<s>[INST] $\frac{1}{n}+\ln {n}<1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}<1+\ln{n}$ for $n\ge2$. How to prove the following bound:
$\frac{1}{n}+\ln {n}<1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}<1+\ln{n}$ for $n\ge2$.

My Attempt:

We have:$$\int_1^n\frac1xdx=\ln n$$So,$$\sum_{i=1}^n\frac{1}{i+1}\le\int_1^n\frac1xdx=\ln n\le\sum_{i=1}^n\frac1i$$
Adding $1$ on both sides to get the inequality: $1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}<1+\ln{n}$
How Can I prove the inequality given on the LHS? Also, in this case: Why is the sum $\sum_{x=1}^n\frac{1}{x}$ more than the value of the integral $\int_1^{n}\frac{1}{x}dx$? Thanks for the help.
 [/INST] Since $e^x>1+x$ for all $x>0$, we have for $k\geq 1$ that
$$
e^{\frac{1}{k}}  > 1 + \frac{1}{k} = \frac{{k + 1}}{k}.
$$
Thus
$$
e^{\sum\limits_{k = 1}^{n - 1} {\frac{1}{k}} }  > \prod\limits_{k = 1}^{n - 1} {\frac{{k + 1}}{k}}  = n,
$$
$$
\sum\limits_{k = 1}^{n - 1} {\frac{1}{k}}  > \log n,
$$
$$
\sum\limits_{k = 1}^n {\frac{1}{k}}  > \log n + \frac{1}{n}.
$$
To prove the upper bound note that
$$
\sum\limits_{k = 1}^n {\frac{1}{k}}  = 1 + \sum\limits_{k = 2}^n {\frac{1}{k}}  < 1 + \sum\limits_{k = 2}^n {\int_{k - 1}^k {\frac{{dt}}{t}} }  = 1 + \int_1^n {\frac{{dt}}{t}}  = 1 + \log n.
$$
</s>