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1. (a) Show that $a_n\geq n$.  (b) Find $N$ s.t. $|\frac{a_n}{a_n+1}-1|\textless \varepsilon$.

[Solution] (a) We prove it by induction. Since $a_1$ is a positive integer, we know that $a_1\geq 1$. Assume that $k\in \mathbb{Z}_+$ and $a_k\geq k$. Since $(a_n)$ is a strictly increasing sequence of positive integers, we get $a_{k+1}-a_k\geq 1$. So 
\begin{align}
a_{k+1} = (a_{k+1}-a_k) + a_k \geq 1+ k = k+1\notag
\end{align}
Therefore, induction tells us that $a_n \geq n$ for positive $n$.

(b) We are asked to solve the inequality $|\frac{a_n}{a_n+1}-1|\textless \varepsilon$. The solution is 
\begin{align}
a_n\textgreater \frac{1}{\varepsilon}-1
\end{align}
We hope to find suffuciently large $n$'s satisfying (1). It suffices to make $n$ large enough so that 
\begin{align}
n\textgreater \frac{1}{\varepsilon} - 1 \notag
\end{align}
We might choose $N_\varepsilon=[\frac{1}{\varepsilon}]+1$. Then for $n\geq N_\varepsilon$, by (a) we find that 
\begin{align}
&a_n \geq n \geq N_\varepsilon = [\frac{1}{\varepsilon}]+1 \geq \frac{1}{\varepsilon} \textgreater \frac{1}{\varepsilon}-1 \notag\\
&\Rightarrow -\varepsilon \textless 0 \textless \frac{1}{a_n +1} \textless \varepsilon\notag
\end{align}
Hence, this $N_\varepsilon$ does work. 


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[Note] 

By (a) we're expected to use the form $a_{k+1} = (a_{k+1} - a_{k}) + a_{k}$, and to be familiar with properties about positive integers, i.e. the fact that $|n-m|\geq 1$ whenever $n\ne m$. The proof by mathematical induction is an ideal way due to preciseness. 

(b) Many people might  find that we have no idea about what $a_n$ is  at the first sight, but part (a) is a helpful message. In fact, we are suggested to solve it on  $a_n$, and then find relations between $a_n$ and $n$. 

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2. Show $|\sin(nx)|\leq n\sin x$.

[Solution]
According to the expression $\sin \theta = \frac{e^{i\theta} - e^{-i\theta}}{2i}$. Substititing $nx$ to $\theta$, we get 
\begin{align}
|\sin ( nx)| &=  | \frac{e^{inx} - e^{-inx}}{2i} | =  | \frac{(e^{ix})^n - (e^{-ix})^n}{2i} |\notag\\
& = |  \frac{e^{ix} - e^{-ix}}{2i} | |  (e^{ix})^{n-1}+ (e^{ix})^{n-1}  (e^{-ix})^{1} + .... + (e^{-ix})^{n-1}|\notag\\
&\leq \sin x \cdot (|  (e^{ix})^{n-1}|+| (e^{ix})^{n-1}  (e^{-ix})^{1}| + .... + |(e^{-ix})^{n-1}|)\notag\\
& = \sin x \cdot (1+1+...+1)= n\sin x\notag
\end{align}

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[note] The key point is to properly use the  exponential expression, and to factorize. What to prove will emerage after application of triangular inequality. 
[Note] If you think the "notation" $e^{ix}$ cumbersome, you might as well alternatively define the function  $e(x)$ by
\begin{align}
e(x) = \cos x  + i\sin x \notag
\end{align}
Then we have (i) $(e(x))^n = e(nx)$, and (ii) $\sin\theta = \frac{e(x) - e(-x)}{2i}$. Hence we may do the same thing to prove this problem without application of "complex exponentiation". 





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