Find the following limit: $$\lim_{n\to\infty} \left(1+\frac{1}{2}+...+\frac{1}{n}\right)\frac{1}{n}$$

My intuition says that this goes to zero, because $1/n$ goes much faster to zero than the harmonic series go to infinity, but how can I prove this?


Hint: compare the harmonic series with $\log n$, by comparing it with $\int_1^{n+1}\frac 1 x dx$. Then use l'Hopital's.

  • $\begingroup$ why should i compare it with that integral? $\endgroup$ – Badshah Oct 25 '12 at 21:01
  • $\begingroup$ @Badshah Calculate the integral, what do you get? Now, draw the graph and approximate the area under it by $n$ rectangles of equal width, using the left end point of each rectangle to determine the height. What is the total area of the rectangles? The area of the rectangles is an approximation of the area under the curve, but more importantly, which is bigger? $\endgroup$ – Graphth Oct 25 '12 at 21:11
  • $\begingroup$ you get log(n+1), should I draw log(n+1) or 1/x? $\endgroup$ – Badshah Oct 25 '12 at 21:13
  • 1
    $\begingroup$ @Badshah I mean, draw the graph of $1/x$ between $x=1$ and $x=n+1$. Then, approximate this area with the rectangles I described. If you recall, this is the idea used when the definite integral is first defined. You just told me the actual area is $\log(n+1)$. When you use rectangles to approximate the area, you should get $1 + 1/2 + \ldots + 1/n$. And, you should be able to clearly tell whether the sum or $\log(n+1)$ is bigger. And, you can use this fact to finish the problem. $\endgroup$ – Graphth Oct 25 '12 at 21:15
  • $\begingroup$ log(n+1) is bigger, so when you take the limit, the sum of the rectangles comes closer and closer to log(n+1) so, $\lim_{n\to\infty}1+1/2+...+1/n=\lim_{n\to\infty}log(n+1)$. is this correct? $\endgroup$ – Badshah Oct 25 '12 at 21:26

A slightly easier way out is prove that $$1 + \dfrac12 + \dfrac13 + \cdots + \dfrac1n < 2\sqrt{n}$$ using induction. Hence, you have that $$ 0 < \left(1 + \dfrac12 + \dfrac13 + \cdots + \dfrac1n\right) \dfrac1n < \dfrac2{\sqrt{n}}$$ Now use squeeze theorem to get what you want.


If you know that $1+\frac12 +\cdots+\frac1n\approx \ln n$ and $\frac{\ln n}n\to 0$, you are done.

More elementary, your limit is the Cesáro sum of the sequence $(\frac 1n)$, hence has the same limit $0$. That is: $$\lim_{n\to\infty} a_n= a \qquad\Rightarrow\qquad \lim_{n\to\infty} \frac{a_1+\cdots+a_n}n= a$$

  • $\begingroup$ why is the harmonic series estimated with logn? $\endgroup$ – Badshah Oct 25 '12 at 21:02
  • $\begingroup$ @Badshah: That's why I said "if you know". You don't need it if you follow the Cesáro idea. For a proof of the log compare Bruno's answer, which uses integration - I still think the Cesáro stuff is the most elementary approach. $\endgroup$ – Hagen von Eitzen Oct 25 '12 at 21:04
  • $\begingroup$ In my real analysis class, we proved the Cesaro idea. You could do it too. It's just a basic delta epsilon thing. $\endgroup$ – Graphth Oct 25 '12 at 21:10
  • $\begingroup$ oke,I get the cesaro idea, so I am going to try and prove it. thanks for the answer! $\endgroup$ – Badshah Oct 25 '12 at 21:11
  • $\begingroup$ The result about Cesaro mean is shown, for example, here: math.stackexchange.com/questions/207910/… (And probably other posts on this site with the same result could be found.) $\endgroup$ – Martin Sleziak Apr 24 '14 at 13:02

What you want to do here is find an upper bound for $1 + \frac{1}{2} + \frac{1}{3} + \ldots$. That is, you want to replace every term in that series with a larger term in such a way that the sums are easier to do. One clever way to do this (which I thought of by trying to modify the classic Oresme proof of divergence by exhibiting a lower bound) is to note that it's smaller than $$1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{8} \ldots.$$

The sum of the first $2^k-1$ terms of this sequence is $k$. Which should be enough for you to prove this limit directly.


$$\sum_{k=1}^n\frac{1}{k}\sim \log n+\gamma$$

$$\lim_{n\to\infty} \frac{1}{n}\left(\log n+\gamma\right)=0$$


Here is a completely elementary and self-contained proof (subject to accepting that $\frac{k}{2^k} \to 0$ as $k \to \infty$).

Let $a_n = \frac{1}{n}\sum_{k=1}^n \frac1{k} $. Then $na_n = \sum_{k=1}^n \frac1{k} $.

Therefore $\frac1{n+1} =(n+1)a_{n+1}-na_n =n(a_{n+1}-a_n)+a_{n+1} $ or $a_{n+1}-a_n =\frac1{n}(\frac1{n+1}-a_{n+1}) $.

Since, for $n > 1$, $a_n \ge \frac1{n}(1+\frac12) = \frac{3/2}{n} $, $a_{n+1}-a_n <-\frac{1}{2n(n+1)} $ so $a_n$ is a decreasing sequence.

We also have

$\begin{array}\\ (2n)a_{2n}-na_n &=\sum_{k=1}^{2n} \frac1{k}-\sum_{k=1}^n \frac1{k}\\ &=\sum_{k=n+1}^{2n} \frac1{k}\\ \text{so}\\ (2n)a_{2n}-na_n &< 1\\ \text{and}\\ (2n)a_{2n}-na_n &> \frac12\\ \end{array} $

Therefore $a_{2n} < \frac1{2n}(1+na_n) = \frac1{2n}+\frac12 a_n $ or

$\begin{array}\\ a_n &>2a_{2n}-\frac1{n}\\ &>2(2a_{4n}-\frac1{2n})-\frac1{n}\\ &=4a_{4n}-\frac1{n}-\frac1{n}\\ &=4a_{4n}-\frac{2}{n}\\ &=4(2a_{8n}-\frac1{4n})-\frac{2}{n}\\ &=8a_{8n}-\frac1{n}-\frac{2}{n}\\ &=8a_{8n}-\frac{3}{n}\\ \end{array} $

By induction we can show that $a_n > 2^ka_{2^kn}-\frac{k}{n} $ or $a_{2^kn} <2^{-k}(a_n+\frac{k}{n}) $.

Since both $\frac{k}{2^k} \to 0 $ and $\frac{a_n}{2^k} \to 0 $ as $k \to \infty$, $a_{2^kn} \to 0 $ as $k \to \infty$.

Since $a_n$ is decreasing and positive, $a_n \to 0$ as $n \to \infty$.


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