Series 2

Algebra Level 3

We define f ( x ) = ∑ n = 0 ∞ 1 x n \displaystyle f(x)=\sum_{n=0}^{\infty}\frac{1}{x^{n}} such that it is finite. Find ∑ x = 2 ∞ f ( x ) x 2 \displaystyle\sum_{x=2}^{\infty}\frac{f(x)}{x^{2}} .


The answer is 1.

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Shivam Jadhav by Shivam Jadhav , 22, India

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1 solution

f ( x ) = ∑ n = 0 ∞ 1 x n ⇒ f ( x ) = 1 + f ( x ) x ⇒ f ( x ) = x x − 1 f(x)=\sum_{n=0}^{\infty}\frac{1}{x^{n}}\Rightarrow f(x)=1+\frac{f(x)}{x}\Rightarrow f(x)=\frac{x}{x-1} Hence, we get ∑ x = 2 ∞ f ( x ) x 2 = ∑ x = 2 ∞ 1 x ( x − 1 ) = 1 1 − 1 2 + 1 2 − 1 3 + ⋯ = 1 \sum_{x=2}^{\infty}\frac{f(x)}{x^{2}}=\sum_{x=2}^{\infty}\frac{1}{x(x-1)}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdots=1

5 Helpful 0 Interesting 1 Brilliant 0 Confused

You might want to show that the last term in each partial sum approaches 0, as an additional detail. A great solution though (+1).

Shaun Leong - 5 years, 4 months ago

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