How this sequence is decreasing?

Question

Given sequence $a_n = \dfrac{1}{n}+\dfrac{1}{n+1}+\dfrac{1}{n+2}...+\dfrac{1}{2n-1}$

$a_{n+1}-a_{n} <0$

Which means its an decreasing sequence. But when i calculate first three terms they are not seems to portray a decreasing sequence, may be i am computing them wrong. Someone correct me please.
$a_1=1$

$a_2=1+\frac{1}{3}=1.33$

$a_3=1+\frac{1}{3}+\frac{1}{5}=1.53$

Answer 1

Note that:

\begin{align} a_n = \frac1{n} +\frac1{n+1} +\frac1{n+2} + \ldots+ \frac1{n+(n-1)}\\ a_{n+1}=\frac{1}{n+1} + \frac1{n+2} +\ldots +\frac1{2n+1}\\ a_{n+1}-a_n = \frac1{2n}+\frac1{2n+1}-\frac1{n}=\frac1{2n+1}-\frac1{2n}=-\frac1{2n(2n+1)}<0 \end{align}

Note that this aligns with your intuitions, but you have made some computational mistakes. Note that $$a_2=\frac12+\frac13 \neq 1 +\frac13$$ and $$a_3=\frac13+\frac14+\frac15=\frac{47}{60}$$ $$\neq 1+\frac13+\frac15$$

Answer 2

Your computations are a bit off: $$ a_1=\frac11 $$ $$ a_2=\frac12+\frac13 $$ $$ a_3=\frac13+\frac14+\frac15 $$

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