Error term in an infinite series

Question

Well, I've the following series:

$$\frac{1}{\left(1+\epsilon\cos\left(x\right)\right)^2}=\sum_{\text{k}=0}^\infty\epsilon^\text{k}\left(1+\text{k}\right)\cos^\text{k}\left(x\right)\tag1$$

This series converges for all values of $x$, because:

$$\lim_{\text{k}\to\infty}\left|\frac{\epsilon^{\text{k}+1}\left(1+\left(\text{k}+1\right)\right)\cos^{\text{k}+1}\left(x\right)}{\epsilon^\text{k}\left(1+\text{k}\right)\cos^\text{k}\left(x\right)}\right|=\epsilon\left|\cos\left(x\right)\right|<1\tag2$$

For all $x$.

Question: What is the consequence for the error to the true value of the series, when I add a term in the series. So when I only use $\text{k}=0$ the error is $\text{E}_1$ and when I use $\text{k}=0,\text{k}=1$ the error is $\text{E}_2$, but what will the error term look like when I use $\text{n}$ terms (so $\text{E}_\text{n}$)?

Answer 1

Let $f(t)=(1+t^2)^{-1}$. Then $$ f(t)=\sum_{n=0}^\infty(-1)^n(n+1)\,t^n,\quad|t|<1. $$ The Lagrange form of the error when approximating $f(t)$ by the $n$-th Taylor polynomial is $$ e_n=\frac{f^{(n+1)}(\tau)}{(n+1)!}\,t^{n+1}, $$ where $\tau$ is between $0$ and $t$. Since $f^{(k)}(t)=(-1)^k(k+1)!(1+t)^{-(k+2)}$, we obtain the upper bound $$ |e_n|\le(n+2)\,|t|^{n+1}. $$ When applied with $t=\epsilon\cos x$ we obtain $$ |E_n|\le(n+2)\,\epsilon^{n+1}|\cos^{n+1}x|. $$