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$$S_n =\frac1n \left(1 + \frac{1}{2} + \frac{1}{3} + \cdots+ \frac{1}{n}\right)$$ 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).

koodeji
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  • Your attempt...? – Mithlesh Upadhyay Apr 25 '16 at 10:54
  • I'm not used to this..but I'll try to type that.. – koodeji Apr 25 '16 at 10:58
  • Sn+1 - sn, I subtracted the nth term from the n+1th term and I Got..

    (N/n+1 - (1 + 1/2 + 1/3 +...+ 1/n))/n(n+1)

     – koodeji Apr 25 '16 at 11:02
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    This might help you: http://math.stackexchange.com/questions/207910/prove-convergence-of-the-sequence-z-1z-2-cdots-z-n-n-of-cesaro-means http://math.stackexchange.com/questions/210681/if-a-n-to-ell-then-hat-a-n-to-ell http://math.stackexchange.com/questions/248116/arithmetic-mean-of-a-sequence-converges http://math.stackexchange.com/questions/930436/a-result-on-sequences – Martin Sleziak Apr 25 '16 at 11:02
  • I have tried to add to your post what you wrote in a comment. I am not sure whether I understood correctly what you meant, so please, edit it further. For some basic information about writing math at this site see e.g. here, here, here and here. – Martin Sleziak Apr 25 '16 at 11:20
  • What you have done so far is fine, can you use it to show the original sequence is decreasing? It is obviously bounded below (by zero), so that's all you need. The intuition is, $S_n$ is the average of $n$ numbers, and as $n$ increases you are adding smaller and smaller numbers, so their average decreases. This is closely related to what Joanpemo told you in his answer. (And Martin as well). –  Apr 25 '16 at 12:03
  • Your question was put on hold, the message above (and possibly comments) should give an explanation why. (In particular, this link might be useful.) You might try to edit your question to address these issues. Note that the next edit puts your post in the review queue, where users can vote to reopen this. (Therefore it would be good to avoid minor edits and improve your question as much as possible with the next edit.) – Martin Sleziak Apr 25 '16 at 15:10
  • Another question about the same limit: http://math.stackexchange.com/questions/221114/find-lim-n-to-infty-1-frac12-frac1n-frac1n – Martin Sleziak Oct 16 '16 at 10:45
  • Possible duplicate of Find $\lim_{n\to\infty} (1+\frac{1}{2}+...+\frac{1}{n})\frac{1}{n}$ – Guy Fsone Jan 01 '18 at 10:22

4 Answers4

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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 $$ 0<\frac1n\left(1+\frac12+\frac13+\cdots+\frac1n\right)<\frac1n+\frac{\log n}n. $$

Olivier Oloa
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    Overkill. One shouldn't need integrals in an easy sequences problem; sequences are studied much earlier than integrals. –  Apr 25 '16 at 11:57
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    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. – Clement C. Apr 25 '16 at 20:51
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You can check here how can this be solved:

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

DonAntonio
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  • I'm sorry, I don't see why the implication. – Bernard Apr 25 '16 at 11:24
  • @Bernard Did you read the link I post in my answer? – DonAntonio Apr 25 '16 at 11:47
  • I know, this is Cesáro's lemma, but you can't use it here: the harmonic series diverge. Or am I missing something? – Bernard Apr 25 '16 at 12:00
  • @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: https://en.wikipedia.org/wiki/Ces%C3%A0ro_mean – DonAntonio Apr 25 '16 at 12:03
  • @Bernard And it has nothing to do with series, whether convergent or not. – DonAntonio Apr 25 '16 at 12:09
  • I'm sorry I don't see. $\sum\frac1k$ doesn't have a limit. – Bernard Apr 25 '16 at 12:52
  • @Bernard In both links I gave it is written that $$;\lim a_n=L\implies\lim\frac1n\sum_{k=1}^n a_k=L;$$ Can you see this? – DonAntonio Apr 25 '16 at 13:21
  • Oh! yes, of course. Stupidly, all the time of this discussion, I was thinking series instead of sequence. – Bernard Apr 25 '16 at 13:40
  • @Bernard Jeje, I know. Sometimes things are in front of my nose and I look for them down in my feet. – DonAntonio Apr 25 '16 at 13:41
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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.

Bernard
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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 $0$.

πr8
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