<s>[INST] How to study this sequence $u_n=\sum_{k=1}^{n}\frac{1}{n+2k}$ Please is there any way to prove that sequence is increasing ?
I do: $u_{n+1}-u_n=\sum_{k=1}^{n+1}\frac{1}{(n+1)+2k}-\sum_{k=1}^{n}\frac{1}{n+2k}=\left[\frac{1}{n+3}+\frac{1}{n+5}+\ldots+\frac{1}{3n+1}+\frac{1}{3n+3}\right]-\left[\frac{1}{n+2}+\frac{1}{n+4}+\ldots+\frac{1}{3n}\right]$
i don't know how to continue 
 [/INST] OP, a good start:
$$u_{n+1}-u_n=\sum_{k=1}^{n+1}\frac{1}{(n+1)+2k}-\sum_{k=1}^{n}\frac{1}{n+2k}$$
Let's continue.  First,
$$ \sum_{k=1}^n
    \frac 1{(n+2\!\cdot\! k-\frac 12)
  \cdot(n+2\!\cdot\!k+\frac 32)}\,
    \, =\,\, \frac 12\cdot\left(
\, \frac 1{n+2-\frac 12}\, -\, \frac 1{3\cdot n+\frac 32}
\right) $$
$$ =\, \frac 1{2\cdot n+3}\, -\, \frac 1{6\cdot n+3} $$
Next,
$$ u_{n+1}-u_n\quad =\quad \frac 1{3n+3}\,\, -\,
 \, \sum_{k=1}^n\frac 1{(n+2\cdot k)
                \cdot(n+2\!\cdot\!k+1)} $$
$$ =\,\, \frac 1{3n+3}\,\, -\,\, \left(
\frac 1{2\cdot n+3}\, -\, \frac 1{6\cdot n+3}\right)\, + $$
$$ \sum_{k=1}^n
    \frac 1{(n+2\!\cdot\! k-\frac 12)
  \cdot(n+2\!\cdot\!k+\frac 32)}
    \, -\, \sum_{k=1}^n\frac 1{(n+2\cdot k)
                \cdot(n+2\!\cdot\!k+1)}   $$
$$ =\,\, \frac 1{3n+3}\,\, -\,\, \left(
\frac 1{2\cdot n+3}\, -\, \frac 1{6\cdot n+3}\right)\, + $$
$$ \frac 34\cdot\sum_{k=1}^n\frac 1
{(n+2\!\cdot\! k-\frac 12)\cdot(n+2\!\cdot\!k+\frac 32)
   \cdot(n+2\cdot k)\cdot(n+2\!\cdot\!k+1)} $$
$$ >\,\, \frac 1{3n+3}\,\, -\,\, \left(
\frac 1{2\cdot n+3}\, -\, \frac 1{6\cdot n+3}\right)\ =
$$ $$ \frac{(2\!\cdot\! n+1)\!\cdot\!(2\!\cdot\! n+3)
 + (n+1)\!\cdot\!(2\!\cdot\! n+3) -
    3\!\cdot\!(n+1)\!\cdot\!(2\!\cdot\! n+1)}
{3\cdot(n+1)\cdot(2\cdot n+1)\cdot(2\cdot n+3)} 
$$ $$ =\, \frac{4\cdot n+3}
{3\cdot(n+1)\cdot(2\cdot n+1)\cdot(2\cdot n+3)} $$
This means that the following theorem holds,
Theorem
$$ \forall_{n=1\, 2\, \ldots}\quad u_n<u_{n+1} $$
Great!
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