$$1-\frac{1}{4}+\cdots +\frac{(-1)^{n+1}}{3n-2}+\cdots$$
Sorry for being silly. I thought this $\frac{1}{3n-2}=\frac{1}{n+1}$ so if $n=3/2$.
Then $$\sum_{n=1}^{\infty } (-1)^{n+1} \int_{0}^{1} \frac{1}{2}x^{\frac{3}{2}n} dx=\sum_{n=1}^{\infty }\frac{(-1)^{n+1}}{3n-2}$$
But, I don't know if they are asking me this and I don't know how to fix it.
Transform $$\frac{1}{3n-2} = \int_0^1 x^{3n-3} \mathrm d x$$ and $$(-1)^{n+1}=(-1)^{3n-3}$$ Thus $$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{3n-2} = \sum_{n=1}^{\infty} (-1)^{n+1} \int_0^1 x^{3n-3} \mathrm d x = \int_0^1 \sum_{n=1}^{\infty} (-1)^{n+1} x^{3n-3} \mathrm d x =$$ $$=\int_0^1 \sum_{n=1}^{\infty} (-x)^{3n-3} \mathrm d x = \int_0^1 \frac{1}{1+x^3}\mathrm d x$$ note that we can swap sum and integral because the geometric series is uniformly convergent on compact sets inside $(-1; 1)$.
Now, you have to compute this integral, which is not that easy. WA says it evaluates $$\frac{\pi}{3 \sqrt 3} + \frac{\log 2}{3} \approx 0.835$$