# Taylor Series to the Power 1/z

I am attempting to find the Taylor Series for $(\frac{\sin{z}}{z})^{\frac{1}{z^2}}$. While I can plug this into Wolfram and use the output, I want to understand how to calculate the Taylor Series myself. I can find the Taylor Series for $\frac{\sin{z}}{z}$ easily enough, but I cannot figure out for the power $\frac{1}{z^2}$ plays into things.

I just need to find the first few terms so that I can evaluate the limit, I do not need the nth term.

Write $$\Bigl(\frac{\sin z}{z}\Bigr)^{1/z^2}=\exp\Bigl(\frac{1}{z^2}\,\log\Bigl(\frac{\sin z}{z}\Bigr)\Bigr)$$ I would first find a few terms of $$\frac{1}{z^2}\,\log\Bigl(\frac{\sin z}{z}\Bigr)=\frac{1}{z^2}\,\log\Bigl(1-\frac{z^2}{3!}+\frac{z^4}{5!}-\frac{z^6}{7!}+\dots\Bigr)$$ using $$\log(1+w)=w-\frac{w^2}{2}+\frac{w^3}{3}-\frac{w^4}{4}+\dots$$ and then use the Taylor expansion of $e^w$.

1. The formula 4.3.71 on page 75 in the handbook [1] below reads that $$\ln\frac{\sin z}{z}=\sum_{n=1}^\infty\frac{(-1)^n2^{2n-1}B_{2n}}{n(2n)!}z^{2n}, \quad |z|<\pi.$$ Hence, we have $$\frac1{z^2}\ln\frac{\sin z}{z}=\sum_{n=0}^\infty\frac{(-1)^{n+1}2^{2n+1}B_{2n+2}}{(n+1)(2n+2)!}z^{2n}, \quad |z|<\pi.$$

[1] M. Abramowitz and I. A. Stegun (Eds), Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, National Bureau of Standards, Applied Mathematics Series 55, 10th printing, Dover Publications, New York and Washington, 1972.

1. Let $$\begin{equation*} D(x)=\sum_{k=0}^\infty d_kx^k \end{equation*}$$ be a power series expansion. Then the function $$E(x)=e^{D(x)}$$ has the power series expansion $$\begin{equation*} E(x)=\sum_{k=0}^\infty e_kx^k, \end{equation*}$$ where the coefficients $$e_k$$ for $$k\in\{0\}\cup\mathbb{N}$$ satisfy \begin{align} e_0&=e^{d_0},\\ e_k&=\frac1k\sum_{\ell=1}^k\ell d_\ell e_{k-\ell} =\frac1k\sum_{\ell=0}^{k-1}(k-\ell)d_{k-\ell}e_{\ell},\quad k\in\mathbb{N},\\ e_n&=e^{d_0}\left(d_n+\sum_{j=1}^{n-1}\sum_{\substack{\sum_{i=0}^jm_i=n,\\ m_i\ge1, 0\le i\le j}} \prod_{i=0}^j\frac{m_id_{m_i}}{n-\sum_{q=0}^{i-1}m_q}\right), \quad n\in\mathbb{N}, \end{align} and $$$$\label{alpha-k-power-eq} e_k= e^{d_0}\sum_{j=1}^k\frac1{j!} \sum_{\substack{\sum_{\ell=1}^ji_\ell=k,\\ i_\ell\ge1,1\le\ell\le j}} \prod_{\ell=1}^jd_{i_\ell}, \quad k\in\mathbb{N}.$$$$ These conclusions can be found in the paper [2] below.

[2] Feng Qi, Xiao-Ting Shi, and Fang-Fang Liu, Expansions of the exponential and the logarithm of power series and applications, Arabian Journal of Mathematics 6 (2017), no. 2, 95--108; available online at https://doi.org/10.1007/s40065-017-0166-4.

1. Consequently, we finally can obtain \begin{align} \biggl(\frac{\sin z}{z}\biggr)^{1/z^2} &=\exp\Biggl(\sum_{n=0}^\infty\frac{(-1)^{n+1}2^{2n+1}B_{2n+2}}{(n+1)(2n+2)!}z^{2n}\Biggr)\\ &=\frac{1}{\sqrt[6]{e}}-\frac{z^2}{180 \sqrt[6]{e}}-\frac{17 z^4}{50400 \sqrt[6]{e}}-\frac{6007 z^6}{244944000 \sqrt[6]{e}}-\frac{26263141 z^8}{13579695360000 \sqrt[6]{e}}-\dotsm \end{align}