Answer
$\dfrac{1}{9}$
Work Step by Step
Let us consider an infinite series $\Sigma_{n=k}^{n=\infty} a_n$. This series will converge to the sum $S$ when the sequence of its partial sums $\{S_n\}$ converge to $S$.
That is, $S=\lim\limits_{N \to \infty} S_n$ and $S=\Sigma_{n=k}^{n=\infty} a_n$.
This series will diverge when the limit does not exist and the series will diverge to infinity when the limit is infinite.
Here, we have the sum $S$ is the limit of the sequence of partial sums which is equal to:
$S=\lim\limits_{n \to \infty} S_n\\=\lim\limits_{n \to \infty} (\dfrac{1}{9} -\dfrac{1}{4n+1})\\=\lim\limits_{n \to \infty} \dfrac{1}{9}-\lim\limits_{n \to \infty} \dfrac{1}{4n+1}$
Because for all $n \gt 0$ , $\lim\limits_{x \to \infty} x^{-n}=0$ and $\lim\limits_{x \to -\infty} x^{n}=\infty$
Thus, $S=\dfrac{1}{9}$