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Prove using the basic properties of a sequence that $\lim \frac{n}{2^n} = 0$
I tried to prove this using the standard epsilon-definition of a limit but the math got really hard, so I stopped. Then, I tried to represent the sequence as a quotient or a product of two sequences, but those sequences were not convergent. Thus, I've got no more ideas on how to tackle this problem. Any suggestions would be most appreciated.

Aemilius
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3 Answers3

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By the binomial theorem $2^n = (1 + 1)^n > n(n-1)/2$ for all $n \ge 1$, so $$\frac{n}{2^n} < \frac{2}{n-1}$$

for all $n \ge 1$. Given $\epsilon > 0$, choose a positive integer $N > \dfrac{2}{\epsilon} + 1$, then use the above inequality to show that $\dfrac{n}{2^n} < \epsilon$ for all $n \ge N$.

kobe
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As the comment suggests, you can notice that $2^{n}$ grows exponentially, and thus, eventually, we would have $2^{n}>n^{2}$, and thus the limit becomes sandwiched between $0$ and $\lim \frac{1}{n}$, and hence it becomes $0$.

Otherwise, you could also note that $2^{n}=e^{nlog(2)}=1+nlog(2)+\frac{1}{2!}(nlog(2))^{2}+..$, and so $\frac{n}{2^{n}}=\frac{1}{\frac{1}{n}+log(2)+n(..)}$, and when $n\rightarrow \infty$, the terms with a factor of $n$ in the denominator dominate and so the limit becomes $0$.

Aritro Pathak
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  • What I don't understand is why you use $n^2$. Could you, please, elaborate a little bit more on the explanation? – Aemilius Oct 18 '17 at 18:37
  • Because you now get $lim \ 0\leq lim \ n/2^{n} \leq lim \ n/n^{2}$, since $2^{n}>n^{2}$. Both $lim \ 0$ and $lim \ n/n^{2}$ is 0; hence the sandwiched limit is also 0. – Aritro Pathak Oct 18 '17 at 18:41
  • Thank you, now it started to make sense to me. But one more thing - why did you use $<$ instead of $\le$ ? – Aemilius Oct 18 '17 at 18:44
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    It doesn't matter in the limit; you can have $a_{n}<b_{n}<c_{n} \ \forall n$, and if $\lim a_{n}=\lim c_{n}=L$, then $\lim b_{n}=L$ as well. (I edited the previous comment to have the $\leq$ sign instead of the $<$ sign ). – Aritro Pathak Oct 18 '17 at 18:46
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Hint:

$\dfrac{u_{n+1}}{u_n}\to \dfrac12$, hence $(u_n)$ is bounded from above by a geometric sequence of ratio $<1$ is $n$ is large enough (actually you should find a geometric series of ratio $\dfrac34$ if $n>1$).

Bernard
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