{(-1)^n}{10h\ln(10)^n} \frac{\partial^n}{\partial \lambda ^n} \frac{1}{1-10^{-\lambda h}} \]

Answer 46

@ganeshie8 I actually don't know if I could generalize it to step sizes of complex numbers... I might try that out but, at any rate here we go:

Let's say s is the step size in our summation. I think the important part is that we have our end point and starting point be evenly divisible by this number, otherwise it gets weird... But interesting, kind of like how if x^2 is x multiplied by itself twice then that means x^(1/2) is x multiplied by itself a half a time... But I digress back to the summation.

So it builds up recursively and is telescoping.
\[\sum_{0}^{a} [(n+s)-n] = [(0+s)-0] +[(s+s)-s] +[(2s+s)-2s] +...\] and it eventually gets to a by taking each term s higher than the last. Think 1 where s is if that helps since that's what you normally would do. The top then simplifies as a telescoping series with the next term subtracting the next just leaving the ends. \[\sum_0^a[(n+s)-n] =a+s\] but this is obviously just \[\sum_0^as =a+s\] pretty straight forward, s is a constant so we can pull it out. So if we plug in numbers as a sanity check, this would be the same as \[\sum_0^a1 =a+1\] which makes sense, we counted 0 right? Also note that if we want a lower bound that's not 0 we can just subtract off one of these sums from itself. That's gonna be more useful for higher order sums. Notice if we plug in 0 then the s in a sense is like dx. \[\sum_0^adx =a\] ok so this looks like an integral, awesome. \[\int\limits _0^adx=a\] ok let's keep going now that we have this.

Answer 47

A substitution can be made to make it slightly simpler - let
\[ 10^{-\lambda h} = e^{x} \rightarrow x = -\lambda h\ln(10)\]
Which yields the formula

\[ \frac{1}{10^h} \left. \left(\frac{d^n}{dx^n} \frac{1}{1-e^x}\right) \right|_{x = -h\ln(10)}\]

Answer 48

Straightforward, this is a telescoping series so it leaves us with this: \[\sum_0^a[(n+s)^2-n^2]=(a+s)^2\] now we can simplify it on the inside of the sum. \[\sum_0^a 2ns +s^2=\sum_0^a 2ns +\sum_0^as^2=(a+s)^2\] pull out the constants and solve for the one sum with n in it... \[2s \sum_0^a n +s(a+s)=(a+s)^2\]\[s \sum_0^a n =\frac{1}{2}[-s(a+s)+(a+s)^2]\] plug in 1 for s you get \[ \sum_0^a n =\frac{1}{2}[-(a+1)+(a+1)^2]=\frac{(-1+(a+1))(a+1)}{2}=\frac{a(a+1)}{2}\] which is gauss's formula for adding up the numbers to n. Cool! But now let's plug in s=0 to get:
\[dx \sum_0^a n =\frac{a^2}{2}\] pretty cool, we are getting the integral all in one! We could also plug in 1/2 and get the sum of all the numbers between 0 and a counting by .5 lol. Continue this process for higher polynomials, pretty interesting.

Answer 49

Some kind of interesting questions, what if your step wasn't a number anymore, but is a function? What if you're stepping into the complex plane and sweeping out a polygon or something? I have no clue, but it works and is pretty fun to play with. But it's sort of a hassle to derive higher order ones since they will all rely on lower order polynomial sums.

Answer 50

You can replace ten by any base you choose (we could imagine base A, for example) in which case
\[ \frac{1}{A^h} \left. \left( \frac{d^n}{dx^n} \frac{1}{1-e^x}\right) \right|_{x=-h\ln(A)} \]
That's about as general as I can possibly imagine being. Note that this works if and only if \(|A^{-h}|\) < 1 , or \(|A^h |> 1\).

@Kainui if you define the function \(F(x,n) = \frac{d^n}{dx^n} \frac{1}{1-e^x}\), then the sum becomes

\[ G(h) = \frac{F(-h\ln(A),n)}{A^h }\]
Which can be viewed as a function of h. Analytic continuation into the complex plane is trivial, and would yield a complex function \(G(z)\).

Answer 51

you folks really make me see my true level in math lol
It would definitely take few days for me to comprehend these stuff I guess ! thank you both @Kainui @Jemurray3 for the cool ideas xD

Answer 52

Haha, you're not too far from us I think; I'm just a baby in the math world. If you have any other ideas or thoughts going on you should ask them I'm curious and now I'm in the mood to do more math! lol

Answer 53

is the formula incorrect

Answer 54

@ganeshie8

Answer 55

why i am ignored, did i say something incorrect or is anyone not here

Answer 56

it is the erlang series , it is somewhat unpopular

Answer 57

I think it's just cause OS is really glitchy right now.

Answer 58

Ok

Answer 59

yeah OS is very laggy :/ lets start a new thread @No.name :)