Maclaurin series expansion for $\dfrac{1}{2t}\left(1+t-\dfrac{(1-t)^2}{2 \sqrt {t}}\log \left(\dfrac{1+\sqrt{t}}{1-\sqrt{t}}\right)\right)$

Question

Show that the expansion of

$$\dfrac{1}{2t}\left(1+t-\dfrac{(1-t)^2}{2 \sqrt{t}} \log \left(\dfrac{1+\sqrt{t}}{1-\sqrt{t}}\right)\right)$$

as a Maclaurin series in power of $t$ to obtain $$\dfrac{4}{3}-4\sum_{n=1}^{\infty}\dfrac{t^n}{(4n^2-1)(2n+3)}$$

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I am not sure how to do this. Below are some steps I have managed to get which gets more complicated and am not sure how to continue. Help is appreciated.

Expand into: $$\dfrac{1}{2t}+\dfrac{1}{2} - \dfrac{(1-t)^2}{4t\sqrt{t}}\log\left(\dfrac{1+\sqrt{t}}{1-\sqrt{t}}\right)$$

$$=\dfrac{1}{2t}+\dfrac{1}{2}- \dfrac{\log\left(\dfrac{1+\sqrt{t}}{1-\sqrt{t}}\right)}{4t\sqrt{t}}(-t+1)^2$$ Multiply by congugate:

$$=\dfrac{1}{2t}+\dfrac{1}{2}-\dfrac{\log\left(\dfrac{(1+\sqrt{t})^2}{-t+1}\right)}{4t\sqrt{t}}(-t+1)^2$$

Answer 1

We have $\frac12\log\frac{1+z}{1-z}=\sum_{n=0}^\infty\frac{z^{2n+1}}{2n+1}$, hence $\frac{1}{2\sqrt{t}}\log\frac{1+\sqrt{t}}{1-\sqrt{t}}=\sum_{n=0}^{\infty}\frac{t^n}{2n+1}$ and $$\frac{(1-t)^2}{2\sqrt{t}}\log\frac{1+\sqrt{t}}{1-\sqrt{t}}=(1-2t+t^2)\sum_{n=0}^{\infty}\frac{t^n}{2n+1}\\=1+\left(\frac13-2\right)t+\sum_{n=2}^{\infty}\left(\frac{1}{2n+1}-\frac{2}{2n-1}+\frac{1}{2n-3}\right)t^n\\=1-\frac53t+\sum_{n=2}^{\infty}\frac{8t^n}{(2n+1)(2n-1)(2n-3)}=1-\frac53t+8\sum_{n=\color{blue}{1}}^{\infty}\frac{t^{n\color{blue}{+1}}}{(4n^2-1)(2n+3)}.$$ Now substitute and get the expected result.

Answer 2

Considering $$A=\dfrac{1}{2t}\left(1+t-\dfrac{(1-t)^2}{2 \sqrt {t}}\log \left(\dfrac{1+\sqrt{t}}{1-\sqrt{t}}\right)\right)$$ just to make life easier, start making $t=x^2$ to get $$A=\frac{2 x(1+x^2)-\left(1-x^2\right)^2 \log \left(\frac{1+x}{1-x}\right)}{4 x^3}$$ and use first, as @metamorphy commented, the expansion of $\log\left(\frac{1+x}{1-x}\right)$ and expand. You should arrive at $$\frac{4}{3}-\frac{4 x^2}{15}-\frac{4 x^4}{105}-\frac{4 x^6}{315}-\frac{4 x^8}{693}-\frac{4 x^{10}}{1287}+O\left(x^{12}\right)$$ Back to $t$ $$A=\frac 43-4\left(\frac{ t}{15}+\frac{t^2}{105}+\frac{t^3}{315}+\frac{ t^4}{693}+\frac{t^5}{1287} \right)++O\left(t^{6}\right)$$ The denominators seem to be what you need.

I must confess that I did not find a way to prove it without using a truncated series and finding a pattern for the coefficients.