Using $$\int_{0}^{1}{x^{x}}dx = \sum_{n = 1}^\infty{\frac{{(-1)^{n-1}}}{{n^{n}}}}.$$
Estimate the Integral with an error of magnitude $<10^{-3}$
What i try
Put $n=1$.Then sum is $\displaystyle 1$
Put $n=2$. Then sum is $\displaystyle 1-\frac{1}{4}=\frac{3}{4}.$
Put $n=3$. Then sum is $\displaystyle 1-\frac{1}{4}+\frac{1}{3^3}=\frac{3}{4}+\frac{1}{27}\approx 0.78$
Put $n=4$. Then sum is $\displaystyle 1-\frac{1}{4}+\frac{1}{3^3}-\frac{1}{256}=\frac{3}{4}+\frac{1}{27}\approx 0.77$
How do i solve it Help me please, Thanks