It's easier to take their difference

Algebra Level 3

1 2 1 + 1 2 2 + 1 2 3 + 1 2 4 + ⋯ , 1 2 1 + 2 2 2 + 3 2 3 + 4 2 4 + ⋯ \dfrac1{2^1} + \dfrac1{2^2} + \dfrac1{2^3} + \dfrac1{2^4} + \cdots , \quad \quad \dfrac1{2^1} + \dfrac2{2^2} + \dfrac3{2^3} + \dfrac4{2^4} + \cdots

I have written two series above. What is the ratio of values between the larger series and the smaller series?


The answer is 2.

Pi Han Goh by Pi Han Goh , 30, Malaysia

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1 solution

Chew-Seong Cheong
Nov 22, 2016

Let S 1 = 1 2 1 + 1 2 2 + 1 2 3 + 1 2 4 + . . . = ∑ n = 1 ∞ 1 2 n \displaystyle S_1 = \frac 1{2^1} + \frac 1{2^2} + \frac 1{2^3} + \frac 1{2^4} + ... = \sum_{n=1}^\infty \frac 1{2^n} and S 2 = 1 2 1 + 2 2 2 + 3 2 3 + 4 2 4 + . . . = ∑ n = 1 ∞ n 2 n \displaystyle S_2 = \frac 1{2^1} + \frac 2{2^2} + \frac 3{2^3} + \frac 4{2^4} + ... = \sum_{n=1}^\infty \frac n{2^n} . Now consider

S 2 = ∑ n = 1 ∞ n 2 n = ∑ n = 0 ∞ n 2 n = ∑ n = 1 ∞ n − 1 2 n − 1 = 2 ∑ n = 1 ∞ n 2 n − 2 ∑ n = 1 ∞ 1 2 n ⟹ S 2 = 2 S 2 − 2 S 1 S 2 = 2 S 1 ⟹ S 2 S 1 = 2 \begin{aligned} S_2 & = \sum_{n=\color{#3D99F6}1}^\infty \frac n{2^n} \\ & = \sum_{n=\color{#D61F06}0}^\infty \frac n{2^n} \\ & = \sum_{n=\color{#3D99F6}1}^\infty \frac {\color{#3D99F6}n-1}{2^{\color{#3D99F6}n-1}} \\ & = 2 \sum_{n=1}^\infty \frac n{2^n} - 2 \sum_{n=1}^\infty \frac 1{2^n} \\ \implies S_2 & = 2S_2 - 2S_1 \\ S_2 & = 2S_1 \\ \implies \frac {S_2}{S_1} & = \boxed{2} \end{aligned}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Great. That's one way to evaluate an arithmetic-geometric progression !

Try this harder problem if you got the time .

Pi Han Goh - 4 years, 6 months ago

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