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So I need to prove that the infinite sum $\frac{1}{(\ln(n)^p)}$ diverges for all values of $p$. I managed to prove it for $p\leq 1$ via comparison test with $1/n$. but for this I can't seem to find a way to prove it diverges for $p>1$.

Martin Sleziak
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dvd280
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  • $\sum_{n= 2^k}^{2^{k+1}} \ln^p( n) \approx \sum_{n= 2^k}^{2^{k+1}} \ln^p( 2^k) =2^k k^p \ln^p (2)$ – reuns Mar 23 '17 at 22:11

4 Answers4

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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

$$\frac{x-1}{x}\le \log(x)\le x-1 <x\tag 1$$

for $0<x$.

Using the property $\log(x^a)=a\log(x)$ in $(1)$, we obtain for $a>0$

$$\log(x)<\frac{x^a}{a}\tag2 $$

For $x>n>1$, we have from $(2)$

$$\frac{1}{\log^p(n)}>\frac{a}{n^{ap}}\tag 1$$

For any $p>1$, we can take $a=1/p$ so that $ap=1$. Hence, the series $\sum_{n=1}\frac{1}{\log^p(n)}$ dominates the series, $\frac1p\sum_{n=2}^\infty\frac{1}{n}$, which diverges. The comparison test guarantees that the series $\sum_{n=1}\frac{1}{\log^p(n)}$ diverges also.

Community
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Mark Viola
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    One typo, but otherwise great! [Note: (2) $\Rightarrow$ $(\log(x))^p < (\frac{x^a}{a})^p =\frac{x^{ap}}{a^p}$ so putting $a=1/p$ gives $(\log(x))^p < \frac{x}{(\frac{1}{p})^p}$. Rearranging/taking reciprocals gives $(\frac{1}{p})^p(\frac{1}{x})<(\frac{1}{\log(x)})^p$ $\Rightarrow \frac{1}{p^p}\frac{1}{n}<\frac{1}{\log^p(n)}$.] – Selrach Dunbar Mar 27 '21 at 20:42
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Use that for $n$ large enough $\ln^p n < n$ and compare your series with $\sum 1/n$.

Martín-Blas Pérez Pinilla
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You can prove it using the following theorem

Theorem (the integral test): Let $f(x)$ be a continuous function of real numbers that is monotonic decreasing. Then for any $N\in\mathbb N$, $\sum_{n=N}^\infty f(n)$ converges if and only if $\int_N^\infty f(x)dx$ is finite.

Stella Biderman
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  • I cant use it to prove stuff yet since we havent proven that theorem yet in class unfortunately.Thanks though! – dvd280 Mar 23 '17 at 22:20
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Use the theorem (Cauchy condensation principle): if a_n stricly decreasing posite sequence, then \sum a_n behaves like \sum 2^na_{2^n}, but \sum 2^n (1\over \ln 2^n)^p=\sum 2^n \over {n^p \cdot \ln^p 2} and, hence the general term 2^n \over {n^p \cdot \ln^p 2} goes \infinity as n\to \infty we have the series divergent

Szabó Tamás
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