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 
 A: We can write
$$u_n = \int \limits_0^1 (x^{n+1} + x^{n+3} + \cdots + x^{3n-1}) dx = \int \limits_0^1 x^{n+1} \left( \frac{1-x^{2n}}{1-x^2} \right) dx.$$
Now, replace $n$ by a continuous variable $t$ and try to prove that $u_t$ is monotonous (Hint: Differentiate with respect to $t$).
A: 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!
A: The early original OP's question was how to study this sequence.
You could think in terms of the logarithmic function. Already, the crudest approximation of $\ \frac 1{m+2\cdot k}\ $ gives about
$$ \frac 12\cdot\log\frac{m+2\cdot k+2}{m+2\cdot k} $$
Now you can approximate (crudely)
$\ \sum_{k=1}^n\frac 1{n+2\cdot k}.\ $ You get roughly
$$ \frac 12\cdot\log\left(\frac{3\cdot n+2}{n+2}\right) $$
The above expression increases (slowly; but I can't make any claim yet about the actual OP's series), and you may expect that the limit will slowly shoot up to reach $\ \frac{\log(3)}2.$
Of course, one needs to be more careful than this. One should use two-sided estimates, and one may enjoy finer approximations of
 $\ \log.$

Remark Axioms
  
  
*
  
*
*
  
*$ \qquad\qquad \log(1)=0;$
  
  
*
  
*$\ \,\forall_{x\ y>0}\quad \log(x\cdot y)\ = \log(x)+\log(y); $
  
  
*
  
*$\ \,\forall_{x>-1}\quad \log(1+x)\ \le\ x; $
characterize logarithm. This means that all sophisticated logarithmic
  inequalities can be derived from the above one simple inequality (with the help of the other two axioms).

For instance:
$$ -\log(1+x)\ =\ \log\left(\frac 1{1+x}\right)\ =
\ \log\left(1-\frac x{1+x}\right)
\ \le\ \frac{-x}{1+x} $$
This, together with axiom 3, gives:
$$ \forall_{x>-1}\quad \frac x{1+x} \le\ \log(1+x)\ \le\ x $$

Now, one gets logarithmic approximations on both sides.

A: Making the problem more general and assuming you know about harmonic numbers
$$u_n=\sum_{k=1}^{n}\frac{1}{n+ak}=\frac 1a\sum_{k=1}^{n}\frac{1}{\frac n a+k}=\frac 1a\left(H_{n+\frac{n}{a}}-H_{\frac{n}{a}} \right)$$ Now, at least for large values of $n$, using the asymptotics
$$H_p=\gamma +\log \left({p}\right)+\frac{1}{2 p}-\frac{1}{12
   p^2}+\frac{1}{120
   p^4}+O\left(\frac{1}{p^6}\right)$$ apply it and continue with Taylor series to get
$$u_{n+1}-u_n=\frac{a}{2 (a+1) n^2}-\frac{a(a^2+5 a+3)}{6 (a+1)^2
   n^3}+O\left(\frac{1}{n^4}\right)$$
