# Find $\lim_{n\to\infty} \left(\frac{1+n}{2n}\right)^n$.

by squenshl   Last Updated August 14, 2019 00:20 AM

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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If $$\frac 3 4 then $$\frac {1+n} {2n} 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 $$(\frac {1+n} {2n})^{n} = (1+\frac 1 n)^{n} \frac 1 {2^{n}}$$.

Kavi Rama Murthy
August 13, 2019 23:45 PM

$$\left(\frac {1+n}{2n}\right)=\left(\frac{1}{2}\right)\left(1+\frac{1}{n}\right)$$

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

August 13, 2019 23:51 PM

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

fleablood
August 14, 2019 00:16 AM