I need to bound the k-th partial sum or the Harmonic series. i.e.
$$ln(k+1)<\sum_{m=1}^{k}\frac{1}{m}<1+ln(k)$$
I'm triying to integrate in $[m,m+1]$ in the relation $\frac{1}{m+1}<\frac{1}{x}<\frac{1}{m}$ for all $x\in[m,m+1]$ and I get: $$\int_{m}^{m+1}\frac{1}{m+1}dx<ln(m+1)-ln(m)<\int_{m}^{m+1}\frac{1}{m}dx$$ then $$\sum_{m=1}^{k}\frac{1}{m+1}<ln(k)<\sum_{m=1}^{k}\frac{1}{m}$$ but I don know how conclude or continue... please help.
Noting that $$ \frac1{x+1}<\frac1m<\frac{1}{x},x\in(m-1,m)$$ one has $$ \sum_{m=1}^k \frac1m=\sum_{m=1}^k\int_{m-1}^m\frac{1}{m}dx<1+\sum_{m=2}^k\int_{m-1}^m\frac{1}{x}dx=1+\int_1^k\frac1xdx=1+\ln k$$ and $$ \sum_{m=1}^k \frac1m=\sum_{m=1}^k\int_{m-1}^m\frac{1}{m}dx>\sum_{m=1}^k\int_{m-1}^m\frac{1}{x+1}dx=\int_0^1\frac1{x+1}dx+\int_1^2\frac1{x+1}dx+\dots+\int_{k-1}^k\frac1{x+1}dx=\int_0^k\frac1{x+1}dx=\ln (k+1). $$ So $$ \ln(k+1)<\sum_{m=1}^k \frac1m<1+\ln k. $$
For $m\ge 2$, $1/u\le 1/m \le 1/(u-1)$ for $u \in [m,m+1]$
By integration on this interval
$$\ln(m+1)-\ln m \le 1/m \le \ln m - \ln(m-1)$$
You then just have to sum those inequalities.
An intuitive approach here can be found by thinking pictorally about the sums and functions at hand. We know that the area underneath the curve $\frac{1}{x}$ from $1$ to $k$ is given by $\ln(k)$. The partial sums $\sum_{m=1}^n \frac{1}{m}$ can be visualized by laying out a block of height and length $1$ starting at $x = 1$, then of height $\frac{1}{2}$ and length $1$ starting at $x = 2$, of height $\frac{1}{3}$ and length $1$ at $x = 3$, and so on up to a block of height $\frac{1}{k}$ and length $1$ at $x = k$. If we draw out a few blocks and then think about how to bound above or below by graphs that look like possibly shifted versions of $\frac{1}{x}$, why this bound holds should become much easier to see. A rigorous argument with integrals then follows fairly naturally.
Continuing your original thinking with a few modifications:
$$ln(m+1)-ln(m)<\int_{m}^{m+1}\frac{1}{m}dx$$ then $$\sum_{m=1}^{k}ln(m+1)-ln(m)<\sum_{m=1}^{k} \int_m^{m+1}\frac{1}{m} dx$$ and $$ln(k+1)<\sum_{m=1}^{k}\frac{1}{m}$$
which gives you the LHS of the inequality.
The RHS comes from:
$$\frac{1}{m-1} < x < \frac{1}{m}$$ for $x\in [\frac{1}{m}, \frac{1}{m - 1}]$
Therefore:
$$\int_{m-1}^m \frac{1}{m} dx< ln(m) - ln(m-1)$$
and
$$\sum_{m=2}^{k}\int_{m-1}^m \frac{1}{m} dx< \sum_{m=2}^{k} ln(m) - ln(m-1)$$
then
$$\sum_{m=2}^{k}\frac{1}{m} < ln(k)$$
and finally,
$$\sum_{m=1}^{k}\frac{1}{m} < ln(k) + 1$$
Where we added $1$ to each side.
In order to arrive at this inequality, we can use an approach that uses your ideas. We have for every $m$,
$$\dfrac{1}{m+1} < \ln(m+1) - \ln(m) < \dfrac{1}{m}$$
Now consider the right hand side:
$$ \ln(m+1) - \ln(m) < \dfrac{1}{m}$$
Putting $m=k$, we can arrive at the left hand side inequality as follows:
$$ \begin{align}\ln(k+1) &< \dfrac{1}{k} + \ln(k) \\ &< \dfrac{1}{k}+ \dfrac{1}{k-1} + \ln(k-1) \tag{putting $m=k-1$} \\ &\vdots \\ &< \sum^{k}_{i=1} \dfrac{1}{i} \end{align}$$
This gives the lower bound of the required inequality.
Now consider,
$$\dfrac{1}{m+1} < \ln(m+1) - \ln(m)$$ $$ \implies \dfrac{1}{m+1} + \ln(m) < \ln(m+1)$$
Here we will start with $m=k-1$:
$$\begin{align} \ln(k) &> \dfrac{1}{k} + \ln(k-1) \\ &> \dfrac{1}{k}+ \dfrac{1}{k-1}+\ln(k-2) \tag{putting $m=k-2$} \\ &\vdots \\ &> \sum^{k}_{i=2} \dfrac{1}{i} + \ln(1) \\ &= \sum^{k}_{i=2} \dfrac{1}{i} \end{align}$$
Thus, adding 1 on both sides gives us the desired upper bound of the inequality.
QED.
