If $y = \frac{2}{5}+\frac{1\cdot3}{2!} \left(\frac{2}{5}\right)^2+\frac{1\cdot3\cdot5}{3!} \left(\frac{2}{5}\right)^3+\cdots$, find $y^2+2y$

Question

If $$y = \frac{2}{5}+\frac{1\cdot3}{2!} \left(\frac{2}{5}\right)^2+\frac{1\cdot3\cdot5}{3!} \left(\frac{2}{5}\right)^3+\cdots$$ what is $y^2+2y$?

Attempt:

We know that for negative and fractional indices,

$$(1+x)^n = 1 + nx + n(n-1)/2!\cdot x^2 + n(n-1)(n-2)/3!\cdot x^3 + \cdots$$

Rewriting the series in question, we get:

$$\frac{2}{5} \left(1 + \frac{1\cdot3}{2!}\cdot \frac{2}{5}+\frac{1\cdot3\cdot5}{3!} \left(\frac{2}{5}\right)^2+\cdots\right)$$

I know this looks like the binomial expansion above, but I have no idea how to proceed further.

Answer 1

You can calculate $y$ using a "well known" binomial series expansion for $$(1-4x)^{-\frac 12} = \sum_{n=0}^{\infty}\binom{2n}nx^n$$

To see this rewrite the coefficients

$$\prod_{k=1}^n\frac{2k-1}{k} = \prod_{k=1}^n\frac{(2k-1)\cdot 2k}{k\cdot 2k}= \frac 1{2^n}\binom{2n}n$$

So in your case the series becomes

$$y = \sum_{\color{blue}{n=1}}^{\infty}\frac 1{2^n}\binom{2n}n \frac{2^n}{5^n} = \sum_{\color{blue}{n=1}}^{\infty} \binom{2n}n\frac 1{5^n}$$ $$ = \left.\frac 1{\sqrt{1-4x}}-1\right|_{x = \frac 15} = \sqrt 5 - 1$$

Answer 2

Substitute $x=\frac{2}{5}$: $$y=\sum_{i=1}^\infty\frac{x^i}{i!}(\prod_{k=1}^{i}(2k-1))$$ $$y=\sum_{i=1}^\infty\frac{x^i}{i!}((2i-1)!!)$$

For $|x|<\frac{1}{2}$, this is the Taylor series for $\frac{1}{\sqrt{1-2x}}-1$

$$y=\frac{1}{\sqrt{1-2x}}-1$$ $$y=\frac{1}{\sqrt{1-0.8}}-1$$ $$y=\frac{1}{\sqrt{0.2}}-1$$ $$y=\sqrt{5}-1$$

$$y^2+2y=(\sqrt{5}-1)^2+2(\sqrt{5}-1)$$ $$y^2+2y=5+1-2\sqrt{5}+2\sqrt{5}-2$$ $$y^2+2y=4$$