I'm currently learning about mathematical induction right now and I'm stuck on this question:
Use induction to prove $\frac{1}{2}+...+\frac{1}{n} \leq \ln(n)$
I know $n=2$ is true, then for $n=k$, $\frac{1}{2}+...+\frac{1}{k} \leq \ln(k)$
Now for $n=k+1$,
$\frac{1}{2}+...+\frac{1}{k}+\frac{1}{k+1} \leq \ln(k)+\frac{1}{k+1}$
This is where I am unsure how to proceed. Should subtract $\ln(k)$ from both sides as $\ln(k)>0, k>1$?
Use the definition of $\ln(n)$ instead. To be clear here: Let $k'$ be an integer. Then $\ln (k')$ is as defined as follows: $$\ln (k') = \int_1^{k'} \frac{dx}{x}.$$
So from this definition, for all positive integers $k$: $$\ln (k+1) = \ln (k) + \int_k^{k+1} \frac{dx}{x} \ \ge \ \ln (k) + \frac{1}{k+1}.$$ [The last inequality following because $\frac{1}{x} \ge \frac{1}{k+1}$ for all $x \in [k,k+1]$.] Now are you able to plug this in to your inductive hypothesis to finish.
You want to prove that$$\frac12+\frac13+\cdots+\frac1k+\frac1{k+1}\leqslant\log(k+1)\tag1$$assuming that$$\frac12+\frac13+\cdots+\frac1k\leqslant\log k.\tag2$$Note that, in order to prove $(1)$, it is enough to prove that $\log(k)+\frac1{k+1}\leqslant\log(k+1)$, since you are assuming $(2)$. But\begin{align}\log(k)+\frac1{k+1}\leqslant\log(k+1)&\iff\frac1{k+1}\leqslant\log(k+1)-\log(k)\\&\iff\frac1{k+1}\leqslant\int_k^{k+1}\frac{\mathrm dx}x,\end{align}which is true, since$$(\forall x\in[k,k+1]):\frac1{k+1}\leqslant\frac1x.$$
Then note that,
$$\ln k+ \frac{1}{k+1} ≤\ln(k+1).$$
$\large{\underline{\text{Proof:}}}$
$$\frac {1}{k+1}≤\ln \left(\frac{k+1}{k}\right)$$
Let $\dfrac{k+1}{k}=1+\dfrac 1k=e^t, t>0$
$$k+1-ke^t=0 \Longrightarrow k(1-e^t)=-1$$
$$k(e^t-1)=1 \Longrightarrow k=\frac{1}{e^t-1}$$
$$\frac{1}{k+1}=\frac{1}{\frac{1}{e^t-1}+1}$$
$$\frac{e^t-1}{e^t}≤t$$
$$f(t)=te^t-e^t+1≥0$$
$f(t)$ is the strictly increasing function for $∀t>0$, because $f'(t)=te^t>0, ∀ t>0$ and if $t=0$, then $f(t)=0$. Therefore, for all $t>0$, we have
$$\begin{align}&f(t)=te^t-e^t+1≥0 \\ &\Longleftrightarrow \ln k+ \frac{1}{k+1} ≤\ln(k+1).\end{align}$$