Infinite Limit Problems Can some one help me to solve the problems?


*

*$$\lim_{n\to\infty}\sum_{k=1}^n\left|e^{2\pi ik/n}-e^{2\pi i(k-1)/n}\right|$$

*$$ \lim_{x\to\infty}\left(\frac{3x-1}{3x+1}\right)^{4x}$$

*$$\lim_{n\to\infty}\left(1-\frac1{n^2}\right)^n $$
(Original scan of problems at http://i.stack.imgur.com/4t12K.jpg)
This questions are from ISI kolkata Computer science Phd exam entrance.The link of the original questions http://www.isical.ac.in/~deanweb/sample/MMA2013.pdf
 A: The first problem is a simple geometry problem in disguise. 
For the second one, rewrite our expression as
$$\left(\frac{(1-\frac{1}{3x})^x}{(1+\frac{1}{3x})^x}       \right)^4.$$
Note that $(1-\frac{1}{3x})^x\to e^{-1/3}$, and $(1+\frac{1}{3x})^x\to e^{1/3}$. 
For the third problem, rewrite the expression as
$$\left(\left(1-\frac{1}{n^2}\right)^{n^2}     \right)^{1/n}.$$
A: For the first one, consider a geometric interpretation.  Recall that when $a$ and $b$ are complex numbers, $|a-b|$ is the distance in the plane between the points $a$ and $b$. Peter Tamaroff says in a comment that the limit is $2\pi$, which I believe is correct.
Addendum: The points $e^{2\pi ik/n}$ are spaced evenly around the unit circle.  The distance $\left|e^{2\pi ik/n} - e^{2\pi i(k-1)/n}\right|$ is the distance from one point to the next.  So we are calculating the perimeter of a regular $n$-gon which, as $n$ increases, approximates the perimeter of the unit circle arbitrarily closely.  The answer is therefore $2\pi$. 
For the third one, factor $$\left(1-\frac1{n^2}\right)$$ as 
$$ \left(1-\frac1{n}\right) \left(1+\frac1{n}\right)$$ and then apply the usual theorem about $\lim_{n\to\infty}\left(1+\frac an\right)^n$.
A: To solve the first one, note $$\begin{align}
  \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\left| {{e^{2\pi ik/n}} - {e^{2\pi i\left( {k - 1} \right)/n}}} \right|}  &= \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\left| {{e^{2\pi ik/n}} - \frac{{{e^{2\pi ik/n}}}}{{{e^{2\pi i/n}}}}} \right|}  \cr 
   &= \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\left| {1 - \frac{1}{{{e^{2\pi i/n}}}}} \right|}  \cr 
   &= \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\left| {\frac{{{e^{2\pi i/n}} - 1}}{{{e^{2\pi i/n}}}}} \right|}  \cr 
   &= \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\left| {{e^{2\pi i/n}} - 1} \right|}  \cr 
   &= \mathop {\lim }\limits_{n \to \infty } n\left| {{e^{2\pi i/n}} - 1} \right| \cr 
   &= \left| {\mathop {\lim }\limits_{n \to \infty } \frac{{{e^{2\pi i/n}} - 1}}{{1/n}}} \right| \cr 
   &= \left| {\frac{d}{{dx}}{{\left. {{e^{2\pi ix}}} \right|}_{x = 0}}} \right| \cr 
   &= \left| {2\pi i} \right| \cr 
  & = 2\pi  \end{align}$$
A: Some slightly different approaches: For the first one
$$e^{2\pi ik/n}-e^{2\pi i(k-1)/n}=\cos\frac{2\pi k}n+i\sin\frac{2\pi k}n-\cos\frac{2\pi(k-1)}n-i\sin\frac{2\pi(k-1)}n=$$
$$=-2\sin\frac{(2k-1)\pi}n\sin\frac{\pi}n-2i\sin\frac\pi n\cos\frac{(2k-1)\pi}n\implies$$
$$\left|e^{2\pi ik/n}-e^{2\pi i(k-1)/n}\right|=\sqrt{4\sin^2\frac{(2k-1)\pi}n\sin^2\frac{\pi}n+4\sin^2\frac\pi n\cos^2\frac{(2k-1)\pi}n}=$$
$$=2\sin\frac\pi n\implies$$
$$\sum_{k=1}^n\left|e^{2\pi ik/n}-e^{2\pi i(k-1)/n}\right|=2n\sin\frac\pi n=2\pi\frac{\sin\frac\pi n}{\frac\pi n}\xrightarrow[n\to\infty]{}2\pi\cdot 1=2\pi$$
$$\color{red}{-------------o--------------}$$
For the second one we have
$$\left(\frac{3x-1}{3x+1}\right)^{4x}=\left[\left(1-\frac2{3x+1}\right)^{3x+1}\right]^{4/3}\cdot\left(1-\frac2{3x+1}\right)^{-4/3}\xrightarrow[x\to\infty]{}\left(e^{-2}\right)^{4/3}\cdot 1=e^{-8/3}$$
$$\color{red}{-------------o--------------}$$
For the last one we use the straightforward (try to produce the easy proof)
Lemma: If $\,\{a_n\}_{n\in \Bbb N}\;$ is a positive real sequence (enough to require "positive except perhaps a finite number of elements) and
$$\lim_{n\to\infty}a_n=L>0\;,\;\;\text{then}\;\;\lim_{n\to\infty}\sqrt[n]{a_n}=1$$
Thus we get
$$\left(1-\frac1{n^2}\right)^n=\sqrt[n]{\left(1-\frac1{n^2}\right)^{n^2}}\xrightarrow[n\to\infty]{}1$$
because
$$\left(1-\frac1{n^2}\right)^{n^2}\xrightarrow[n\to\infty]{}e$$
The above last problem, of course, is just what André wrote.
