Since that is an alternating series, you know that the "n+1" sum lies between the "n-1" sum and "n" sum. One of the things that means is that the infinite sum lies between any two consecutive terms. If two terms have a difference of less than .01, either of them is within .01 of the infinite sum.
If $\displaystyle a_n>0$ and $\displaystyle \sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^n}{a_n}} = S$ and $\displaystyle S_n=\sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^n}{a_n}} $ is a partial sum then $\displaystyle \left|S_n-S\right|<a_{n+1}$.
So make $\displaystyle \frac{\sqrt{n+1}}{2n+3}<0.01$