Prove that $|\int_{0}^{1}e^{x^2}dx-\sum_{k=0}^{n-1}\frac{1}{(2k+1)k!}| \leq \frac{3}{n!}$

Question

Prove that $$|\int_{0}^{1}e^{x^2}dx-\sum_{k=0}^{n-1}\frac{1}{(2k+1)k!}| \leq \frac{3}{n!}$$ for all $n \in \mathbb{N}$

I find the Maclaurin expansion of $$e^{x^2}=\sum_{k=0}^\infty \frac{x^{2k}}{k!}$$ and $$\int_{0}^{1}e^{x^2}dx=\int_{0}^{1}\sum_{k=0}^\infty \frac{x^{2k}}{k!}dx=\sum_{k=0}^\infty \frac{1}{(2k+1)k!}=\sum_{k=0}^{n-1} \frac{1}{(2k+1)k!}dx+\frac{1}{(2n+1)n!}$$ so $$\int_{0}^{1}e^{x^2}dx-\sum_{k=0}^{n-1}\frac{1}{(2k+1)k!}=\frac{1}{(2n+1)n!}$$ How can I explain that $|\frac{1}{(2n+1)n!}| \leq \frac{3}{n!}$?

Answer 1

The last equality is wrong : $$\int_{0}^{1}e^{x^2}dx=\int_{0}^{1}\sum_{k=0}^\infty \frac{x^{2k}}{k!}dx=\sum_{k=0}^\infty \frac{1}{(2k+1)k!}=\sum_{k=0}^{n-1} \frac{1}{(2k+1)k!}dx+\sum_{k=n}^{\infty}\frac{1}{(2k+1)k!}$$ So : $$\int_{0}^{1}e^{x^2}dx-\sum_{k=0}^{n-1}\frac{1}{(2k+1)k!}=\sum_{k=n}^{\infty}\frac{1}{(2k+1)k!}=\sum_{k=0}^{\infty}\frac{1}{(2k+2n+1)(k+n)!}\\=\frac1{n!}\sum_{k=0}^{\infty}\frac{1}{(2k+2n+1)(n+1)(n+2)\dots(n+k)}$$ But : $$(n+1)(n+2)\dots(n+k)\geq1.2\dots k=k!$$ $$(2k+2n+1)(n+1)(n+2)\dots(n+k)\geq k!$$ So : $$\frac1{n!}\sum_{k=0}^{\infty}\frac{1}{(2k+2n+1)(n+1)(n+2)\dots(n+k)}\leq \frac1{n!}\sum_{k=0}^{\infty}\frac{1}{k!}=\frac{e}{n!}$$ Finally : $$\int_{0}^{1}e^{x^2}dx-\sum_{k=0}^{n-1}\frac{1}{(2k+1)k!}\leq \frac{e}{n!}\leq\frac3{n!}$$

Answer 2

$$\int_{0}^{1}e^{x^2}\,dx-\sum_{k=0}^{n-1}\frac{1}{(2k+1)k!}=\color{red}{\sum_{k\geq n}\frac{1}{(2k+1)k!}}\tag{A}$$ and since $$ \frac{1}{2k!}-\frac{1}{2(k+1)!} = \frac{k}{(2k+2)k!}\geq\frac{1}{(2k+1)k!}\tag{B} $$ for any $k\geq 2$, we have $$ \color{red}{0} \leq \left[\int_{0}^{1}e^{x^2}\,dx-\sum_{k=0}^{n-1}\frac{1}{(2k+1)k!}\right] \leq \color{red}{\frac{1}{2n!}}\tag{C} $$ for any $n\geq 2$.

Answer 3

That is clear since $1 \le 3 (2n+1)$.