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Calculus Level 4

$1+\dfrac{1}{2} +\dfrac{2}{4} + \dfrac{3}{8} + \dfrac{5}{16} + \dfrac{8}{32} +\ldots = \ ?$

The answer is 4.

1 solution

Parth Kohli
Jul 31, 2015

Nice problem! It involves the observation that the differences of consecutive terms of the Fibonacci sequence also form the Fibonacci sequence. Thus, if we let $S = \frac{1}{1}+ \frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\frac{5}{16}+\frac{8}{32}+\cdots$ Then $\frac{S}{2}= \frac{1}{2}+\frac{1}{4} + \frac{2}{8}+\frac{3}{16} + \frac{5}{32}+\frac{8}{64}+\cdots$ Subtract the equations. $\frac{S}{2}= \frac{1}{1}+ \underbrace{\left(\frac{1}{4}+\frac{1}{8} + \frac{2}{16} + \frac{3}{32} + \frac{5}{64}+\cdots\right)}_{\dfrac{S}{4}}$ And so, finally, we get the equation $\frac{S}{2}=1+\frac{S}{4}$ $\therefore \boxed{S=4}$

Moderator note:

Prasun is right. You need to first show that the series converge else you will be performing operation on infinities.

Even though you got the correct answer, your method is partly unjustified since you're assuming convergence instead of showing that the sum indeed converges.

You can simply observe that the sum is $\sum\limits_{k=0}^\infty\frac{F_k}{2^{k-1}}$ where $F_k$ is the $k^{\textrm{th}}$ Fibonacci number and then use Binet's Fibonacci Formula (which can be proved by induction) noting that the conditions for convergence are indeed satisfied since the sum turns out to be the difference of two convergent geometric series.

Prasun Biswas - 5 years, 10 months ago

Why can't he just use the ratio test?

Akiva Weinberger - 5 years, 10 months ago

How is that much different from my argument? If you use the ratio test, you have to compute the limit $L=\lim\limits_{k\to\infty}\left|\frac{T_{k+1}}{T_k}\right|$ where $T_k=\frac{F_k}{2^{k-1}}$ . Since, as $k\to\infty$ , all the terms inside the limit are positive, there's no need for the absolute value sign here. The limit evaluates to,

$L=\lim_{k\to\infty}\frac{F_{k+1}\cdot 2^{k-1}}{F_k\cdot 2^k}=\frac 12\lim_{k\to\infty}\frac{F_{k+1}}{F_k}$

Now, to show that $L\lt 1$ , or more precisely $L=\frac{\phi}{2}$ , where $\phi$ is the golden ratio, you need to use Binet's fibonacci formula after all.

Are you suggesting that this is simpler than my argument?

P.s. : Sorry for the late reply. Since my college classes started, I'm out of town every week and get time to reply to comments only on the weekends.

Prasun Biswas - 5 years, 10 months ago

What's going on?

The answer is 4.

1 solution

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