Divergence of Harmonic Series and Sequence Behavior at Infinity
Let $(H_n)$ the sequence defined for $n\in \mathbb{N}$ by
$$H_n= \sum_{k=1}^n \frac 1k.$$
- Prove that $H_n\xrightarrow[n\to+\infty]{}+\infty$.
- Let $(u_n)$ a sequence such that $n(u_{n+1}-u_n)\xrightarrow[n\to+\infty]{}1$. Prove that $u_n\xrightarrow[n\to+\infty]{}+\infty$.
Show Answer:
1. Knowing $\ln(1+x)\le x$, we have
$$\frac1k\ge \ln\left(1+\frac1k\right)=\ln(k+1)-\ln k$$
so
$$H_n\ge \sum_{k=1}^n \ln(k+1)-\ln k= \ln(n+1)$$
and then $H_n\xrightarrow[n\to+\infty]{}+\infty$.
2. There exists $N\in \mathbb{N}$ such that for all $n\ge N$,
$$n(u_{n+1}-u_{n})\ge 1/2.$$
Hence
$$ u_{n+1}-u_{N}\ge \sum_{k=N}^n u_{k+1}-u_{k}\ge \frac12 \sum_{k=N}^n\frac1k= \frac12(H_{n}-H_{N-1})\xrightarrow[n\to+\infty]{}+\infty$$
and then $u_{n}\xrightarrow[n\to+\infty]{}+\infty$.