Calculating Square Root of 2

I had so much fun deriving that series which converges to pi the other day that I thought it’d be fun to occasionally whip up such series for other irrational numbers. Today I’ll do \sqrt{2}.

There are many different series that would do the job, the one I’ll derive is found by expanding \sqrt{x} around x=1 as a power series.

Indeed, for x satsfying |x-1| < 1 we have

\begin{array}{ccl} \sqrt{x} & = & 1 + \frac{1}{2}(x-1) - \frac{1}{8}(x-1)^2 + \frac{1}{16}(x-1)^3-\frac{5}{128}(x-1)^4+ ... \\ & = & 1+ \frac{1}{2}\sum_{n=0}^\infty (-\frac{1}{4})^n\frac{(2n)!}{(n!)^2(n+1)}(x-1)^{n+1}\end{array}.

This series is guaranteed to converge for all 0<x<2, and to diverge for all x<0 and x>2.   The number 2 itself sits right on the boundary of the circle of convergence, so although we would like to say

\sqrt{2} = 1 + \sum_{n=0}^\infty (-\frac{1}{4})^n\frac{(2n)!}{(n!)^2(n+1)},

we first need to show that the series is convergent. Consider the sequence a_n = \frac{(2n)!}{4^n(n!)^2(n+1)}. This sequence is monotonically decreasing, and we will show that for all n>0, a_n < \frac{1}{n}. Firstly,

a_1 = \frac{2!}{4\cdot 2} = \frac{1}{4} < 1.

We now assume that a_k < \frac{1}{k} for some integer k and show that a_{k+1} < \frac{1}{k+1}. This is true, since

\begin{array}{ccl}\frac{(2k+2)!}{4^{k+1}((k+1)!)^2(k+2)} & = & \frac{(2k+1)(2k+2)}{4(k+1)(k+2)}\frac{(2k)!}{4^k(k!)^2(k+1)} \\ &=& \frac{(2k+1)(2k+2)}{4(k+1)(k+2)}a_k\\& < & \frac{(2k+1)(2k+2)}{4k(k+1)(k+2)}\\ & = & \frac{1}{4}(\frac{1}{k}+\frac{3}{k+2}) \\ & \leq & \frac{1}{k+1}\end{array},

where the last inequality holds for k\geq1. Thus a_{k} < \frac{1}{k} implies a_{k+1}<\frac{1}{k+1}, and so by induction a_n < \frac{1}{n} for all n.

To show convergence of our series, we note we have shown enough to invoke the alternating series test. So it is true that

1 + \sum_{n=0}^\infty (-\frac{1}{4})^n\frac{(2n)!}{(n!)^2(n+1)}

converges, and by continuity  it must equal \sqrt{2}. Truncating the series at n=10,000,000 gives

\sqrt{2}\approx 1.4142135623775118

which is correct up 11 decimal places.

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Author: ficetea

I'll write more here later!

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