Limit of a sequence of fractions

Calculus Level 5

Define a 0 ( n ) = 2 n 1 2 n a_0(n) = \dfrac{2n-1}{2n} and a k + 1 ( n ) = a k ( n ) a k ( n + 2 k ) a_{k+1}(n) = \dfrac{a_{k}(n)}{a_{k}(n+2^k)} for k 0 k \geq 0 .

The first several terms in the sequences a k ( 1 ) a_k(1) for k 0 k \geq 0 are 1 2 \dfrac{1}{2} , 1 / 2 3 / 4 \dfrac{1/2}{3/4} , 1 2 / 3 4 5 6 / 7 8 \dfrac{\frac{1}{2}/\frac{3}{4}}{\frac{5}{6}/\frac{7}{8}} , 1 / 2 3 / 4 / 5 / 6 7 / 8 9 / 10 11 / 12 / 13 / 14 15 / 16 \dfrac{\frac{1/2}{3/4}/\frac{5/6}{7/8}}{\frac{9/10}{11/12}/\frac{13/14}{15/16}} , \cdots

What limit do the values of these fractions approach?

Use 5 significant digits.


The answer is 0.70711.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

0 solutions

No explanations have been posted yet. Check back later!

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...