An inequality with infinite sum If $p>1$, and $k$ is a positive integer more than $1$, 
show that $$\sum_{n=2}^{\infty}\frac{(\ln n)^k}{n^p} \le \frac{k!}{(p-1)^{k-1}}$$ 
At first, I thought many ideas such as Cauchy-Schwarz inequality, Taylor expansion, Induction on k, etc..
But they does not work efficiently.....
 A: Notice that $\dfrac{\ln^k x}{x^p}$ is a decreasing function with limit $0$, and so we may use integrals to understand its convergence and size. Then
$$ \sum_{n \geq 1} \frac{\ln ^kn}{n^p} < \int_1^\infty \frac{\ln ^k x}{x^p} \mathrm{d}x,$$
so we need to approximate the integral. Perform the substitution $x \mapsto e^x$ (or $u = \ln x$) to see that our integral is
$$ \int_0^\infty u^{k+1}e^{-(p-1)u}\frac{\mathrm{d}u}{u},$$
which is almost a Gamma function. Perform the substitution $u \mapsto \frac{u}{p-1}$ to see that our integral is now
$$\frac{1}{(p-1)^{k+1}} \int_0^\infty u^{k+1}e^{-u}\frac{\mathrm{d}u}{u} = \frac{\Gamma(k+1)}{(p-1)^{k+1}} = \frac{k!}{(p-1)^{k+1}},$$
which is very similar but just a little smaller than your bound. 
A: Too long for a comment:
Euler's famous expression for the Gamma function can be rewritten $($using the fact that $a\ln b=$ $=\ln b^a$ for $a=-1$, and then substituting $x\to\frac1x\Big)$ as follows:
$$k!=\int_0^1(-\ln x)^kdx=\int_0^1\ln^k\left(\dfrac1x\right)dx=\int_1^\infty\dfrac{\ln^kx}{x^2}dx$$
Then, by changing the power of x in the denominator to p, we get
$$\int_1^\infty\dfrac{\ln^kx}{x^p}dx=\frac{k!}{(p-1)^{k+1}}$$
At the same time, since $\displaystyle\zeta(p)=\sum_{n=1}^\infty\frac1{n^p}$ , by differentiating k times with respect to p and ignoring the resulting $(-1)^k$ factor, we get
$$\left|\zeta^{(k)}(p)\right|=\sum_{n=1}^\infty\frac{\ln^kn}{n^p}=\sum_{n=2}^\infty\frac{\ln^kn}{n^p}$$
since $\ln1=0$, and $(a^x)'=a^x\cdot\ln a$. For $p=2$ we have $\left|\zeta^{(k)}(2)\right|\simeq k!$ These links between the two functions, $\Gamma$ and $\zeta$, should not come as too much of a surprise, especially given the fact that $\Gamma\left(\frac1p\right)$ is linked, via Newton's binomial theorem, with geometric shapes of the form $x^p+y^p=r^p$, or, in other words, with bounded sums of powers, while on the other hand, the $\zeta$ function is also related to convergent $($and therefore bound$)$ sums of powers: hence the numerous mathematical identities between the two.
So your sum is nothing more than $|\zeta^{(k)}(p)|$, and the term on the right is nothing else than what one would obtain if one were to replace the summation sign on the left with an integral sign with the limits of integration $1$ and $\infty$, and their link to the factorial function is due to the reasons laid out above. Why replace the summation sign with an integral one ? Because integrals are nothing else than continuous sums, and the sign itself comes from medieval Latin s, just like the one for discrete sums comes from the Greek letter sigma, representing the same sound.
