# Find the limit of $\frac{n^{x}}{(1 + x)^{n}}$

Let $$x>0$$. Find $$\lim_{n\to\infty}\frac{n^{x}}{(1 + x)^{n}}.$$

What I was thinking involved an inequality, by reason of considering the fact that $$(1 + x)^{n} > C(n, k) x^{k}$$ for some $$k$$ such that $$n>k$$. Expanding the binomial coefficient and taking the $$n$$ common from each bracket, we have something like $$(1 + x)^{n} > C(n, k) x^{n} > \frac {x^{k}n^{k}}{k! 2^{k}}.$$ But now, I am totally clueless. By intuition, I know that the limit has to be zero, so I just want to prove using the concept of inequalities that what I thought was the right approach (being the sandwich theorem). Any help would be appreciated!

• Look up Bernoulli inequality. – Jakobian Aug 1 '19 at 14:25
• You have $(1+x)^n>\binom{n}{k}x^k,$ not $\binom{n}{k}x^n.$ – Thomas Andrews Aug 1 '19 at 14:28
• Fix some $x$. Then, taking $n-$th roots, you have: $$\sqrt[n]{\frac{n^x}{(1+x)^n}}=\frac{\sqrt[n]{n}^x}{1+x}\to\frac{1}{1+x}<1,$$ for $x>0$ (we need $x>0$ so as to define $n^x$). Then, by $n-$th root test you get the result. – Βασίλης Μάρκος Aug 1 '19 at 14:42
• Thanks for pointing that out @Thomas Andrews – Aditya Garg Aug 1 '19 at 15:14
• @Jakobian what you suggest seems not right to me...I tried using the Bernoulli inequality first but it is not satisfying the my purpose...kindly post a more insightful answer so that I can think about your idea – Aditya Garg Aug 1 '19 at 15:16

@Aditya Garg. You can still go ahead with your proof that does not involve exponential or logarithmic functions. First, fix a $$k$$ to be determined later. Then, for all $$n>2k$$ \begin{aligned} (1+x)^n\geq\binom{n}{k}x^k=\frac{n(n-1)\cdot\ldots\cdot(n-k+1)}{k!}x^k>\frac{n^k}{2^k}\frac{x^k}{k!} \end{aligned} This is because each factor in $$n(n-1)\cdots(n-k+1)$$ is of the form $$n-j$$ for $$0\leq j\leq k<\frac{n}{2}$$. From this, you obtained that \begin{aligned} \frac{n^x}{(1+x)^n}\leq \frac{2^kk!}{x^k}n^{x-k} \end{aligned} Then, you can see that by taking a fix $$k$$ with $$k>x$$ you get that $$\lim_{n\rightarrow\infty}\frac{n^x}{(1+x)^n}=0,\qquad (x>0)$$ The argument is a beautiful one. You can replace $$n^x$$ by $$n^a$$ for a fix number $$a$$ and you get the notable limit $$\lim_{n\rightarrow\infty}\frac{n^a}{(1+x)^n}=0,\qquad (x>0)$$ The advantage of this type of argument is that you do not need to define yet (properly) the exponential and/or logarithmic function and requires very few properties of numbers. Rudin's baby book page 57, has other notable limits that used this type of arguments.

Of course, if you have already gone through Calculus and just need to determine the limit, by all means, use the properties of the exponential, logarithms, L'Hospital rules, etc.

• A quick question: it is possible to choose $k> x$ be the Archimedian property and the above method works because the limit of the sequence does not depend on the starting index, right? – Karthik Kannan Aug 1 '19 at 16:40
• Yes, the only nuanced order property of numbers that we are using is that given any real number $x$, there is an positive integer $k$ such that $k>x$. – Oliver Diaz Aug 1 '19 at 17:17
• Hmmm nice !!! A sure +1 – Aditya Garg Aug 1 '19 at 18:10

Take the logarithm, $$\ln \frac{n^x}{(1+x)^n} = x \ln n - n \ln (1+x) \to_n -\infty,$$ thus, $$\frac{n^x}{(1+x)^n} \to_n 0.$$

• Or simply, $n^x$ is polynomial, compared to $(1+x)^n$ which is exponential. – Olivier Massicot Aug 1 '19 at 15:26
• The OP asked for help with a specific approach, not for an entirely different solution. – Mars Plastic Aug 1 '19 at 15:27
• Not the thing I asked for but still great ! I Appreciate your work ;-) but I already did that. You see I don't like logarithms though ! Could you improvise? – Aditya Garg Aug 1 '19 at 15:31
• This is technically equivalent to saying that $n^x/(1+x)^n$ goes to $0$, and doesn't really introduce anything. Just poses the problem in a different shade. – Jakobian Aug 1 '19 at 15:33
• Well technically yes, but both facts are well-known compared-growths results. For a quick proof, we immediately get from the integral definition of $\ln$ that $\ln x \leq x$ for all $x \geq 0$, then, $$\frac{\ln n}{n} = \frac{2\ln(\sqrt{n})}{n} \leq \frac{2\sqrt{n}}{n} \to_n 0.$$ Therefore, $$x \ln n - n \ln(1+x) \sim_n - n \ln(1+x) \to_n -\infty.$$ Let me think of a quick one with the exponentials. – Olivier Massicot Aug 1 '19 at 15:59

Just for fun, here is another elementary answer that does not involve logarithms not exponentials. It only requires the fact that $$a^n\xrightarrow{n\rightarrow\infty}0$$ for any $$0.

Let $$a_n=\frac{n^\alpha}{(1+p)^n}$$, where $$\alpha>0$$ and $$p>0$$. Then $$\frac{a_{n+1}}{a_n}=\big(1+\frac{1}{n}\Big)^\alpha\frac{1}{1+p}$$

Since $$\Big(1+\frac{1}{n}\Big)^\alpha \xrightarrow{n\rightarrow\infty}1$$, there is $$N\in\mathbb{N}$$ such that $$n\geq N$$ implies that $$\big(1+\frac{1}{n}\Big)^\alpha<1+\frac{p}{2}$$. Let

Then, for all $$n\geq N$$

$$a_{n+1}\leq \frac{2+p}{2+2p} a_n$$

This implies that $$a_{n+N}\leq \Big(\frac{2+p}{2+2p}\Big)^na_N\xrightarrow{n\rightarrow\infty}0$$ since $$0<\frac{2+p}{2+2p}<1$$.