Prove by induction that $\sum_{k=1}^n \frac 1k \ge \ln (n+1)$ for all $n \ge 1, n \in \Bbb N$ I'm working on the following question:
Prove by induction that $$\sum_{k=1}^n \frac 1k \ge \ln (n+1)$$ for all $n \ge 1,        n \in \Bbb N$
Here's what I managed to achieve so far:
Base case for $n=1$
For n=1, we are left with the inequality $1 \ge \ln (2)$, which is obviously true.
Induction assumption
We assume that $\sum_{k=1}^n \frac 1k \ge \ln (n+1)$ is true for some natural number $n \ge 1$.
Proof
We must now prove the following: $\sum_{k=1}^{n+1} \frac 1k \ge \ln (n+2)$, given the assumption.
Now here's what I managed to do in this part:
$$\sum_{k=1}^{n+1} \frac 1k = \sum_{k=1}^n \frac 1k + \frac 1{n+1}$$
Using the assumption $\sum_{k=1}^n \frac 1k \ge \ln (n+1)$ we may write $$\sum_{k=1}^{n+1} \frac 1k \ge \ln (n+1) + \frac 1{n+1}$$
It is here that I'm stuck. I would greatly appreciate it if anyone could point out the next steps or at least give me a hint as to where to continue.
Thanks a lot in advance!
 A: You're almost done, it remains to prove that $$\ln (n+1) + \frac 1{n+1}\geq \ln(n+2)$$
This is equivalent to $$\frac 1{n+1}\geq  \ln\left(1+\frac{1}{n+1}\right)$$
which is a consequence of the concavity inequality $\ln(1+x)\leq x$ for $x>-1$.
A: it must be $$\frac{1}{n+1}\geq \ln\left(1+\frac{1}{n+1}\right)$$ with $$x=\frac{1}{n+1}$$ you must prove that $$x\geq \ln(1+x)$$ can you finish?
A: Just out of curiosity, how does this comparison to ∫dt/t work? 
I'm not familiar with this method. 


The curve $y=\dfrac 1x$ in the interval $[k,k+1]$ is stuck between $y=\dfrac 1k$ above and $y=\dfrac 1{k+1}$ below.
You are typically in the case of squeezing the integral of $\frac 1x$ between the upper and lower rectangles area approximation.
On the interval $[k,k+1]\begin{cases}
\color{red}{\mathcal U_k}=((k+1)-k)\times\frac 1k=\frac 1k\\
\color{green}{\mathcal L_k}=((k+1)-k)\times\frac 1{k+1}=\frac 1{k+1}\end{cases}\quad$ and $\quad\displaystyle\color{green}{\mathcal L_k}\le\color{blue}{\int_k^{k+1} \dfrac{\mathop{dt}}t}\le \color{red}{\mathcal U_k}$
Now sum everything to get :
$\displaystyle \sum\limits_{k=1}^{n} \dfrac 1{k+1}\le\int_1^{n+1} \dfrac{\mathop{dt}}t\le \sum\limits_{k=1}^{n} \dfrac 1{k}$
In particular you get the desired inequality :
$\displaystyle \sum\limits_{k=1}^{n} \dfrac 1{k}\ge\int_1^{n+1} \dfrac{\mathop{dt}}t=\ln(n+1)$
