What is the technique to determine $\sum_{n=1}^{\infty} M_n$ if the given series of function is a trigonometric series with $n$ as power?

Question

Two series of functions are given in which I cannot figure out how to find $M_n$ of the second problem. $$1.\space \sum_{n=1}^{\infty} \frac{1}{1+x^n}, x\in[k,\infty)\\ 2. \space \sum_{n=1}^{\infty} (\cos x)^n, x\in(0,\pi)$$..

I have determined the $M_n$ for problem no. $1.$ [$\space|\sum_{n=1}^{\infty} \frac{1}{1+x^n}|<|\sum_{n=1}^{\infty} \frac{1}{1+k^n}|<\sum_{n=1}^{\infty} \frac{1}{k^n}$]

From problem no. $2.$, since $-1\leq \cos x\leq1$, therefore for higher $n$ the values of $\cos x$ will lie between $[-1,1]$ and in $(0,\pi)$ $\cos x$ is decreasing. But is it correct to choose $n$ as $M_n$, so that $$|f_n(x)|=|(\cos x)^n|<n,$$ where $n$ is decreasing.

I am not sure what the $M_n$ should be. Any help or suggestion please? Any help is greatly appreciated.

Answer 1

I presume you are looking for bounds on the $n$-th term, in the hope of applying Weierstrass's M-test. Alas, the best bound you can get is $M_n=1$.

Answer 2

Geometric series? \begin{align*} \sum_{n=1}^{\infty}(\cos x)^{n}=\dfrac{\cos x}{1-\cos x}. \end{align*} Note that $|\cos x|<1$ for $x\in(0,\pi)$.