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Infinite Sum of a Geometric Series

  • Thread starter BraedenP
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  • #1
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Homework Statement



I feel bad asking another question after I just asked one yesterday, but I'm really close this time, I think!

I have:

[tex]\sum_{n=2}^{\infty}\frac{n^2-n}{2^n}[/tex]

And need to find the sum.

Homework Equations



[tex]\sum_{n=1}^{\infty}nx^{n-1}=\frac{1}{(1-x)^2}[/tex]

The Attempt at a Solution



I have refectored this sum into the form:

[tex]\sum_{n=1}^{\infty}\frac{n^2+n}{2^{n+1}}[/tex]

and can then split it into its two terms.

When finding the sum of the term [itex]\frac{n}{2^{n+1}}[/itex] by factoring out 1/4 and using the formula above, I get 1/4, however, when the sum should be 1. Am I not applying this formula properly?

Additionally, how can I apply the above formula to the term [itex]\frac{n^2}{2^{n+1}}[/itex]? I can again factor out 1/4, but then I'm left with an n2 rather than n.

Guidance for any of these steps would be awesome!
 

Answers and Replies

  • #2
192
3

Homework Statement




Homework Equations



[tex]\sum_{n=1}^{\infty}nx^{n-1}=\frac{1}{(1-x)^2}[/tex]

...
Guidance for any of these steps would be awesome!
Try taking the relevant equation and finding the derivative with respect to x. Then take whatever steps are necessary to relate this to the problem, setting x=1/2.
 
  • #3
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Try taking the relevant equation and finding the derivative with respect to x. Then take whatever steps are necessary to relate this to the problem, setting x=1/2.
I've seen this "set x=1/2" technique used both in class and online. How are we just able to set this value of x? Does it derive from some theorem or something?

I'll definitely do that, but I'd just like to know the logic behind it, and why it works (or that there is a theorem that this comes from.)

Thanks!
 
  • #4
192
3
I've seen this "set x=1/2" technique used both in class and online. How are we just able to set this value of x? Does it derive from some theorem or something?

I'll definitely do that, but I'd just like to know the logic behind it, and why it works (or that there is a theorem that this comes from.)

Thanks!
Well, if you have an expression that is true for any x, and you're trying to find a value for an expression that corresponds to the specific value x=1/2, then what else would you do? In your case you are trying to find an finite sum with 2n in the denominator, using an expression with xn in the numerator, so it seems pretty obvious that you will eventually substitute x=1/2 because then xn = 1/2n....
 
  • #5
121
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The answer's 4, but I cheated to get it. If you write a program calculating the sums from n=0 to infinity of (n^k)/(2^n) for k=0,1,2,3..., you'll obtain the sequence 1, 2, 6, 26, 150, 1082...

Looking up that sequence at http://oeis.org/ leads you to Sloan's A000629. The 4th entry under the FORMULA section confirms the connection. I have no idea how to get there from 1st principles, but perhaps one the entries in the REFERENCES or LINKS sections might help.
 
  • #6
192
3
The answer's 4, but I cheated to get it. If you write a program calculating the sums from n=0 to infinity of (n^k)/(2^n) for k=0,1,2,3..., you'll obtain the sequence 1, 2, 6, 26, 150, 1082...

Looking up that sequence at http://oeis.org/ leads you to Sloan's A000629. The 4th entry under the FORMULA section confirms the connection. I have no idea how to get there from 1st principles, but perhaps one the entries in the REFERENCES or LINKS sections might help.
Doing this in my head, I also get 4. The idea is to take the sum that appears as the OP's Relevant Equation, and take the derivative with respect to x. For the value of x=1/2, you get precisely 22=4 times the sum that is being asked in the problem. On the other side, you get 2(1-x)-3=16....
 
  • #7
Ray Vickson
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In 2, what do you get if you take the x-derivative on both sides? Now, you are allowed to substitute any number strictly between -1 and +1. One particular value of x is especially relevant to your original problem.

RGV
 

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