Answer
$ \dfrac{-1}{4}$
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:
$S=\lim\limits_{n \to \infty} \dfrac{-1}{4} (1+ \dfrac{1}{4n+3}) \\=\dfrac{-1}{4} [
\lim\limits_{n \to \infty} 1+\lim\limits_{n \to \infty} \dfrac{1}{4n+3})]\\=\dfrac{-1}{4} [
\lim\limits_{n \to \infty} 1+\lim\limits_{n \to \infty} \dfrac{1}{4+3/n})]$
Because for all $n \gt 0$ , $\lim\limits_{x \to \infty} x^{1/n}=0$ and $\lim\limits_{x \to -\infty} x^{n}=\infty$
Thus, $S= \dfrac{-1}{4}-0= \dfrac{-1}{4}$