Test for convergence: I'm studying a chapter of my Analysis book, it's on convergence and divergence tests for series. The chapter concludes with some series to test, and these are the few I am having problem with:

Test for convergence:
$$\sum \frac{\ln{k}}{k^2}$$
$$\sum (\sqrt[k]{k}-1)$$
$$\sum \frac{k!}{k^k}$$

Now for the first series. It's obvious to by writing out terms that $\ln{k}\le \sqrt{k}$ and that the comparison test therefore yields that the given series converges. However I am failing to prove the $\ln{k}\le \sqrt{k}$ without using derivatives and other calculus-related methods, and I'm interested in a rather neat way of proving the given series converges.

For the second one.. It's clear that $\sqrt[k]{k}$ goes to $1$ as $k$ get's really large, (I've proven that in an earlier chapter), but I'm not sure how to test this series.

The third series seems to converge when I write out the first terms. I notice that the terms are less or equal than those of $\sum \frac{1}{2^k}$, which I know converges. However here again I'm having trouble proving these inequalities (at least in a neat way).
Herein, we show using non-calculus based tools that $\log(k)\le \sqrt{k}$ for all integer $k\ge 1$.  We begin with a primer on an elementary inequalities for the logarithm and exponential functions.


  PRIMER:
  
  In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm and exponential functions satisfy the inequalities
  
  $$\frac{x-1}{x}\le \log(x)\le x-1\tag1$$
  
  for $x&gt;0$
  
  and 
  
  $$1+x\le e^x\le \frac{1}{1-x} \tag 2$$
  
  for $x&lt;1$.




PROBLEM $(1)$:

Let $f(x)=\frac x2-\log(x)$.  Then, using $(1)$ we find for $h&gt;0$ that

$$\begin{align}
f(x+h)-f(x)&amp;=\frac h2-\log\left(1+\frac hx\right)\\\\
&amp;\ge \frac h2-\frac hx\\\\
&amp;\ge 0
\end{align}$$

for all $x\ge 2$.  So, $f(x)$ is monotone increasing for $x\ge 2$. And since $f(2)=1-\log(2)&gt;0$ we have 

$$\log(x)&lt;x/2 \tag 3$$

for $x\ge 2$.  

Now, setting $x= \sqrt{k}$ in $(3)$ reveals

$$\log(k)\le \sqrt k$$

for $k\ge 4$.

We also have from $(1)$, that $\log(x)\le 2(\sqrt x-1)$.  When $x\le 4$, we see that $2(\sqrt x-1)\le \sqrt x$.

Hence, for all $x&gt;0$, we find that $\log(x)\le \sqrt x$.  Certainly then for all integer $k\ge 1$, $\log(k)\le \sqrt{k}$ as was to be shown!



PROBLEM $(2)$:

We can write $k^{1/k}=e^{\frac1k \log(k)}$.  Using the left-hand side inequality in $(2)$ reveals

$$k^{1/k}-1\ge \frac{\log(k)}{k}&gt;\frac{1}{k}$$

for $k\ge 3$.  The series $\sum_{k=1}^\infty (k^{1/k}-1)$ diverges by comparison with the harmonic series. 



PROBLEM $(3)$

Note that we can write

$$ \frac{k!}{k^k} =\frac{1\cdot 2\cdot 3\cdots k}{k\cdot k\cdot k\cdots k}\le \frac 2{k^2}$$

The series converges by comparison with the series $\sum_{k=1}^\infty \frac{1}{k^2}$.
Why downvoting. it is correct .

Hint  for he first
$x\mapsto \ln(x)-\sqrt{x}$ is decreasing $\implies \ln(x)\leq \sqrt{x}$  for large enough $x$.

we have

$$\lim_{k\to+\infty}k^{\frac{3}{2}}\frac{\ln(k)}{k^2}=0$$

since the power grows quickly than logarithm.

$$\implies k^{\frac{3}{2}}\frac{\ln(k)}{k^2}\leq 1$$ for large enough $k$.

$$\implies \frac{\ln(k)}{k^2}\leq \frac{1}{k^{\frac{3}{2}}}$$

and you can conclude.


  for the second


$$\sqrt[k]{k}-1=e^{\frac{\ln(k)}{k}}-1\sim \frac{\ln(k)}{k}\;\;(k\to+\infty)$$

use the integral comparison test.