Infinite series question requiring no explanation Determine if the statement is true or false. No explanation needed.

$$\sum_{n=0}^{\infty}\frac{\sin n}{n!}\leq e$$

Although no explanation is needed I was wondering how you would approach this problem in the first place. Could I possibly use a comparison test of the infinite series? Would possibly a start would be,
$$\sum_{n=0}^{\infty}\frac{\sin n}{n!}\leq \sum_{n=1}^{\infty}\frac{(-1))^{n}}{n}$$ ?
We have $\sin x\le1$ so it follows: $$\sum_{n=0}^\infty\frac{\sin n}{n!}\le\sum_{n=0}^\infty\frac1{n!}$$
Given that we know $\sum_{n=0}^\infty\frac1{n!}=e$ we're done.



If you didn't know that, consider the Maclaurin series expansion of $e^x$ at $x=0$:
$$\begin{align*}e^x&amp;=e^0+e^0x+\frac{e^0x^2}2+\frac{e^0x^3}{3!}+\dots\\&amp;=1+x+\frac{x^2}2+\frac{x^3}{3!}+\dots\\&amp;=\sum_{n=0}^\infty\frac{x^n}{n!}\end{align*}$$... since $\frac{d^n}{dx^n}e^x=e^x$. Now observe at $x=1$:
$$e^1=\sum_{n=0}^\infty\frac{1^n}{n!}\\e=\sum_{n=0}^\infty\frac1{n!}$$
It's very easy show that
$$\sum_{n=1}^{\infty}\dfrac{\sin{nx}}{n!}=e^{\cos{x}}\sin{(\sin{x})}$$
proof:
\begin{align}
\sum_{n=1}^{\infty}\dfrac{\sin{nx}}{n!}&amp;=Im\left[\sum_{n=1}^{\infty}\dfrac{(e^{ix})^n}{n!}\right]\\
&amp;=Im\left[\sum_{n=0}^{\infty}\dfrac{(e^{ix})^n}{n!}\right]=Im(e^{e^{ix}})\\
&amp;=Im(e^{\cos{x}+i\sin{x}})\\
&amp;=e^{\cos{x}}\sin{\sin{x}}
\end{align}
so
$$\sum_{n=1}^{\infty}\dfrac{\sin{n}}{n!}=e^{\cos{1}}\sin{\sin{1}}&lt;e^{1}\cdot 1=e$$