Solve this: $\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}>\frac{2}{3}$; without induction

Question

How can I solve this $\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}>\frac{2}{3}$; using this relation: $AM>HM$?

Answer 1

$$\frac{\sum_{i=n}^{2n}\frac1i}{n+1}\geqslant \frac{n+1}{\sum_{i=n}^{2n}i}=\frac{n+1}{\frac{3}2\cdot n(n+1)}=\frac{2}{3n}\implies \sum_{i=n}^{2n}\frac1i\geqslant \frac23\cdot \frac{n+1}n>\frac23$$

Answer 2

Since $\dfrac1{n} \gt \int_n^{n+1} \dfrac{dt}{t} $,

$\begin{array}\\ \sum_{k=n}^{2n} \dfrac1{k} &\gt \sum_{k=n}^{2n} \int_k^{k+1} \dfrac{dt}{t}\\ &= \int_n^{2n+1} \dfrac{dt}{t}\\ &= \ln(2n+1)-\ln(n)\\ &=\ln(2+\frac1{n})\\ &\gt \ln(2)\\ &= 0.693...\\ &\gt \dfrac23\\ \end{array} $

Answer 3

Dr. Mathva has given an AM>HM answer. Here is another approach based on "reflecting" the sum and grouping pairs of terms together.

$$\begin{align} \sum_{k=0}^n{1\over n+k}&={1\over2}\sum_{k=0}^n\left({1\over n+k}+{1\over2n-k}\right)\\ &=\sum_{k=0}^n\left({1\over3n-(n-2k)}+{1\over3n+(n-2k)} \right)\\ &=\sum_{k=0}^n{6n\over9n^2-(n-2k)^2}\\ &\ge\sum_{k=0}^n{6n\over9n^2}\\ &={2(n+1)\over3n}\\ &\gt{2\over3} \end{align}$$

Answer 4

Here's an elementary proof that the sum is between $\dfrac23$ and $\dfrac{3(1+\frac1{n})}{4} $.

$\begin{array}\\ s(n) &=\sum_{k=0}^{n} \dfrac1{n+k}\\ &=\sum_{k=0}^{n} \dfrac1{2n-k}\\ \text{so}\\ 2s(n) &=\sum_{k=0}^{n} (\dfrac1{n+k}+\dfrac1{2n-k})\\ &=\sum_{k=0}^{n} \dfrac{n+k+2n-k}{(n+k)(2n-k)}\\ &=\sum_{k=0}^{n} \dfrac{3n}{(n+k)(2n-k)}\\ &=3n\sum_{k=0}^{n} \dfrac1{2n^2+nk-k^2}\\ &=3n\sum_{k=0}^{n} \dfrac1{2n^2+k(n-k)}\\ &\lt 3n\sum_{k=0}^{n} \dfrac1{2n^2}\\ &= 3n\dfrac{n+1}{2n^2}\\ &= \dfrac{3(n+1)}{2n}\\ \text{so}\\ s(n) &\lt \dfrac{3(1+\frac1{n})}{4}\\ \text{and}\\ 2s(n) &=3n\sum_{k=0}^{n} \dfrac1{2n^2+kn-k^2}\\ &=3n\sum_{k=0}^{n} \dfrac1{2n^2+n^2/4-n^2/4+kn-k^2}\\ &=3n\sum_{k=0}^{n} \dfrac1{2n^2+n^2/4-(k-n/2)^2}\\ &\gt 3n\sum_{k=0}^{n} \dfrac1{2n^2+n^2/4}\\ &\gt 3n\sum_{k=0}^{n} \dfrac{4}{9n^2}\\ &= \dfrac{4(n+1)}{3}\\ &\gt \dfrac43\\ \text{so}\\ s(n) &\gt \dfrac23\\ \end{array} $