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Evaluate $$\sum_{n=0}^\infty (n+1)\left(\frac13\right)^n.$$

$$1 + 2\left(\frac{1}{3}\right) + 3\left(\frac{1}{3}\right)^2 + 4\left(\frac{1}{3}\right)^3 + 5\left(\frac{1}{3}\right)^4 +\cdots$$

Show using ‘Techniques of Convergence of Geometric Series’ only that it converges to 9/4?

QC_QAOA
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rmigdow
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    I don’t know what ‘Techniques of Convergence of Geometric Series’ means, but you can simply differentiate $\sum_{k=0}^\infty x^k=\frac1{1-x}$ and plug $x=\frac13$ to get your answer. – ViHdzP Dec 23 '19 at 00:49
  • What have you tried? – Ninad Munshi Dec 23 '19 at 00:51
  • 1/1-(1/3) is 3/2, but that’s not the answer, the correct answer is 9/4 or (3/2)^2, how do you address the (n+1)? – rmigdow Dec 23 '19 at 00:53
  • I understand it can be shown by many methods to equal 1/(1-x)^2 but the directions for this problem are specific as this problem comes from an algebra 2 book where they have only just learned about convergent geometric series … so how in this context only knowing about convergent geometric series can we show that this tends to 9/4? – rmigdow Dec 23 '19 at 01:10

1 Answers1

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We have $$S =1 + 2\left(\frac{1}{3}\right) + 3\left(\frac{1}{3}\right)^2 + 4\left(\frac{1}{3}\right)^3 + 5\left(\frac{1}{3}\right)^4 +\cdots$$ so $$\frac13S = 1\left(\frac13\right) + 2\left(\frac{1}{3}\right)^2 + 3\left(\frac{1}{3}\right)^3 + 4\left(\frac{1}{3}\right)^4 + 5\left(\frac{1}{3}\right)^5 +\cdots$$

Subtracting,

$$\frac23S = 1 +\frac13 +\left(\frac13\right)^2+\left(\frac13\right)^3+\cdots$$

Take it from here.

saulspatz
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  • Perfect, I needed a simpler Algebraic way to get there for Algebra 2 HS students. This did the trick. TY! – rmigdow Dec 23 '19 at 03:45