Find a partial sum of the series $\sum_{n=0}^{+\infty}{(-1)^n\frac{1}{(2n+1)!(4n+3)}}$ which approximates the value of the integral $\int_{0}^{1}\sin(x^2)\,dx$ with 3 decimal digits accuracy.
My thought: $(-1)^n\frac{1}{(2n+1)!(4n+3)}<10^{-3}$ and then you solve for $n$ to find how many terms of the sum you need.
On
I'm not sure that the term "to three decimal digits accuracy" has a standard, commonly accepted definition in math, although I can tell you for certain that you need to take the absolute value of the error. I would take it to mean that the absolute error is $< 5 \times 10^{-4}.$
Since the sum is an alternating series whose terms decrease in absolute value and go to $0$, the error does not exceed the first omitted term. Your general idea is correct, but be sure to specify how you are interpreting the phrase "to three decimal digits accuracy," or ask your teacher, if you have an opportunity before the assignment is due.
$$\int_{0}^{1}\sin(x^2)\,dx=\sum_{n\geq 0}\int_{0}^{1}\frac{(-1)^n}{(2n+1)!}x^{4n+2}\,dx =\sum_{n\geq 0}\frac{(-1)^n}{(4n+3)\cdot(2n+1)!}$$ and $(4n+3)\cdot(2n+1)!$ at $n=3$ is already larger than $75\cdot 10^3$, so the LHS is well approximated by $$ \sum_{n=0}^{2}\frac{(-1)^n}{(4n+3)(2n+1)!}=\frac{2867}{9240}=\color{green}{0.3102}81\ldots $$ within an error of $5\cdot 10^{-5}$. An alternative approach is to manipulate the integral directly. We have $$ \mathcal{J}=\int_{0}^{1}\sin(x^2)\,dx = \int_{0}^{1}\frac{\sin(z)}{2\sqrt{z}}\,dz $$ and by integration by parts $$ \mathcal{K}=\int_{0}^{1}x^4(1-x^4)^3\sin(x^2)\,dx = \frac{3}{256}\left[539107\sin(1)-426722\cos(1)-718999 \mathcal{J}\right].$$ Since $\mathcal{K}\leq\frac{1}{50}$ due to $\sin(x^2)\leq x^2$, we have $$\mathcal{J}\approx \frac{539107\sin(1)-426722\cos(1)}{718999}=\color{green}{0.3102}70\ldots $$ still with four correct figures.