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For a given sequence $a_1,a_2,\ldots,a_n$ if $\lim\limits_{n\to \infty}a_n=a$, then $\lim\limits_{n\to \infty}\dfrac{1}{\ln(n)}\sum\limits_{r=1}^{n}\dfrac{a_r}{r}$ is:

(A)zero $\hspace{150pt}$ (B)a

(C)$\dfrac{a}{2}$ $\hspace{157pt}$ (D)None of these

My approach: I'm trying to bring this limit into this form: $\lim\limits_{n\to \infty}\dfrac{1}{n}\sum\limits_{r=1}^{n}f\left(\dfrac{r}{n}\right)$ as, $$\lim\limits_{n\to \infty}\dfrac{1}{n}\sum\limits_{r=1}^{n}f\left(\dfrac{r}{n}\right)=\int_{0}^{1}f(x)\ dx$$ so here it is what I'm doing: \begin{align*} \lim\limits_{n\to \infty}\dfrac{1}{\ln(n)}\sum\limits_{r=1}^{n}\dfrac{a_r}{r} &=\lim\limits_{n\to \infty}\dfrac{1}{n}\cdot\dfrac{n}{\ln(n)}\sum\limits_{r=1}^{n}\dfrac{a_r}{r}\\ &=\lim\limits_{n\to \infty}\dfrac{1}{n}\sum\limits_{r=1}^{n}\dfrac{a_r}{r/n}\cdot\dfrac{1}{\ln(n)} \end{align*} but I can't make the limit in the above form, is there any way of doing it?

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4 Answers 4

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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \lim_{n \to \infty}\bracks{{1 \over \ln\pars{n}}\sum_{r = 1}^{n}{a_{r} \over r}} & = \lim_{n \to \infty}\bracks{{1 \over \ln\pars{n + 1} - \ln\pars{n}} \pars{\sum_{r = 1}^{n + 1}{a_{r} \over r} - \sum_{r = 1}^{n}{a_{r} \over r}}} \\[5mm] & = \lim_{n \to \infty}{a_{n + 1}/\pars{n + 1} \over \ln\pars{1 + 1/n}} \qquad\pars{~Stolz-Ces\grave{a}ro Theorem~} \\[5mm] & = \lim_{n \to \infty}\bracks{% {1/n \over \ln\pars{1 + 1/n}}\,{n \over n + 1}\,a_{n + 1}} = \bbx{\ds{a}} \end{align}

See this link about Stolz-Cesàro theorem Theorem.

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  • $\begingroup$ Can you please explain how you simplified the limit in your second line i.e. how you removed the summation sign.thanks. $\endgroup$
    – Navin
    Apr 16, 2017 at 17:10
  • $\begingroup$ @navinstudent I added some details to my answer. Thanks. $\endgroup$ Apr 17, 2017 at 22:12
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We can easily show that

$$\lim_{n\to \infty}\frac{1}{\log(n)}\sum_{r=1}^n\frac{a}{r}=a$$

by using the bounds

$$\frac{1}{\log(n)}\int_1^n \frac{a}x\,dx\le \frac{1}{\log(n)}\sum_{r=1}^n\frac{a}{r}\le \frac{1}{\log(n)}\left(1+\int_{1}^n\frac{a}{x}\,dx\right)$$

and applying the squeeze theorem.


Note that for any $\epsilon>0$ there exists a number $N$ such that $|a_r-a|<\epsilon/2$ whenever $n>N$. Then, we have$$\begin{align}\left|\frac{1}{\log(n)}\sum_{r=1}^n\frac{a_r-a}{r}\right|&=\left|\frac{1}{\log(n)}\sum_{r=1}^N\frac{a_r-a}{r}+\frac{1}{\log(n)}\sum_{r=N+1}^n\frac{a_r-a}{r}\right|\\\\&\le \frac{1}{\log(n)}\sum_{r=1}^N\frac{|a_r-a|}{r}+\frac{\epsilon/2}{\log(n)}\sum_{r=N+1}^n \frac{1}{r}\tag1\end{align}$$For fixed $N$, the first sum on the right-hand side goes to $0$ as $n\to \infty$.

For the second term, we know that $\sum_{r=N+1}^n \frac{1}{r}\le \int_{N}^n \frac1x\,dx=\log(n)-\log(N)$. Therefore, the entire right-hand side of $(1)$ can be made smaller than any pre-assigned number $\epsilon$ by selecting $N$ large enough.

Therefore, we find that

$$\lim_{n\to \infty}\frac{1}{\log(n)}\sum_{r=1}^n\frac{a_r}{r}=\lim_{n\to \infty}\frac{1}{\log(n)}\sum_{r=1}^n\frac{a}{r}=a$$

and we are done!

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  • $\begingroup$ thanks but can't get it, can you explain a little bit about this formula? $\endgroup$
    – k.Vijay
    Apr 3, 2017 at 17:56
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    $\begingroup$ @k.Vijay You're welcome. If you're not familiar with the asymptotic expansion of the Harmonic number, then perhaps my edited solution offers another way forward with which you have the requisite tools. $\endgroup$
    – Mark Viola
    Apr 3, 2017 at 18:21
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I would start like this.

If $\lim_{n\rightarrow\infty} a_n = a$, then for any $\epsilon > 0$ (pick an $\epsilon$ close to 0) there is some large enough $N$ so that every $a_n$ is within $\epsilon$ of $a$ when $n>N$. That is, as $n$ gets large, you can just say that $a_n$ is approximately $a$, and so $a_n/n$ is approximately $a/n$.

Then, since you know the technique of integral approximation, use it to show that $\sum_1^n a_r/r$ is approximately $a \ln n $ plus a constant (or look up partial sum of harmonic series, I'm suggesting show the result in Dr. MV's answer) and take your limit from there.

I put in "plus a constant" because it can take care of the error from your choice of $\epsilon$ and the fact that the first $N$ terms of $a_n$ that might not be close to $a$ at all.

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Hint: What if $a_n = 1$ for all $n?$

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