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This is a problem taken from an exam of Analysis 1 course I attend.

$$\lim_{n\to \infty}\left(\frac{1}{5n-2}+\frac{1}{5n-1}+\cdots+\frac{1}{8n+2}\right)$$

I tryed to solve this limit and I've got the following results

$$\underbrace{\lim_{n\to \infty}\left(\frac{3n+5}{5n-2}\right)}_{{\longrightarrow}\frac{3}{5}}\geq\lim_{n\to \infty}\sum_{k=1}^{3n+5}\frac{1}{5n+k-3}\geq\underbrace{\lim_{n\to \infty}\left(\frac{3n+5}{8n+2}\right)}_{{\longrightarrow}\frac{3}{8}}$$

I was going for Squeeze theorem, but I got that $\text{LHS}\left(\frac{3}{5}\right)\space\ne\space \text{RHS}\left(\frac{3}{8}\right)$. Since the squeeze theorem works only in one direction, it's not guaranteed it diverges.

So, what did I do wrong or what other method to use on this problem ?

Reminder: The limit is supposed to be solved only with the knowledge prior to derivatives and integrals.

EDIT: Are these kinds of problems called truncated sums or series, similar problem here?

Thanks in advance

powerline
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  • Who told you the limit diverges? – Jacky Chong Jan 25 '20 at 07:46
  • Nobody, I should change the title, since only afterwards it came to my mind that LHS≠RHS doesn't mean anything. – powerline Jan 25 '20 at 07:47
  • I'm sure it converges to $4.700648791973119 \times 10^{-1}$ when you put $n=10^6$ and do the sum for example. – K.K.McDonald Jan 25 '20 at 07:49
  • You need integrals or specialized estimates to evaluate the sum, see here So I retracted the vote to close as a duplicate of that particular thread. Anyway, a study of the differences between two consecutive sums together with the estimates you already have suggest that the sequence is monotone and bounded. – Jyrki Lahtonen Jan 25 '20 at 08:06
  • @JyrkiLahtonen I used the proposed rule and I got the result $log\frac{8}{5}$, what is the same Dr Zafar Ahmed Dsc got using integrals below. So I guess that works. What bothers me is, I don't know the name of that property and I can't recall using it at lectures(i.e. is it a consequence of some theorems which come e.g. after derivatives?). Also I got the mentioned result when I neglected free terms everywhere, and I don't know the reason why I could neglect them. Anyway, thanks for the answer, that tool will surely find its place, if not now, someday. – powerline Jan 25 '20 at 08:18
  • @JyrkiLahtonen In the same thread I saw this, Method 2. We learned that Lagrange consequence, so I can use it. But I am not sure if it applies to my problem. What do you think ? – powerline Jan 25 '20 at 08:24
  • @GrigoriPerelman, do you have the exam on Monday? – PinkyWay Jan 25 '20 at 14:41
  • @VerkhovtsevaKatya No, why? – powerline Jan 25 '20 at 16:29
  • @GrigoriPerelman, I thought you might because this is how my studying looks like most of the time. – PinkyWay Jan 25 '20 at 17:59
  • @VerkhovtsevaKatya Does it work ? :) – powerline Jan 25 '20 at 18:02
  • So far, so good,MSE is an excellent place to become independent on some sense and get used to working on your own. May I ask which literature does this task come from? (: – PinkyWay Jan 25 '20 at 18:38
  • It is a problem from an exam, so I am not sure from which literature it comes. It's possible as well that the teaching assistant made up the problem by himself alone. But generally we follow the V. Zorich Analysis textbook. – powerline Jan 25 '20 at 21:34

4 Answers4

5

In one of your comments you cite a property you are allowed to use:

  • $\frac 1{k+1} < \ln (k+1) - \ln k < \frac 1k$

Using this you get

$$ \sum_{k=-2}^{3n+2} \left(\ln (5n+k+1) - \ln (5n+k)\right) <$$ $$\sum_{k=-2}^{3n+2}\frac 1{5n+k} < $$ $$\sum_{k=-2}^{3n+2}\left(\ln (5n+k) - \ln (5n+k-1)\right)$$

Hence, telescoping gives

$$\underbrace{\ln\left(\frac{8n+3}{5n-2}\right)}_{\stackrel{n\to\infty}{\longrightarrow}\ln \frac 85} < \sum_{k=-2}^{3n+2}\frac 1{5n+k} < \underbrace{\ln\left(\frac{8n+2}{5n-3}\right)}_{\stackrel{n\to\infty}{\longrightarrow}\ln \frac 85}$$

trancelocation
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  • You wrote $\sum_{k=-2}^{3n+2}\left(\ln (5n+k) - \ln (5n+k-1)\right)$ should here instead be $\sum_{k=-2}^{3n+2}\left(\ln (5n+k) - \ln (5n+k+1)\right)$ ? – powerline Jan 25 '20 at 17:50
  • For the RHS of my inequality you need $\frac 1{k+1} < \ln (k+1) - \ln k$. So, the argument of the second logarithm must be the first one minus $1$. – trancelocation Jan 25 '20 at 18:00
  • I can't understand that part at all, could you explain it further please. Besides from that, again a great and clear answer, with tools I can use. I knew it can be this simple. – powerline Jan 25 '20 at 18:06
3

$$\lim_{n \rightarrow \infty}S_n=\lim_{n\rightarrow \infty}\sum_{k=-2}^{3n+2}\frac{n}{5n+k} \frac{1}{n}= \int_{0}^{3} \frac{dx}{5+x}=\ln 8-\ln5=\ln(8/5).$$

Z Ahmed
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1

It's monotone and bounded. Hence the limit exists.

Jacky Chong
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Without any integral but using the properties of harmonic numbers, we have $$S_n=\sum_{i=0}^{3n+4}\frac 1 {5n-2+i}=H_{8 n+2}-H_{5 n-3}$$ Now, for large $p$, consider the asymptotics $$H_p=\gamma +\log \left({p}\right)+\frac{1}{2 p}+O\left(\frac{1}{p^2}\right)$$ Apply it twice and continue with Taylor series to get $$S_n=\log \left(\frac{8}{5}\right)+\frac{13}{16 n}+O\left(\frac{1}{n^2}\right)$$ which shows the limit and how it is approached.

Claude Leibovici
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  • Thanks for the contribution, but this is far more advanced than what we learned so far. We didn't encounter the O notation and the taylor series yet. – powerline Jan 25 '20 at 09:23
  • @GrigoriPerelman. Are you supposed to know that $H_p\sim \gamma+\log(p)$. If yes, you have the limit. – Claude Leibovici Jan 25 '20 at 09:25
  • This also appeared in the linked thread. Can you offer something new and/or more elementary (that does not depend on the ability to integrate $1/x$). – Jyrki Lahtonen Jan 25 '20 at 11:13