Let $K$ be a field. Show that any element of the form $1+m$ with $m\in (t_1,\ldots, t_n)$ is a unit in $K[[t_1,\ldots,t_n]]$
I'd say that $\sum_{n=0}^\infty (-1)^nm^n$ is the inverse of $1+m$, and hence $1+m$ is a unit. But I need to show that $\sum_{n=0}^\infty (-1)^nm^n=\frac{1}{1+m}$. To do this, I was thinking of using the definition of convergence as given below.
Let $A=K[[t_1,\ldots,t_n]]$ consider the ideal $(t_1,\ldots, t_n)$ in $A$. Let $\{a_n\}$ be a collection of elements in $A$. We say that $a_n$ converges to $a\in A$ if for all $N>0$ there exists some $M>0$ such that for any $m>M$ we have $a_m-a\in (t_1,\ldots, t_n)^N$.
I considered the coefficients of $\sum_{n=0}^\infty (-1)^nm^n$ as the collection $\{a_n\}$. So I'd say that for large enough $m$, then $(-1)^m-\frac{1}{1+m}\in (t_1,\ldots,t_n)^N$ since $(-1)^m-\frac{1}{1+m}$ is zero in $K[[t_1,\ldots,t_n]]/(t_1,\ldots,t_n)$, since $m$ is in $(t_1,\ldots,t_n)$. Is the valid, and is the approach correct to show that $1+m$ is a unit?
As pointed out in the comment, you just have to compute \begin{align} (1+m)\sum_{n\geq 0}(-1)^nm^n&=\sum_{n\geq 0}(-1)^nm^n+\sum_{n\geq 0}(-1)^nm^{n+1}\\ &=1+\sum_{n\geq 1}(-1)^nm^n+\sum_{n\geq 0}(-1)^nm^{n+1}\\ &=1+\sum_{n\geq 1}(-1)^nm^n+\sum_{n\geq 1}(-1)^{n-1}m^n\\ &=1+\sum_{n\geq 1}[(-1)^n+(-1)^{n-1}]m^n\\ &=1. \end{align}