$$S_n =\frac1n \left(1 + \frac{1}{2} + \frac{1}{3} + \cdots+ \frac{1}{n}\right)$$ Show the convergence of $S_n$ (the method of difference more preferably)

I just began treating sequences in school, and our teacher taught that monotone increasing sequence, bounded above and monotone decreasing sequences, bounded below converge.

and so using that theorem here.. I found the $$(n+1)_{th} term$$, $$S_{n+1} = \frac1{n+1} \left(1 + \frac{1}{2} + \frac{1}{3} + \cdots+ \frac{1}{n+1}\right)$$

and then subtracted the (n)th term from it

What I was able to get was... $$S_{n+1}-S_n = \frac1{(n+1)^2} - \frac{\left(1+\frac12+\dots+\frac1n\right)}{n(n+1)}.$$ ...but then this is where I get stucked, but i'm trying to prove that the sequence > 0(i.e Converges) or < 0 (i.e diverges).


Hint. As an alternative to a Cesaro-like theorem, one may use the fact that $x \mapsto \dfrac1x$ is decreasing over $[1,\infty)$ to get $$ 0<1+\frac12+\frac13+\cdots+\frac1n<1+\int_1^n\frac{dx}x=1+\log n $$ giving $$ 0<\frac1n\left(1+\frac12+\frac13+\cdots+\frac1n\right)<\frac1n+\frac{\log n}n. $$

  • 4
    $\begingroup$ Overkill. One shouldn't need integrals in an easy sequences problem; sequences are studied much earlier than integrals. $\endgroup$ – mathguy Apr 25 '16 at 11:57
  • 2
    $\begingroup$ Not necessarily -- this can also (and perhaps more naturally) seen as a series, and series are studied around the same time as integrals. Moreover, this is a simple proof, provided one has the basic definition of integrals and of logarithm. It may not be the most elementary or even canonical way, but it is a self-contained alternative -- why not have it written as an answer, somewhere? It does not hurt to have options. $\endgroup$ – Clement C. Apr 25 '16 at 20:51

You can check here how can this be solved:

$$\lim_{n\to\infty}\frac1n=0\implies \lim_{n\to\infty}\frac1n\sum_{k=1}^n\frac1k=0$$

  • $\begingroup$ I'm sorry, I don't see why the implication. $\endgroup$ – Bernard Apr 25 '16 at 11:24
  • $\begingroup$ @Bernard Did you read the link I post in my answer? $\endgroup$ – DonAntonio Apr 25 '16 at 11:47
  • $\begingroup$ I know, this is Cesáro's lemma, but you can't use it here: the harmonic series diverge. Or am I missing something? $\endgroup$ – Bernard Apr 25 '16 at 12:00
  • $\begingroup$ @Bernard I think you're missing something big time, and I don't think this is widely known as Cesaro's Lemma, though it is related to Cesaro sums, certainly.. Please do read the link again, or read 4-6 lines here: en.wikipedia.org/wiki/Ces%C3%A0ro_mean $\endgroup$ – DonAntonio Apr 25 '16 at 12:03
  • $\begingroup$ @Bernard And it has nothing to do with series, whether convergent or not. $\endgroup$ – DonAntonio Apr 25 '16 at 12:09

If you denote $H_n=\displaystyle\sum_{k=1}^n\frac1k$, one knows $H_n\sim_\infty\ln n$, hence $$S_n\sim_\infty \frac{\ln n}n\to 0.$$

A more elementary proof: \begin{align*}&S_n =\frac1n H_n>S_{n+1}=\frac1{n+1}\Bigl(H_n+\frac{1}{n+1}\Bigr)\\ \iff \enspace&(n+1)H_n>nH_n+\frac n{n+1}\iff H_n>\frac n{n+1}, \end{align*} which is true since $H_n>1>\dfrac n{n+1}$. So the sequence $(S_n)$ is decreasing and bounded from below by $0$. Apply the monotone convergence theorem.

  • $\begingroup$ One doesn't need to know the $\ln n$ behavior of the harmonic series, the problem is easier than that. $\endgroup$ – mathguy Apr 25 '16 at 11:58
  • $\begingroup$ @mathguy: It has the advantage of being very short… $\endgroup$ – Bernard Apr 25 '16 at 12:01
  • $\begingroup$ Think of what the OP will learn from the answer. $\endgroup$ – mathguy Apr 25 '16 at 12:04
  • $\begingroup$ @mathguy: OK I've added an elementary proof. $\endgroup$ – Bernard Apr 25 '16 at 12:48

Avoiding appeals to logarithms:

Writing $H_n = 1 +\frac{1}{2}+\cdots + \frac{1}{n}$, note that for $n>M$:

$$H_n=1 +\frac{1}{2}+\cdots + \frac{1}{n}=1 +\frac{1}{2}+\cdots + \frac{1}{m}+\cdots+\frac{1}{n}=H_M+\frac{1}{M+1}+\cdots+\frac{1}{n}$$

Now, each of the terms after $H_M$ is $<\frac{1}{M}$, whence $H_n<H_M+\frac{n-M}{M}$.

Thus, $\frac{H_n}{n}<\frac{H_M}{n}+\frac{n-M}{nM}=\frac{H_m-1}{n}+\frac{1}{M}$ for any positive integer $M$, as long as $n$ is larger than $M$.

From this, we deduce that $\limsup\frac{H_n}{n}\le\frac{1}{M}$ for all positive integers $M$, and thus that $\limsup\frac{H_n}{n}=0$.

As $\frac{H_n}{n}>0$, it is thus trivial to deduce that the sequence converges to $0$.


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