<s>[INST] Taylor expansion of $\frac{ \cos( \ln (n+1))}{(n+1)^{1/2}}$ Taylor Series Expansion of$$\frac{ \cos( \ln (n+1))}{(n+1)^{1/2}}$$
My try-
$$\cos(\ln (n+1))= \cos\left (\ln \ n+ \ln\left(1+\frac{1}{n}\right)\right)$$$$=\cos (\ln\  n) \cos\left(\ln\left(1+\frac{1}{n}\right)\right)-\sin(\ln \ n)\sin\left(\ln \left(1+\frac{1}{n}\right)\right) $$
How to proceed?
 [/INST] Near $x=0$, we have
$$
\log (x + 1) = x - \frac{1}{2}x^2  + \frac{1}{3}x^3  -  \cdots ,
$$
$$
\cos (x) = 1 - \frac{1}{2}x^2  +  \cdots
$$
and
$$
\frac{1}{{\sqrt {x + 1} }} = 1 - \frac{1}{2}x + \frac{3}{8}x^2  -\frac{5}{16}x^3 + \cdots .
$$
Thus
\begin{align*}
& \frac{{\cos (\log (x + 1))}}{{\sqrt {x + 1} }} = \\ &= \left( {1 - \frac{1}{2}\left( {x - \frac{1}{2}x^2  + \frac{1}{3}x^3  -  \cdots } \right)^2  +  \cdots } \right)\left( {1 - \frac{1}{2}x + \frac{3}{8}x^2  - \frac{5}{{16}}x^3  +  \cdots } \right)
\\ &
 = 1 - \frac{1}{2}x - \frac{1}{8}x^2  + \frac{7}{{16}}x^3  +  \cdots .
\end{align*}
</s>