Why does this series converge?(log)

Question

I have the following series: $\dfrac{\log n}{n^{a-1}}$ why does it converge? I know $\dfrac{1}{n^{a-1}}$ is a convergent series because I know from the problem that $a-1>1$? What criterion do I use?

Answer 1

Let $$ p=a-1=1+\varepsilon $$ for some $\varepsilon>0$. Then because $$ \log n\leq n^{\varepsilon/2} $$ for sufficiently large $n$, it follows that $$ \frac{\log n}{n^p}\leq \frac{1}{n^{1+\varepsilon/2}} $$ from which the result follows.

Answer 2

Since $\log n/n \to 0$, $\log(n^a)/n^a =a\log n/n^a \to 0$, so $\log n/n^a \to 0$.

Answer 3

(other solution)You can use the Cauchy condensation test

Let $(x_n)$ be a decrescent sequence of positive terms, then $\sum x_k$ converge $ \Leftrightarrow $ $\sum 2^k.x_{2^k}$ converges.

We can show that $\frac{\ln(k)}{k^r}$ is decrescent, and $$\sum 2^k\frac{\ln (2^k)}{2^{kr}} =\sum 2^kk\frac{\ln (2)}{2^{kr}}=\ln(2)\sum \frac{k}{2^{k(r-1)}} $$ converges if $r>1$. (In your case r=a-1>1)

Answer 4

You only need $a>1$.

Theorem: If $b>0$ then: $$\frac{\log n}{n^b}\to 0$$

Proof:

Let $m>\frac{1}{b}$, $m$ a positive integer. Then we have, for $x\geq 0$ that $$e^{x}>\frac{1}{m!}x^m$$

Then $$n=e^{\log n}> \frac{1}{m!}\left(\log n\right)^m$$

So $$n^{b}=n^{b-\frac{1}{m}}n^{1/m}>n^{b-1/m}\frac{1}{m!^{1/m}}\log n$$ or:

$$0<\frac{\log n}{n^b}<\frac{m!^{1/m}}{n^{b-1/m}}$$

Since $m,b$ are constants and $b-1/m>0$, we can apply the squeeze theorem and we are done.