Radius of Convergence without Ratio Test I've reached my wit's end with this problem. We have not been taught about Ratio Tests or anything involving Radius of Convergence, but we have a problem asking for it.
For what values of $p$ does the sum converge?
$$\sum_{n=1}^\infty{\ln(n) \over{n^p}}$$
I notice that it is "similar" to
$$\sum_{n=1}^\infty{\frac {1}{n^p}}$$
And we know from the Integral Test of Convergence that this series converges when $\lvert p\rvert > 1$. Does this have any connection to the answer? Because I also notice that (via WolframAlpha) that when $p=1$ the series diverges, but when $p=1.1$ it converges.
 A: Note that $\ln n<n$ so
$$\sum_{n=1}^\infty \frac{\ln n}{n^p}<\sum_{n=1}^\infty\frac{n}{n^p}=\sum_{n=1}^\infty \frac{1}{n^{p-1}}$$
which is the familiar $p$-series. Thus the series converges by the comparison test if $p-1>1$, or if $p>2$.
This is not a full proof since it remains to show whether the series converges for $1< p\le 2$. Note that divergence for $p\le1$ is given by the same comparison test above.
A: For any $d>0$ we have $\lim_{x\to \infty} (\ln x)/x^d=0.$
For $p>1$ let $d=(p-1)/2.$ Then for all but finitely many $n$ we have $0<(\ln n)/x^d<1.$ So for all but finitely many $n$ we have $$0<(\ln n)/x^p<x^d/x^p=1/ x^{p-d}=1/x^{(p+1)/2}.$$ And $(p+1)/2>1.$ So by the Comparison test, $\sum_n (\ln n)/n^p$ converges.
For $p\leq 1$ we have $$n\geq 3\to (\ln n)/n^p>1/n^p >1/n>0,$$ so  again by Comparison, $\sum_n (\ln n)/n^p$ diverges. 
A: In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequalities
$$\frac{x-1}{x}\le \log(x)\le x-1 \tag 1$$
for $x>0$.  Then, for any number $\alpha >0$, we have from $(1)$
$$\frac{x^{\alpha}-1}{x^{\alpha}}\le \log(x^\alpha)\le x^{\alpha}-1 \tag 2$$
Using $\log(x^{\alpha})=\alpha \log(x)$ in $(2)$ reveals that for any $x>0$ and any $\alpha >0$
$$\frac{x^{\alpha}-1}{\alpha x^{\alpha}}\le \log(x)\le \frac{x^{\alpha}-1}{\alpha} \tag 3$$
Finally, using $(3)$ we find that
$$\frac{n^{-p}-n^{-(\alpha+p)}}{\alpha }\le \frac{\log(n)}{n^p}\le \frac{n^{\alpha-p}-n^{-p}}{\alpha}\tag 4$$



*

*From the right-hand side of $(4)$, for any $p>1$, we can choose $\alpha$ so that $p-\alpha >1$ also.  Therefore, if $p>1$, then we find that

$$\sum_{n=1}^N \frac{\log(n)}{n^p}\le \frac{1}{\alpha}\sum_{n=1}^N\frac{1}{n^{p-\alpha}} \tag 5$$
and the series of interest converges by the comparison test (or the integral test).




*

*From the left-hand side of $(4)$, for any $p\le 1$ and for all $\alpha$, we see that

$$\sum_{n=1}^N \frac{\log(n)}{n^p}\ge \frac{1-2^{-\alpha}}{\alpha}\sum_{n=2}^N \frac{1}{n^p}$$
and the series of interest diverges by the comparison test (or the integral test).

