$a_{n+1}\ge a_n$ for $a_n=\frac1{n+1}+\frac1{n+2}+\dots+\frac1{2n}$

Question

Let $a_n = {1 \ \over n+1} +{1 \ \over n+2}+ \ldots\ +{1 \ \over 2n}. $ Prove that for $ n \ge\ 3 $ one has $ a_{n+1}\ge\ a_n $. and based on this conclude that $ {a_{2019}}>{3 \over 5}$

I try $ a_{n+1}- a_n = {1 \ \over n+2}+{1 \ \over n+3} +{1 \ \over n+4}+\ldots\ + {1 \ \over 2(n+1)} - {1 \ \over n+1} -{1 \ \over n+2}-{1 \ \over n+3}-{1 \ \over n+4}- \ldots\ -{1 \ \over 2n} $ $$ = {1 \ \over 2(n+1)}-{1 \ \over n+1}={1 \ \over (n+1)} \gt\ 0 $$

Answer 1

You are done with $$a_{2019}>a_3={1\over 4}+{1\over 5}+{1\over 6}={37\over 60}>{3\over 5}$$

Answer 2

The function $f(x)=\frac1x$ is convex for $x>0$, by Jensens inequality $$ f\left(\frac{x_1+...+x_n}n\right)\le \frac{f(x_1)+...+f(x_n)}n, $$ which is equivalent to the inequality of harmonic and arithmetic mean.

Now setting $x_k=n+k$ results in $$ \frac{n}{(n+1)+(n+2)+...+(2n)}\le \frac{\frac1{n+1}+\frac1{n+2}+...+\frac1{2n}}n $$ or $$ a_n\ge\frac{2n}{3n+1}=\frac23-\frac{2}{3(3n+1)} $$ which is clearly greater than $\frac35=\frac23-\frac1{15}$ for $n=2019$.

Answer 3

It's easy and nice to prove:

THEOREM $$ \sum_{k=n+1}^{2\cdot n}\frac 1k\ = \ \sum_{k=1}^{2\cdot n} \frac {(-1)^{k-1}}k $$

Now $\ a_{n+1}>a_n\ $ is instantly obvious.

Answer 4

Cancelling the common terms: $$ \begin{align} a_{n+1}-a_n &=\frac1{2n+1}+\frac1{2n+2}-\frac1{n+1}\\ &=\frac1{(2n+1)(2n+2)}\\[6pt] &\gt0 \end{align} $$