for which value of $a$ that $ \big(\sum\frac {1}{u_n}\big) $ converges? For any real number $a$ (positive or negative), define a sequence $\{u_n\}$ (depending on $a$) recursively by $u_0=2$ and
$$ \int_{u_n}^{u_{n+1}}(\ln u)^a\,du=1.$$
For which $a\in\mathbb{R}$ does
$$ \sum_{n=0}^\infty \frac {1}{u_n}$$
 converge?
Thank you for your replies 
 A: Note that an alternate definition of $u_n$ ($n\ge0$) is
$$
\int_2^{u_n} (\ln u)^a\,du = n.
$$
By the same techniques used to asymptotically evaluate the logarithmic integral, one can show that
$$
\int_2^{y} (\ln u)^a\,du \sim y(\ln y)^a
$$
(with explicit error terms even), from which it follows that
$$
u_n \sim n(\ln n)^{-a}.
$$
Therefore $\sum_{n=0}^\infty 1/u_n$ diverges if $a\ge-1$ and converges if $a<-1$, by (the limit comparison test and) the integral test.
A: First notice that since $\int_2^{+\infty} (\ln u)^a\,du = +\infty$ for all $a$, the sequence $(u_n)$ is well-defined and increasing.
If $a \geq 0$, then $n = \int_2^{u_n} (\ln u)^{a}\,du \geq (u_n-2) (\ln 2)^a$, so $\sum \frac{1}{u_n} = +\infty$.
If $-1 \leq a < 0$, we obtain similarly $(u_n-2)(\ln u_n)^a \leq n \leq (u_n-2) (\ln 2)^a$, which gives $\ln u_n \sim \ln n$ and $u_n = O(n(\ln n)^{-a})$, so we have again $\sum \frac{1}{u_n} = +\infty$.
If $a < -1$, we write
$$
\frac{u_{n+1}-u_n}{(\ln u_n)^{-a}} \geq \int_{u_n}^{u_{n+1}} \frac{du}{(\ln u)^{-a}} = 1
$$
and we deduce that there exists $c > 0$ such that for all $n$ large enough,
$$
u_n \geq c \sum_{k=1}^n (\ln n)^{-a} \asymp c\, n (\ln n)^{-a}.
$$
Therefore, $\sum \frac{1}{u_n} < \infty$ iff $a < -1$.
