Evaluate $\lim_{n\to \infty}\left(\frac{1}{5n-2}+\frac{1}{5n-1}+\cdots+\frac{1}{8n+2}\right)$ 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
 A: 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
$$\underbrace{\ln\left(\frac{8n+3}{5n-2}\right)}_{\stackrel{n\to\infty}{\longrightarrow}\ln \frac 85} < \sum_{k=-2}^{3n+2}\frac 1{5n+k} <  \underbrace{\ln\left(\frac{8n+2}{5n-3}\right)}_{\stackrel{n\to\infty}{\longrightarrow}\ln \frac 85}$$
A: $$\lim_{n \rightarrow \infty}S_n=\lim_{n\rightarrow \infty}\sum_{k=-2}^{3n+2}\frac{n}{5n+k} \frac{1}{n}= \int_{0}^{3} \frac{dx}{5+x}=\ln 8-\ln5=\ln(8/5).$$
A: It's monotone and bounded. Hence the limit exists.
A: 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.
