Finding if $\sum_{}^{}\frac{\ln k}{k}$ converges or diverge How can I find whether the serie
$$\sum_{}^{}\frac{\ln k}{k}$$
converge or diverge using the basic comparison test or limit test.
The part that is confusing me is how to deal with $\ln$.
 A: Hint: $\dfrac{\ln k}{k} > \dfrac1k$ for $k \geqslant 3$.
A: You know that the harmonic series $\sum_{k = 1}^\infty \dfrac 1k$  diverges. 
$\dfrac{\ln k}{k} > \dfrac 1k$ as $k$ grows, indeed, for all $k \gt 3$. You don't even need to know that when exactly, $\dfrac {\ln k}{k}$ exceeds $\dfrac 1k$, only that as $k$  grows large (indeed, very early on), $\dfrac{\ln k}{k} \gt \dfrac 1k.$ 
So you can use the comparison test for convergence/divergence.

Added for clarification about the harmonic series and its general term $a_n = \dfrac 1n$:
In a follow-up question below, it was asked why the harmonic series diverges even though the limit as $n \to \infty$ of the general term is zero? 
Recall that with respect to the convergence behavior of a given series $\sum_{n = 1}^\infty a_n$,

"IF $\;\lim_{n\to \infty} |a_n| \neq 0,\;$ then it diverges." 

So this theorem tells us definitely that any such series diverges. What it doesn't tell us is what happens when that same limit does approach $0.\;$ In those cases where $\lim_{n\to \infty} |a_n| = 0$, as in the case of the general term of the harmonic series, the associated series may converge or it may diverge. So that particular theorem is inconclusive in this case.
For a nice discussion about the divergence of the harmonic series, with proofs of its divergence (using the comparison test and one using the integral test), see the Wikipedia entry on the divergence of the harmonic series.
A: The Comparison Test: Suppose that $\sum a_n$ and $\sum b_n$ are series with positive terms.
(i) If $\sum b_n$ is convergent and $a_n\le b_n$ for all $n$, then $\sum a_n$ is also convergent.
(ii) If $\sum b_n$ is divergent and $a_n\ge b_n$ for all $n$, then $\sum a_n$ is also divergent.
Consider the Harmonic series $\sum_{k=1}^\infty {1\over k}$. We know that this series diverges. Using the Comparison Test we know that ${\ln k\over k}>{1\over k}$ for $k\ge3$. Thus the given series is divergent by the Comparison Test.
A: Another way would be to use the integral test. The series
$$
\sum_{k = 1}^{\infty} \frac{\ln(k)}{k}
$$
converges if and only if the integral
$$
\int_1^{\infty} \frac{\ln(x)}{x} \,dx
$$
converges. Letting $u = \ln(x)$ we get $du = \frac{1}{x} \,dx$ so we obtain
$$
\int_{0}^{\infty} u \, du = \left[ \frac{u^2}{2} \right]^{\infty}_{0} = \lim_{t \to \infty} \left[ \frac{u^2}{2} \right]^t_0
$$
which tends to infinity, implying that the series is divergent.
A: use limit test where
where numerater is given function and denominater is g(x)=1/x
as n->infinite limit 'l' also tends to infinite.
so if limit is infinite 
denomnater diverges implies numerater diverges.
since g(n)=1/n diverges =>given series diverges
