0

Let $u_n=\frac{1}{1+n}+\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{2n}$ be a sequence (for $n \ge 1$).

Given theses inequality for $n \in \mathbb{N} \ge 2$: $$ \tag{A}\int_{n}^{n+1}\frac{dx}{x} \le \frac{1}{n}\le \int_{n-1}^{n}\frac{dx}{x},$$ how can I prove this inequality? $$ \ln\left(\frac{2n+1}{n+1}\right)\le u_n \le \ln(2) $$

Could I use $(\text{A})$ to start with this for $k \in \mathbb{N} \ge 1$ as following:

$$\int_{k+1}^{k+2}\frac{dx}{x} \le \frac{1}{k+1}\le \int_{k}^{k+1}\frac{dx}{x},$$ and sum up all the term in order to get $u_k$?

Gary
  • 31,845
ZchGarinch
  • 355
  • 2
    You just need to sum the inequalities, this is direct – julio_es_sui_glace Sep 06 '23 at 13:36
  • Related. Some answers there may give you pointers (mine not so much). – Jyrki Lahtonen Sep 06 '23 at 13:55
  • Assuming first term is not there and its the sum $\sum_{k+1}^{2k} \frac{1}{n}$, you are trying to bound, its proved in answer by Mark Viola in Methology 1 of his/her answer in https://math.stackexchange.com/questions/73550/the-limit-of-truncated-sums-of-harmonic-series-lim-limits-k-to-infty-sum-n?noredirect=1&lq=1 (same as above comment.) – Balaji sb Sep 10 '23 at 03:05

0 Answers0