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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

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  • $\begingroup$ Who told you the limit diverges? $\endgroup$ Jan 25 '20 at 7:46
  • $\begingroup$ Nobody, I should change the title, since only afterwards it came to my mind that LHS≠RHS doesn't mean anything. $\endgroup$
    – tau20
    Jan 25 '20 at 7:47
  • $\begingroup$ I'm sure it converges to $4.700648791973119 \times 10^{-1}$ when you put $n=10^6$ and do the sum for example. $\endgroup$ Jan 25 '20 at 7:49
  • $\begingroup$ 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. $\endgroup$ Jan 25 '20 at 8:06
  • $\begingroup$ @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. $\endgroup$
    – tau20
    Jan 25 '20 at 8:18
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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

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

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  • $\begingroup$ You wrote $\sum_{k=-2}^{3n+2}\left(\ln (5n+k) - \ln (5n+k-1)\right)$ should here instead be $\sum_{k=-2}^{3n+2}\left(\ln (5n+k) - \ln (5n+k+1)\right)$ ? $\endgroup$
    – tau20
    Jan 25 '20 at 17:50
  • $\begingroup$ 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 $1$. $\endgroup$ Jan 25 '20 at 18:00
  • $\begingroup$ 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. $\endgroup$
    – tau20
    Jan 25 '20 at 18:06
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$$\lim_{n \rightarrow \infty}S_n=\lim_{n\rightarrow \infty}\sum_{k=-2}^{3n+2}\frac{n}{5n+k} \frac{1}{n}= \int_{0}^{3} \frac{dx}{5+x}=\ln 8-\ln5=\ln(8/5).$$

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

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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.

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  • $\begingroup$ 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. $\endgroup$
    – tau20
    Jan 25 '20 at 9:23
  • $\begingroup$ @GrigoriPerelman. Are you supposed to know that $H_p\sim \gamma+\log(p)$. If yes, you have the limit. $\endgroup$ Jan 25 '20 at 9:25
  • $\begingroup$ 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 $1/x$). $\endgroup$ Jan 25 '20 at 11:13

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