$a_{n+1}\ge a_n$ for $a_n=\frac1{n+1}+\frac1{n+2}+\dots+\frac1{2n}$ 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}      $
$$  = {1 \ \over 2(n+1)}-{1 \ \over n+1}={1 \ \over (n+1)} \gt\ 0 $$
 A: Supppose $n \geq 3$.\begin{align*} a_{n+1} &= \frac{1}{n+2} + \frac{1}{n+3}+ \cdots + \frac{1}{2n+1} + \frac{1}{2n+2}  \\ & =\left(  \frac{1}{n+1} + \frac{1}{n+2}+ \cdots + \frac{1}{2n-1} + \frac{1}{2n} \right) + \left( \frac{1}{2n+1} + \frac{1}{2n+2} - \frac{1}{n+1}   \right) \\ & = a_n + \left( \frac{1}{2n+1} + \frac{1}{2n+2} - \frac{1}{n+1} \right) \end{align*}
Observe $$ \frac{1}{2n+1} + \frac{1}{2n+2} - \frac{1}{n+1} > \frac{1}{2n+2} + \frac{1}{2n+2} - \frac{1}{n+1} =0$$
Thus $a_{n+1} \geq a_n$.
To prove $a_{2019} > \frac{3}{5}$, I recommended you to compute first few values of $\{a_n\}_{n \geq 3}$. Fortunately, at $n=3$ we have 
$$a_{2019} \geq a_3 = \frac{1}{4} + \frac{1}{5} + \frac{1}{6} = \frac{37}{60} >\frac{3}{5}$$
A: 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 $n=2019$.
A: Cancelling the common terms:
$$
\begin{align}
a_{n+1}-a_n
&=\frac1{2n+1}+\frac1{2n+2}-\frac1{n+1}\\
&=\frac1{(2n+1)(2n+2)}\\[6pt]
&\gt0
\end{align}
$$
A: You are done with $$a_{2019}>a_3={1\over 4}+{1\over 5}+{1\over 6}={37\over 60}>{3\over 5}$$
A: 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.
