Prove $\sum\limits_{i=1}^{n}\frac1i > \ln(n+1)$ by induction

Question

Problem

Prove $$\sum\limits_{i=1}^{n}\frac1i > \ln(n+1)$$

Caveat

While I'm familiar with the integral approach, this problems shows up in my textbook before integrals are introduced, but after induction is learned, so I'm trying to prove this by induction.

My attempt

I've only gotten so far as to the premise of the induction step.

$$\begin{split} \sum\limits_{i=1}^{k+1}\frac1i & = \sum\limits_{i=1}^{k}\frac1i + \frac1{k+1} \\ & > \ln(k+1) + \frac1{k+1} \end{split}$$

And somehow, I need to show that this will be greater than $\ln((k+1)+1) = \ln(k+2)$, however I'm completely stuck here.

Any help appreciated!

Answer 1

$$ \ln(k+2) = \ln((k+1)+1) = \ln\left((k+1)\cdot(1+\frac{1}{k+1})\right) = \ln(k+1) + \ln\left(1+\frac{1}{1+k}\right) $$ Now knowing that $\ln(x+1) \leq x$ you can deduce that : $$ \ln(k+2) \leq \ln(k+1) + \frac{1}{k+1} $$ And therefore conclude your proof.

For the proof of the fact I used, simply consider $e^x = \sum_{i=0}^\infty x^i/i! \geq 1+x \implies x \geq \ln(1+x)$.

Answer 2

Note that $\ln t$ is the area under the curve $f(x) = \frac 1x$ from $1$ to $t$.

Draw the graph of $f$ and note that: $\ln(n+1) - \ln(n) \le \max\limits_{x \in [n,n+1]}f(x) = f(n) = \frac 1n$

Answer 3

\begin{split} \sum\limits_{i=1}^{k+1}\frac1i & = \sum\limits_{i=1}^{k}\frac1i + \frac1{k+1} \\ & > \ln(k+1) + \frac1{k+1} \end{split}

All you need now is to prove that $$ \frac 1 {k+1} > \ln(k+2)-\ln(k+1). $$ you have $$ \ln(k+2)-\ln(k+1) = \int_{k+1}^{k+2} \frac{dx} x <\int_{k+1}^{k+2} \frac {dx} {k+1} = \frac 1 {k+1}. $$