24
$\begingroup$

Let's suppose that I have the sequence $a_n = \frac{1}{n^2} + \frac{2}{n^2} + \frac{3}{n^2} + \ldots + \frac{n}{n^2}, n \in \mathbb{N}$. And I have to find the limit of the sequence as $n \rightarrow \infty$. Would the below solution be correct?

The sequence $a_n$ can be rewritten as \begin{align} a_n &= \sum_{k=1}^{n} \frac{k}{n^2} \\ &= \frac{1}{n^2} \sum_{k=1}^{n} k \\ &= \frac{1}{n^2} \cdot \frac{n(n+1)}{2} \\ &= \frac{n+1}{2n}. \end{align}

Thus we have \begin{align} \lim_{n \rightarrow \infty} a_n = \lim_{n \rightarrow \infty} \frac{n+1}{2n} = \lim_{n \rightarrow \infty} \frac{n}{2n} = \frac{1}{2}. \end{align}

$\endgroup$
4
  • $\begingroup$ Well done! If I was your professor, I would just ask you to clarify how you know that $$\lim_{n\to\infty} \frac{n+1}{2n} = \lim_{n\to\infty}\frac{n}{2n}$$ $\endgroup$
    – 5xum
    Commented Feb 2, 2016 at 14:11
  • $\begingroup$ Yes, that's correct, although I would rather make the critical step like this: $$\frac{n+1}{2n} = \frac{1+\frac 1n}2$$ and then conclude the limit is $$\frac{1+\frac 1\infty}2 = \frac{1+0}2 = \frac 12$$ $\endgroup$
    – CiaPan
    Commented Feb 2, 2016 at 14:51
  • 1
    $\begingroup$ We have proved the general case of the limits of fractions with polynomials at the denominator and the numerator. It was the main reason why I left it like that. $\endgroup$
    – Atheridis
    Commented Feb 2, 2016 at 15:22
  • 2
    $\begingroup$ Because $\frac{1}{2n}\to 0$, we have $\lim \frac{n+1}{2n} =\lim\left(\frac{n}{2n}+\frac{1}{2n}\right)=\lim \frac{n}{2n}$ and thus your calculation is perfectly good. No need to write $\frac{1+\frac{1}{n}}{2}$ as has been suggested. $\endgroup$
    – Pedro
    Commented Feb 2, 2016 at 15:41

2 Answers 2

19
$\begingroup$

Your solution is fine. Here is an other way to get the result. I know it is not the easier way to solve it, but the advantage is that you can generalize to $b_n = \sum_{k=1}^{n} \frac{k^2}{n^3}$ for instance.

Notice that $$a_n = \frac{1}{n} \sum_{k=0}^{n} \frac{k}{n}$$ looks like a Riemann sum for $f : x \mapsto x$ since $$a_n = \sum_{k=0}^{n} f\left(\frac{k}{n}\right) \left(\frac{k+1}{n} - \frac{k}{n} \right)$$

Therefore the limit of the sequence $(a_n)_{n≥1}$ is $$\int_0^1 f = \left[\frac{x^2}{2}\right]_0^1 = \frac{1}{2}$$

$\endgroup$
1
  • 1
    $\begingroup$ I accepted this answer, since you also gave another interesting way of solving it. $\endgroup$
    – Atheridis
    Commented Feb 2, 2016 at 15:26
11
$\begingroup$

This seems correct. Personally I would not say $\lim\frac {n+1}{2n}=\lim\frac n{2n}$. Not without explanation at least (e.g. include the step $\lim \frac{n+1}{2n}=\lim \left(\frac{n}{2n}+\frac 1{2n}\right)$). I would rather say $\lim \frac {n+1}{2n}=\lim \frac{1+\frac1n}{2}$. But this is of course a matter of preference.

$\endgroup$
1
  • $\begingroup$ Might I ask who downvoted and why? Is there something that can be improved? $\endgroup$
    – gebruiker
    Commented Feb 2, 2016 at 14:01

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .