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  1. Show that $\frac{1}{2} < \frac{1+n}{2n} < M$, $n\geq 2$, for some $M < 1$, and find $M$.
    So the lower bound is obvious just throw $n = 2$ into the fraction to get $\frac{3}{4} > \frac{1}{2}$ but how do I get $M$?
  2. Hence or otherwise find $\lim_{n\to\infty} \left(\frac{1+n}{2n}\right)^n$.
    Clearly it is $0$ because anything less than $1$ raised to a huge number approaches $0$.

Thank you.

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    $\begingroup$ $${1+n\over2n}={1\over2}+{1\over2n}$$ $\endgroup$ – Gerry Myerson Aug 13 at 23:43
  • $\begingroup$ 1) You can't assume $n=2$ is the lower limit unless you know it is an increasing function. Which it is not. In fact finding the M is the easy part. 2) You can't assume that $\frac {1+n}{2n}$ is constant so your argument that anything less than $1$ raised to a larger values approaches $0$ is not valid as for each $n$ it is a different number being raised to the $n$. $\endgroup$ – fleablood Aug 14 at 0:03
  • $\begingroup$ $\left(1-\frac1n\right)^n$ is something less than $1$ raised to a huge number, but it does not approach $0$. $\endgroup$ – robjohn Aug 14 at 1:21
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1) You can't assume $n=2$ is the lower limit unless you know it is an increasing function. Which it is not. In fact finding the M is the easy part.

2) You can't assume that $\frac {1+n}{2n}$ is constant so your argument that anything less than $1$ raised to a larger values approaches $0$ is not valid as for each $n$ it is a different number being raised to the $n$.

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$\frac {1+n}{2n} = \frac 1{2n} + \frac n{2n} = \frac 12 + \frac 1{2n}$

For $n \ge 2$ we have $2n \ge 4$ so frac $0 < \frac 1{2n} \le \frac 14$ and So $\frac 12 < \frac 12 + \frac 1{2n} \le \frac 12 + \frac 14$

so $\frac 12 < \frac {1+n}{2n} \le frac 34$. (So your $M$ can be any $M: \frac 34 < M < 1$. Say, $M = \frac 78$. I'm honestly not sure why your text asked for $M$ being strictly greater; it won't make any difference.)

So $(\frac 12)^n < (\frac {1+n}{2n})^n \le (\frac 34)^n < M^n < 1$

So $\lim_{n\to \infty} (\frac 12)^n \le \lim_{n\to \infty} (\frac {1+n}{2n})^n \le \lim_{n\to \infty} (\frac 36)^n \le \lim_{n\to \infty}M^n \le 1$

And $0 \le \lim_{n\to \infty} (\frac {1+n}{2n})^n \le 0$ so $\lim_{n\to \infty} (\frac {1+n}{2n})^n=0$.

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  • $\begingroup$ Great thank you! $\endgroup$ – squenshl Aug 14 at 1:47
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$$\left(\frac {1+n}{2n}\right)=\left(\frac{1}{2}\right)\left(1+\frac{1}{n}\right)$$

Thus $$\left(\frac {1+n}{2n}\right)^n\longrightarrow 0\times e =0$$

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  • $\begingroup$ There are two parts in OP's question. $\endgroup$ – Kavi Rama Murthy Aug 14 at 0:14
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If $\frac 3 4 <M <1$ then $\frac {1+n} {2n} <M$ for all $n \geq 2$. [$n \geq 2$ and $2M-1 >\frac 1 2$ implies $n(2M-1) > 2\frac 1 2=1$]. Now use squeeze theorem.

You can also prove the second part by noting that $(\frac {1+n} {2n})^{n} = (1+\frac 1 n)^{n} \frac 1 {2^{n}}$.

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