Infinite series for $ \sqrt 2 $

Question

What is infinite series for $ \sqrt 2 $? I don't mean continued fraction. That kind of series such as like for $e, \pi, $etc.

Answer 1

The generating function for the Central Binomial Coefficients is $$ (1-4x)^{-1/2}=\sum_{k=0}^\infty\binom{2k}{k}x^k\tag{1} $$ We can plug $x=\frac18$ into $(1)$ to get $$ \begin{align} \sqrt2 &=\sum_{k=0}^\infty\binom{2k}{k}\frac1{8^k}\\ &=\sum_{k=0}^\infty\frac{(2k-1)!!}{4^kk!}\tag{2} \end{align} $$

---

Alternatively, we could plug $x=-\frac14$ into $(1)$ and double the result to get $$ \begin{align} \sqrt2 &=2\sum_{k=0}^\infty\binom{2k}{k}\left(-\frac14\right)^k\\ &=2\sum_{k=0}^\infty(-1)^k\frac{(2k-1)!!}{(2k)!!}\tag{3} \end{align} $$ However, the error in the partial sum of $(3)$ is $O\left(\frac1{\sqrt{k}}\right)$. The error in the partial sum of $(2)$ is $O\left(\frac1{2^k\sqrt{k}}\right)$, which yields much faster convergence.

---

Using Continued Fractions, we get rational approximations to $\sqrt2$ that can be used with $(1)$ to get other series for $\sqrt2$: $$ \begin{array}{l} \sqrt2&=&\left(1-\frac48\right)^{-1/2}&=&\sum_{k=0}^\infty\binom{2k}{k}\frac1{8^k}\\ \sqrt2&=&\frac43\left(1-\frac4{36}\right)^{-1/2}&=&\frac43\sum_{k=0}^\infty\binom{2k}{k}\frac1{36^k}\\ \sqrt2&=&\frac75\left(1-\frac4{200}\right)^{-1/2}&=&\frac75\sum_{k=0}^\infty\binom{2k}{k}\frac1{200^k}\\ \sqrt2&=&\frac{24}{17}\left(1-\frac4{1156}\right)^{-1/2}&=&\frac{24}{17}\sum_{k=0}^\infty\binom{2k}{k}\frac1{1156^k}\\ \sqrt2&=&\frac{41}{29}\left(1-\frac4{6728}\right)^{-1/2}&=&\frac{41}{29}\sum_{k=0}^\infty\binom{2k}{k}\frac1{6728^k}\\ \sqrt2&=&\frac{140}{99}\left(1-\frac4{39204}\right)^{-1/2}&=&\frac{140}{99}\sum_{k=0}^\infty\binom{2k}{k}\frac1{39204^k}\\ \sqrt2&=&\frac{239}{169}\left(1-\frac4{228488}\right)^{-1/2}&=&\frac{239}{169}\sum_{k=0}^\infty\binom{2k}{k}\frac1{228488^k}\\ \end{array} $$

Answer 2

As suggested by NovaDenizen, Taylor expansion of $f(x) = \sqrt{x + 1}$ has a general term which write $$\frac{(-1)^{n-1} (2 n-3)\text{!!} x^n}{(2 n)\text{!!}}$$ Setting $x=1$ then leads to $$\sqrt{2}=\sum _{n=0}^{\infty } \frac{(-1)^{n-1} (2 n-3)\text{!!}}{(2 n)\text{!!}}$$

Answer 3

It is too easy to give series with irrational terms. So let us try for rational. One can note that $\sqrt{2}\approx 1.41421356\dots$. Thus an infinite series for $\sqrt{2}$ is $$1+\frac{4}{10}+\frac{1}{10^2}+\frac{4}{10^3}+\frac{2}{10^4}+\frac{1}{10^5}+\frac{3}{10^6}+\frac{5}{10^7}+\frac{6}{10^8}+\cdots.$$ The only issue is with the $\cdots$. We have not given an explicit expression for the $n$-th term.

If we use the Maclaurin series for $(1-x)^{-1/2}$, evaluated at $x=1/2$, we can get an explicit series with rational terms that converges to $\sqrt{2}$.

Answer 4

As indicated in the other answers, you use the binomial series for $\sqrt{1+x}$. However, $x=1$ is at the boundary of the region of convergence, so you first reduce the problem algebraically by observing that, as robjohn has used in his answer, $\sqrt2=(\frac12)^{-1/2}=(1-\frac12)^{-1/2}$ or with even smaller offsets as

$$\sqrt{2}=\frac32\sqrt{\frac89}=\frac32\sqrt{1-\frac19}=\frac32\left(1+\frac18\right)^{-\frac12}$$

or

$$\sqrt{2}=\frac75\sqrt{\frac{50}{49}}=\frac75\sqrt{1+\frac1{49}}=\frac75\left(1-\frac1{50}\right)^{-\frac12}$$

With these smaller values for $x$ under the root in any of those 4 expressions, convergence of the binomial series is much more rapid.

Answer 5

$$\sqrt{2}=\frac{1}{\left(1-\frac{1}{2^2}\right) \left(1-\frac{1}{6^2}\right) \left(1-\frac{1}{10^2}\right) \left(1-\frac{1}{14^2}\right) \cdots}$$ $$\sqrt{2}=\left(1+\frac{1}{1}\right) \left(1-\frac{1}{3}\right) \left(1+\frac{1}{5}\right) \left(1-\frac{1}{7}\right) \cdots$$ $$\sqrt{2}=1+\frac{1}{2}-\frac{1}{2\cdot4}+\frac{1\cdot3}{2\cdot4\cdot6}-\frac{1\cdot3\cdot5}{2\cdot4\cdot6\cdot8} + \cdots$$ $$\sqrt{2}=1+\cfrac{1}{2+\cfrac{1}{2+\cfrac{1}{2+\ddots}}}$$

https://en.wikipedia.org/wiki/Square_root_of_2#Series_and_product_representations

The first one is the true answer. It is in the format of the natural number $e = \left(1+\frac{1}{\infty}\right)^\infty$ except it's a minus sign instead and a reciprocal. Consider it's reduced form here:

$$\sqrt{2}=\frac{1}{\left(1-\frac{1}{4\cdot1^2}\right) \left(1-\frac{1}{4\cdot3^2}\right) \left(1-\frac{1}{4\cdot5^2}\right) \left(1-\frac{1}{4\cdot7^2}\right) \cdots}$$

This is the most reduced true form of the infinite series that is $\sqrt2$. An amazing property of $\sqrt2$ is that the reciprocal is equal to exactly $\frac{1}{2}$ of its value. So

$$\sqrt{2}=2 \left(1-\frac{1}{4\cdot1^2}\right) \left(1-\frac{1}{4\cdot3^2}\right) \left(1-\frac{1}{4\cdot5^2}\right) \left(1-\frac{1}{4\cdot7^2}\right) \cdots$$

Answer 6

A fast series that seems to produce the same result as the Babylonian method is given by

$$\sqrt{2}=\frac{3}{2}-\sum_{k=0}^\infty \frac{2\sqrt{2}}{(17+12\sqrt{2})^{2^k}-(17-12\sqrt{2})^{2^k}}$$

This question asks for a similar one starting from $\dfrac{99}{70}$.

Answer 7

You can use the Taylor Series for $\sqrt{1+x}$ centered at 0, which is $1+\sum_{n=1}^{\infty}\frac{\prod_{m=0}^{n-1}\big(\frac{1}{2}-m\big)}{n!}(x^n)$, and $\sqrt{2}$, which is $\sqrt{1+1}$, will sit right on the cliff of that sum, such that it will still converge. Hence, $\sqrt{2}=1+\sum_{n=1}^{\infty}\frac{\prod_{m=0}^{n-1}\big(\frac{1}{2}-m\big)}{n!}$. But wait! We can still go further than this, and a lot better too.

The Taylor series will actually be very slow, and summing up the first 10,000 terms or so will only give you about 5 decimal places correct. We want to converge faster than this. So, what we can do is actually take an approximation of $\sqrt{2}$ itself, then manage to make an even better series out of it. Here's how; $$ \sqrt2\approx1+\sum_{n=1}^{1}\frac{\prod_{m=0}^{n-1}\big(\frac{1}{2}-m\big)}{n!}=1+\frac{1}{2}=\frac{3}{2} $$ $$\text{Factor out } \frac{3}{2}\text{ from }\sqrt2;$$ $$\sqrt2=\frac{3}{2}\sqrt{\bigg(\frac{2}{3}\bigg)^2\times2}=\frac{3}{2}\sqrt\frac{8}{9}=\frac{3}{2}\sqrt{1-\frac{1}{9}}$$ You'll notice that our new square root will actually fit in with our Taylor series, just like $\sqrt{1+1}$, but this time, because $-\frac{1}{9}$ will be placed in " $x^n$ " as x, it will make each term smaller and thus converge more rapidly. We'll end up with a new infinite series; $$\sqrt2=\frac{3}{2}\Bigg(1+\sum_{n=1}^{\infty}\frac{\prod_{m=0}^{n-1}\big(\frac{1}{2}-m\big)}{n!}\Big(-\frac{1}{9}\Big)^n\Bigg)$$ This series turns out to be MUCH more helpful than the last, as just adding up the first 10 terms will give you the first 12 decimal places! In turn, you can make even another better series with this one, giving you even more approximations, and you never have to stop!

I'll keep replacing the "$\infty$" in the series with $1$, and I'll show you what series you end up with going along (You don't have to choose just $1$ though; go higher if you want!). $$\sqrt2=\frac{3}{2}\Bigg(1+\sum_{n=1}^{\infty}\frac{\prod_{m=0}^{n-1}\big(\frac{1}{2}-m\big)}{n!}\Big(-\frac{1}{9}\Big)^n\Bigg)$$ $$\sqrt2=\frac{17}{12}\Bigg(1+\sum_{n=1}^{\infty}\frac{\prod_{m=0}^{n-1}\big(\frac{1}{2}-m\big)}{n!}\Big(-\frac{1}{289}\Big)^n\Bigg)$$ $$\sqrt2=\frac{577}{408}\Bigg(1+\sum_{n=1}^{\infty}\frac{\prod_{m=0}^{n-1}\big(\frac{1}{2}-m\big)}{n!}\Big(-\frac{1}{332929}\Big)^n\Bigg)$$ $$\sqrt2=\frac{665857}{470832}\Bigg(1+\sum_{n=1}^{\infty}\frac{\prod_{m=0}^{n-1}\big(\frac{1}{2}-m\big)}{n!}\Big(-\frac{1}{443365544449}\Big)^n\Bigg)$$ $$\sqrt2=\frac{886731088897}{627013566048}\Bigg(1+\sum_{n=1}^{\infty}\frac{\prod_{m=0}^{n-1}\big(\frac{1}{2}-m\big)}{n!}\Big(-\frac{1}{786292024016459316676609}\Big)^n\Bigg)$$ As you would guess, the last series is absolutely MONSTROUS; you can just use the fraction from the start itself and already get 24 decimal places correct. Use the first four terms of the series, and you'll be accurate to 120 digits!

As I've said, you can reapply this over and over again. As far as I know, this method works for any positive rational numbers/roots (Ex. to try 5th roots, just replace the "$\frac{1}{2}$" in the series with $\frac{1}{5}$). Have fun with these!

Answer 8

Hint: You might one to consider f(x) = (x+2)^(1/2) and find the Taylor series of f about x = 0. The series you obtained gives you a series for 2^(1/2).