Prove that $\sum_{k=1}^n (1/k) > \ln(n+1)$. I have been trying to do this for some time now, but I cannot figure it out. It is on the study guide for my final exam, which is tomorrow so I am trying to figure it out. Thanks
So I know that $\sum_{k=1}^n (1/k) = 1+1/2+1/3+1/4+1/5+\dots$, but I am having a tough time really figuring out how to prove it is greater than ln(n+1). Can someone help me please? Thanks so much
We have $$\ln(n+1) = \int_1^{n+1} \dfrac{dx}x = \underbrace{\sum_{k=1}^n \int_k^{k+1} \dfrac{dx}x < \sum_{k=1}^n \int_k^{k+1} \dfrac{dx}k}_{\frac1x < \frac1k \text{ for all }x \in (k,k+1)} = \sum_{k=1}^n \dfrac1k$$
Hint: $$\log x= \int_1^n \frac{1}{x} dx=\int_1^2 \frac{1}{x}dx+\int_2^3 \frac{1}{x}dx+\cdots+\int_{n-1}^n\frac{1}{x}dx$$
A proof similar to the others, but using derivatives instead of integrals:
First, we show $\log(n+1)-\log n\le\frac1n$. Given this, we can sum to get $$\sum_{k=1}^n(\log(k+1)-\log k)=\log(n+1)\le\sum_{k=1}^n\frac1n.$$
Note that $\log'(x)=\frac1x$, so that on the interval $[n,n+1]$ the derivative of $\log(x)$ is bounded by $\frac1n$. This implies Lipschitz continuity of $\log$, specifically $|\log x-\log y|\le\frac1n|x-y|$ for all $x,y\in[n,n+1]$, and choosing $x=n+1$ and $y=n$ we get $$\log(n+1)-\log n\le|\log(n+1)-\log n|\le\frac1n.$$
