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I am currently studying the Lebesgue Dominated Convergence Theorem and, to apply it, I must calculate the following. $$ \lim_{n\to\infty} \frac{1+xn}{(1+x)^n}$$

where $x\in (0,1]$.

I know the limit exists using the ratio test however I thought the limit was 0, which my lecturer has queried. Would someone be able to indicate how to calculate the limit please.

user284331
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Student A
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3 Answers3

1

It should be valid only for $x\in(0,1]$.

Use that $(1+x)^{n}\geq 1+nx+\dfrac{n(n-1)}{2}x^{2}$, then \begin{align*} \dfrac{1+nx}{(1+x)^{n}}\leq\dfrac{1+nx}{1+nx+\dfrac{n(n-1)}{2}x^{2}}=\dfrac{\dfrac{1}{n}+x}{\dfrac{1}{n}+x+\dfrac{n-1}{2}x^{2}}\rightarrow 0. \end{align*}

user284331
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1

We will show an alternative approach.

As others have pointed out the limit $l$ equals $0$ when $x\in(0,1]$. In that case it is true that $1+x\gt1$ and thus $$\lim_{n\rightarrow\infty}(1+x)^n=\infty$$ It now remains to show that $(1+x)^n$ is an increasing sequence for every $n$ so that we can use the Cesàro-Stolz lemma to calculate the limit: $$\lim_{n\rightarrow\infty}\frac{1+nx}{(1+x)^n}=\lim_{n\rightarrow\infty}\frac{(1+(n+1)x)-(1+nx)}{(1+x)^{n+1}-(1+x)^n}=\lim_{n\rightarrow\infty}\frac{x}{(1+x)^n((1+x)-1)}=\\x\lim_{n\rightarrow\infty}\frac{1}{(1+x)^n}=0$$ When $x=0$ it holds $(1+x)^n=1 \ \forall n\in\mathbb{N}$ and $1+nx=1$ so the limit $l=1$ when $x=0$

Summand
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1

If dominated convergence is what you want, you may want to use the following well known result: $\lim_n\frac{n^p}{(1+q)^n}=0$ for all $p\in\mathbb{R}$ and $q>0$.

To apply dominated convergence then, it is enough to find an appropriate dominating function, for $$f_n(x)=\frac{1+nx}{(1+x)^n}=\frac{1-n}{(x+1)^n}+\frac{n}{(1+x)^{n-1}}\xrightarrow{n\rightarrow\infty}0\qquad \text{if} \quad x\neq0,$$ that is, $f_n(x)\xrightarrow{n\rightarrow\infty}0$ a.s. in $(0,1]$.

The binomial theorem is handy here: for $n>2$, $$(1+x)^n\geq 1+\binom{n}{1}x=1+nx$$ Hence $$ |f_n(x)|\leq1$$ and $\int^1_0\,dx=1<\infty$. Consequently $\lim_n\int^1_0f_n(x)\,dx=0$.

Notice also that limit of $\int^1_0f_n(x)\,dx$ of the sequence of functions in your problem can be estimated directly:

\begin{align} \int^1_0\frac{1+nx}{(1+x)^n}\,dx&=\int^1_0\frac{dx}{(1+x)^n}+ n\int^1_0\frac{x+1-1}{(x+1)^n}\,dx\\ &=(1-n)\frac{(1+x)^{-n+1}}{-n+1}\Big|^1_0 +n\frac{(x+1)^{-n+2}}{-n+2}\Big|^1_0\\ &=-(2^{-n+1}-1)-\frac{n}{n-2}\big( 2^{-n+2}-1)\\ &=1-\frac{n}{n-2}-2^{-n+1}-\frac{n}{n-1}2^{-n+2}\xrightarrow{n\rightarrow\infty}0 \end{align}

This is one instance in which dominated convergence gives a cleaner answer (once the dominating function and the convergence of the integrands $f_n$ have been established.)

Mittens
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