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Let $\sum_{n=1}^\infty a_nx^{n-1} = 1+x+2x^2+3x^3+5x^4+8x^5+13x^6+21x^7+34x^8+55x^9+89x^{10}+\ldots $ Use the ratio test to prove that f(x) converges if |x|< $\frac{1}{2}$ . Edit: $a_n$ in this case is a Fibonacci sequence

BB8
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  • You have to find out how to express $a_n$ in terms of $n$. That isn't at all clear in this case. Besides the fact that any sequence of real numbers could follow the given coefficients. – vonbrand Mar 21 '16 at 01:08
  • Hint: $a_{n+1} = a_n + a_{n-1}$. – Cameron Williams Mar 21 '16 at 01:11
  • You need a closed form expression for the $n$-th term of the Fibonacci sequence it seems. – Nap D. Lover Mar 21 '16 at 01:11
  • The limit of the ratios will be $|x| \lim_{n \to \infty} \frac{a_{n+1}}{a_n}$. So you will eed to prove that $\lim_{n \to \infty} \frac{a_{n+1}}{a_n}<2$ (it is actually $\frac{1+\sqrt{5}}{2}$, as it turns out). – Ian Mar 21 '16 at 01:13

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Use Fibonacci closed form expression: $$ a_n = \frac{\varphi^n - (-\varphi)^{-n}}{\sqrt{5}} $$ You can also use the fact that $a_{n+1} = a_n + a_{n-1}$, so let $$ r = \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{a_n + a_{n-1}}{a_n} = 1 + \frac{1}{r} $$ Solving for $r$ yields $(1 + \sqrt{5})/2$. The other root is extraneous.

Henricus V.
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