# 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, $$\sum_{n=0}^{\infty}\frac{\sin n}{n!}\leq \sum_{n=1}^{\infty}\frac{(-1))^{n}}{n}$$ ?

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Use the upper bound $\sin x \leq 1$ for all $x$. – A Walker Apr 30 '13 at 1:08
Maybe a little more of an explanation that that please. So we know that $$-1\leq sin(x)\leq 1$$ – EhBabay Apr 30 '13 at 1:12

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&=e^0+e^0x+\frac{e^0x^2}2+\frac{e^0x^3}{3!}+\dots\\&=1+x+\frac{x^2}2+\frac{x^3}{3!}+\dots\\&=\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!}&=Im\left[\sum_{n=1}^{\infty}\dfrac{(e^{ix})^n}{n!}\right]\\ &=Im\left[\sum_{n=0}^{\infty}\dfrac{(e^{ix})^n}{n!}\right]=Im(e^{e^{ix}})\\ &=Im(e^{\cos{x}+i\sin{x}})\\ &=e^{\cos{x}}\sin{\sin{x}} \end{align} so $$\sum_{n=1}^{\infty}\dfrac{\sin{n}}{n!}=e^{\cos{1}}\sin{\sin{1}}<e^{1}\cdot 1=e$$