Please show how to prove that $$\lim_{n\to\infty} n\left(\frac{1}{2}\right)^n = 0$$
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Can you use L'Hôpital's rule? – Ben Grossmann Jun 23 '14 at 19:22
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calculate $\frac{a_{n+1}}{a_n}$ where $a_n = n (\frac{1}{2})^n$ – mm-aops Jun 23 '14 at 19:23
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5By the Binomial Theorem, if $n\ge 2$ we have $(1+1)^n \ge 1+n+\frac{n(n-1)}{2}\gt \frac{n(n-1)}{2}$. – André Nicolas Jun 23 '14 at 19:25
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@AndréNicolas: I used a different approach so you could add an answer; they are not as ephemeral as comments (and you could use the points). – robjohn Jun 23 '14 at 20:48
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Notice that $$ \frac{(n+1)/2^{n+1}}{n/2^n}=\frac12\left(1+\frac1n\right)\tag{1} $$ For $n\ge2$, the ratio in $(1)$ is at most $\frac34$. At $n=2$, $\dfrac{n}{2^n}=\dfrac12$. Therefore, $(1)$ implies $$ \frac{n}{2^n}\le\frac12\left(\frac34\right)^{n-2}\tag{2} $$ for $n\ge2$. Hopefully, it is clearer that $$ \lim_{n\to\infty}\frac12\left(\frac34\right)^{n-2}=0\tag{3} $$
robjohn
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By the Binomial Theorem, if $n\ge 2$ then $$(1+1)^n \ge 1+n+\frac{n(n-1)}{2}\gt \frac{n(n-1)}{2}.$$ It follows that if $n\ge 2$ then $\dfrac{n}{2^n}\lt \dfrac{2}{n-1}$.
André Nicolas
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There... doesn't that 10 points of reputation make you feel better? ;-) At least you haven't capped yet. – robjohn Jun 23 '14 at 21:51
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Indeed it feels very good, it has made my day (I am easily made happy). Liked your solution, another estimate-based approach. – André Nicolas Jun 23 '14 at 21:56
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With Stolz-Cesaro Theorem : $$ \color{#66f}{\large\lim_{n \to \infty}{n \over 2^{n}}} =\lim_{n \to \infty}{\left(n + 1\right) - n \over 2^{n + 1} - 2^{n}} =\lim_{n \to \infty}{1 \over 2^{n}}=\color{#66f}{\Large 0} $$
Felix Marin
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Consider extending the sequence {$n/2^n$} to the function $f(x)=x/2^x$.
Then use L'Hopital's rule: lim$_{x\to\infty} x/2^x$ has indeterminate form $\infty/\infty$. Taking the limit of the quotient of derivatives we get lim$_{x\to\infty} 1/($ln$2\cdot 2^x)=0$. Thus lim$_{x\to\infty} x/2^x=0$ and so $n/2^n\to 0$ as $n\to\infty$.
Forever Mozart
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It's sequences and series... L'Hopital may be too much (however, from the OP's response, perhaps not). – robjohn Jun 23 '14 at 20:45
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I agree that a proof using Binomials is more elegant, but many calculus students see L'Hopital's Rule before sequences and series. – Forever Mozart Jun 23 '14 at 21:40
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Interesting. Sequences and series was a pre-requisite when I taught calculus, but that was a while ago. Perhaps things have changed. – robjohn Jun 23 '14 at 21:48
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Elementary (?) Proof.
$n\geq4$ $~\Longleftrightarrow~$ $\dfrac{n}{n-1}<\sqrt{2}$
$\dfrac{n}{4}=\dfrac{5}{4}\times\dfrac{6}{5}\times\dfrac{7}{6}\cdots\times\dfrac{n}{n-1}<\sqrt{2}\times\sqrt{2}\times\sqrt{2}\cdots\times\sqrt{2}=2^{\left(\frac{n}{2}-2\right)}$
$\therefore$ $n^{2}<2^{n}$$~\Longrightarrow~$ $0<\dfrac{n}{2^{n}}<\dfrac{1}{n}$
$\displaystyle 0=\lim_{n\to\infty}0\leq\lim_{n\to\infty}\dfrac{n}{2^{n}}\leq\lim_{n\to\infty}\dfrac{1}{n}=0$
$\therefore$ $\displaystyle\lim_{n\to\infty}\dfrac{n}{2^{n}}=0$
chloe_shi
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