Evaluate $\lim\limits_{n \to\infty} \left(\frac{n-1} {2n+2}\right)^n$ What is the easiest way to evaluate this limit?
$\displaystyle{\lim_{n \to\infty} \left(n-1 \over 2n+2\right)^n}$
$$
\text{Is this possible ?$\,$:}\quad
\lim_{n \to\infty}\left(n/n - 1/n \over 2n/n + 2/n\right)^n
=
\lim_{n \to\infty}\left(1 - 1/n \over 2 + 2/n\right)^{n}
=
\lim_{n \to\infty} \left(1 \over 2\right)^{n} = 0
$$
 A: The limit can be written as $$\frac{\left(\lim_{n\to\infty}\left(1-\frac1n\right)^{-n}\right)^{-1}}{\lim_{n\to\infty} 2^n\cdot \lim_{n\to\infty}\left(1+\frac1n\right)^n}$$
Now use $$\lim_{m\to\infty}\left(1+\frac1m\right)^m=e$$
A: $0\le\left(\dfrac{n-1}{2n+2}\right)^n=\dfrac{1}{2^n}\times\dfrac{\left(1-\dfrac{1}{n}\right)^n}{\left(1+\dfrac{1}{n}\right)^n}\le\dfrac{1}{2^n}\times\dfrac{1}{\left(1+\dfrac{1}{n}\right)^n}\to0$ since $\left(1+\dfrac{1}{n}\right)^n\to e$
So by squeezing $\displaystyle{\lim_{n \to\infty} \left(n-1 \over 2n+2\right)^n}=0.$
A: $\newcommand{\+}{^{\dagger}}%
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$$
\color{#0000ff}{\large\lim_{n \to\infty} \left(n - 1 \over 2n + 2\right)^{n}}
=
\lim_{n \to\infty}\bracks{{1 \over 2^{n}}\,
{\pars{1 - 1/n}^{n} \over \pars{1 + 1/n}^{n}}}
=
\lim_{n \to\infty}\pars{{1 \over 2^{n}}\,{\expo{-1} \over e}}
= \color{#0000ff}{\large 0}
$$
A: For simple limits like this, the easiest way is often to sandwich the sequence between simpler sequences for which the limit is known.
Notice that $n-1 \leq n$ and $2n+2 \geq 2n$ entail
$$
\frac{n-1}{2n+2} \leq \frac{1}{2}
$$
so that for $n \geq 1$ one has
$$0 \leq \left(\frac{n-1}{2n+2}\right)^n \leq \left(\frac{1}{2}\right)^n \xrightarrow[n\to\infty]{} 0.$$
Conclude with the squeeze theorem.
Remark. This method gives more than just the limit:  you get a good estimate of how quickly the sequence converges. For example, it is now easy to give a $n$ such that $\left(\frac{n-1}{2n+2}\right)^n \leq 10^{-100}$.
A: $$\frac{n-1}{2n+2} = \frac{1}{2}\frac{n+1}{n+1}-\frac{1}{2}\frac{2}{n+1}$$
$$\lim _{n \rightarrow \infty}(\frac{1}{2}-\frac{1}{n+1})^n$$
which is less than your $\frac 1 2$ so it does go to $0$
A: Another approach:
$$\left(\frac{n-1}{2n+2}\right)^n=\frac1{2^n}\left(1-\frac2{n+1}\right)^n=$$
$$=\frac1{2^n}\left(1-\frac2{n+1}\right)^{n+1}\left(1-\frac2{n+1}\right)^{-1}\xrightarrow[n\to\infty]{}0$$
since the second factor's limit above is $\;e^{-2}\;$ , and the third's is $\;1\;$
A: You can calculate this limit easily by using the root test:
First, we have to make sure that the sequence is bigger or equal to 0 for all n, i think it is quite easy to see it and then:
$$a_n = \left(\frac{n-1}{2n+2}\right)^n$$
$$(a^n)^{1/n}=\frac{n-1}{2n+2} \rightarrow \frac{1}{2}$$
Therefore, by the root test:
$$a_n \rightarrow 0$$
You can find about the root test here: http://en.wikipedia.org/wiki/Root_test, though in this link, it looks more complicated.
