Convergence of $\sum_{n=1}^{\infty}\bigg((2\lambda)^{3n}\ \ln\big(\cos(n^{\lambda})\big)\bigg)$, for $\lambda\leqslant0, \lambda\in\mathbb{R}$ Study the convergence of the following series for $\lambda\leqslant0, \lambda\in\mathbb{R}$
$$\sum_{n=1}^{\infty}\bigg((2\lambda)^{3n}\ \ln\big(\!\cos(n^{\lambda})\big)\bigg)$$
I observed that if $-\frac{1}{2}\lt\lambda\lt0$ then $(2\lambda)^{3n}$ is the geometric series with ratio $-1\lt r \lt0$ that converges, whereas for $\lambda\lt -\frac{1}{2}$ we have $r\lt-1$ that diverges. In particular when $\lambda\lt -\frac{1}{2}$, the $(2\lambda)^{3n}$ is an oscillating series.
I am not sure how to handle $b_n=\ln\big(\!\cos(n^{\lambda})\big)$ which tend to $0$ for $n\mapsto\infty$. Giving $k=-\lambda\gt0$, can I say that $b_n= O\bigg(\frac{1}{n^k}\bigg)$ for $n \mapsto \infty$? Or am I wrong, having lost too much informations?
 A: 
I am not sure how to handle $b_n=\ln\left(\cos(n^{\lambda})\right)$ which tends to $0$ for $n\to\infty$. 
  Giving $k=-\lambda\gt0$, can I say that $b_n= O\bigg(\frac{1}{n^k}\bigg)$ for $n \to \infty$?

You rather have $\displaystyle b_n= O\bigg(\frac{1}{n^{2k}}\bigg)$. One may use Taylor series expansions to see this, noticing that, as $n \to \infty$,
$$
\cos \left(\frac1{n^k} \right)=1-\frac1{2n^{2k}}+o\bigg(\frac{1}{n^{2k}}\bigg)
$$$$
\ln \left(\cos \left(\frac1{n^k} \right)\right)=\ln\left(1-\frac1{2n^{2k}}+o\bigg(\frac{1}{n^{2k}}\bigg)\right)=-\frac1{2n^{2k}}+o\bigg(\frac{1}{n^{2k}}\bigg)=O\bigg(\frac{1}{n^{2k}}\bigg).
$$
A: Solution. $\blacktriangleleft$ First determine the radius of convergence for
$$
\sum y^n \log(\cos (n^\lambda)).
$$
By Cauchy-Hadamard formula, we need to calculate the following:
\begin{align*}
\left| \log (\cos (n^\lambda)) \right|^{1/n} &= \exp \left( \frac 1n \log (- \log (\cos (n^\lambda)))\right) \\
&= \exp \left( \frac 1n \log (-\log (1 - n^{2\lambda}/2 + o(n^{2\lambda})))\right)\\
&= \exp \left( \frac 1n \log (n^{2\lambda} + o(n^{2\lambda}))\right).
\end{align*}
For sufficiently large $n$ we can assume that 
$$
\frac 12 n ^{2\lambda} \leqslant n^{2\lambda} + o(n^{2\lambda}) \leqslant \frac 32 n^{2\lambda}.
$$
Since
\begin{align*}
\lim_{x\to +\infty} \frac {\log (x^{2\lambda})} x &= \lim_{x \to+ \infty } \frac {2\lambda x^{2\lambda -1}/x^{2\lambda}} 1 [\mathsf{ L'hopital\ rules}] \\
&= \lim_{x\to +\infty } \frac {2\lambda} x \\
&= 0,
\end{align*}
we deduce that
$$
\lim_n \left| \log (\cos (n^\lambda)) \right|^{1/n} = 1.
$$
Therefore the power series converges when $|y| <1$, diverges when $|y|>1$.
Now return to the original question. Using the convergent radius, we know that the series converges when $|8\lambda^3| < 1$, i.e. $\boldsymbol \lambda \in (-\mathbf 1/\mathbf 2, \mathbf 0]$; diverges when $\boldsymbol \lambda < \mathbf{-1/2}$. When $\lambda = {-1/2}$, the series becomes
$$
\sum (-1)^n \log (\cos(n^{-1/2})) =\colon \sum (-1)^{n-1} a_n
$$
which is a Leibniz series:
$$
\frac 1{\sqrt n} > \frac 1{\sqrt {n+1}}  \implies a_n > a_{n+1} \searrow 0,
$$
hence it converges.
Conclusion: the series diverges when $\boldsymbol \lambda < -\mathbf{1/2}$, converges when $\boldsymbol \lambda \in [\mathbf{-1/2}, \mathbf 0]$. $\blacktriangleright$
