Let $f_n(x) = \frac{1}{nx}$ for $x\in[1, \infty)$. Show that $$\lim\limits_{n\to\infty}f_n(x) = 0$$ but $$\lim\limits_{n\to\infty}\int_1^\infty f_n(x)dx \neq 0.$$

My progress:

The first limit was rather trivial given that x is positive.

For the second limit, I have come as far as to integrate $f_n(x)$, which yielded

$$\lim\limits_{n\to\infty}\left[\frac1n \left[ \lim\limits_{x\to\infty}\ln(x)\right]\right]$$

Right now, I have a $0\cdot\infty$ expression, which I suppose can be said to be not zero, but it seems cheap.

I also contemplated, since both limits tend to infinity, switch x with n, so that we have $$\lim\limits_{n\to\infty}\frac1n \cdot \lim\limits_{n\to\infty}\ln(n)$$

I tried L'Hopital on this, but that gave me 0, which is clearly not what I want.

Any help appreciated!

  • $\begingroup$ $\int_1^{+\infty}1/(nx)\,dx=+\infty$ for all $n=1,2,\ldots$. $\endgroup$
    – mickep
    Commented Feb 16, 2015 at 20:31
  • $\begingroup$ Are you sure that the integral is from $1$ to $\infty$, not to $n$ ? $\endgroup$
    – larry01
    Commented Feb 16, 2015 at 20:33
  • $\begingroup$ @larry01 - Yes, it was infinity explicitly. $\endgroup$
    – Alec
    Commented Feb 16, 2015 at 20:49

3 Answers 3


In the improper Riemann sense you are dealing with:

$$\lim_{n \to \infty} \lim_{b \to \infty} \int_1^b \frac{1}{nx} dx = \lim_{n \to \infty} \left ( \frac{1}{n} \lim_{b \to \infty} \int_1^b \frac{1}{x} dx \right ).$$

The point of the exercise is that you cannot exchange the two limits. But the situation is simpler than you are making it, because $\int_1^b \frac{1}{x} dx = \ln(b)$. So the inner limit is always $+\infty$ regardless of the value of $n$. This means the outer limit is also $+\infty$.

What you did towards the end is a very different calculation; you showed that:

$$\lim_{n \to \infty} \frac{1}{n} \int_1^n \frac{1}{x} dx = 0.$$

This takes the limit in $b$ simultaneously with the limit in $n$.

  • $\begingroup$ I can see that there is a core concept here that I'm not familiar with. Why do you not end up with $0\cdot\infty$ when $n\to\infty$? $\endgroup$
    – Alec
    Commented Feb 16, 2015 at 20:36
  • $\begingroup$ @Alec Do you see the parentheses in Ian's first equation here? You have to do $+\infty$ times $\frac{1}{n}$ before you take the limit $n\to+\infty$. And $+\infty$ times $\frac{1}{n}$ is $+\infty$. (In whatever extended reals we have to do that kind of arithmetic.) $\endgroup$
    – anon
    Commented Feb 16, 2015 at 20:37
  • $\begingroup$ @Alec The indeterminate form $0 \cdot \infty$ arises when you have $a_n$ and $b_n$ which are both finite, such that $a_n \to 0$ and $b_n \to \infty$, and you are considering the limit of $a_n b_n$. But if $a_n$ is finite and never zero and $b_n$ is always $\infty$, then the sequence $a_n b_n$ is just the sequence $\infty,\infty,\infty,\dots$ forever. $\endgroup$
    – Ian
    Commented Feb 16, 2015 at 20:38
  • $\begingroup$ @whacka - Yes, however, I can't entirely grasp it. It looks like he's absorbing the 1/n into the infinity of the integral, which renders $n\to\infty$ pointless. $\endgroup$
    – Alec
    Commented Feb 16, 2015 at 20:39
  • $\begingroup$ @Alec Once you've done the integrals, you're dealing with a sequence of numbers, all of which are $+\infty$. $\endgroup$
    – Ian
    Commented Feb 16, 2015 at 20:40

Hint: $$\lim_{n\to \infty}\frac{1}{n}\underbrace{\color{#05f}{\int_{1}^{\infty} \frac{1}{x} dx}}_{\to \infty} = \infty $$

$$\color{#05f}{\int_{1}^{\infty} \frac{1}{x} dx} = \ln x \Big|_1^\infty \to \infty$$



If the integral converges (which it doesn't), the limit is $0$. Otherwise it doesn' t exist.


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