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Evaluate infinite sum

  • Thread starter ptolema
  • Start date
  • #1
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Homework Statement



evaluate the sum:
sum.jpg


(it may help to think of this as a value of a function defined by a power series.)

Homework Equations



as a function defined by a power series, the function is centered at 0: f(1/3)=
sum.jpg
=1/3+4/9+1/3+16/81+...+n2/3n

The Attempt at a Solution


i'm not entirely clear on how to actually evaluate the value of an infinite series. i could definitely use a few pointers
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
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hi ptolema! :smile:

(have a sigma: ∑ :wink:)

(not sure about the power series, but … )

have you tried ∑ (n+1)2/3n - ∑ n2/3n ? :wink:
 
  • #3
lanedance
Homework Helper
3,304
2
ok so how about considering a taylor series about zero as a start
[tex] f(a) = \sum_{n=0} f^n(a) x^n [/tex]

then re-write your series as
[tex] 0 + 1.\frac{1}{3} + 2^2.\frac{1}{3^2} +3^2.\frac{1}{3^3} +...+ n^2 (\frac{1}{3^n}) [/tex]
 
  • #4
Gib Z
Homework Helper
3,346
5
[tex] \sum_{n=1}^{\infty} \frac{x^n}{3^n} = \frac{x}{3-x} [/tex] for |x| < 3.

Try fiddling around with various manipulations such as differentiating both sides, and try to lead yourself to a series that will drop out your required sum after a suitable substitution for x.
 
  • #5
dextercioby
Science Advisor
Homework Helper
Insights Author
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Going along the hint by tiny-tim, complete the dots below

[tex] \sum_{n=1}^{\infty} \frac{n^2}{3^n} = \sum_{n=0}^{\infty} \frac{(n+1)^2}{3^{n+1}} = \frac{1}{3}\sum_{n=0}^{\infty} \frac{n^2 + 2n +1}{3^n} =...[/tex]
 

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