Integral Test For Convergence - Borderline Between Divergence and Convergence

Borderline Between Divergence and Convergence

The above examples involving the harmonic series raise the question, whether there are monotone sequences such that f(n) decreases to 0 faster than 1/n but slower than 1/n1+ε in the sense that

$\lim_{n\to\infty}\frac{f(n)}{1/n}=0 \quad\text{and}\quad \lim_{n\to\infty}\frac{f(n)}{1/n^{1+\varepsilon}}=\infty$

for every ε > 0, and whether the corresponding series of the f(n) still diverges. Once such a sequence is found, a similar question can be asked with f(n) taking the role of 1/n, and so on. In this way it is possible to investigate the borderline between divergence and convergence.

Using the integral test for convergence, one can show (see below) that, for every natural number k, the series

$\sum_{n=N_k}^\infty\frac1{n\ln(n)\ln_2(n)\cdots \ln_{k-1}(n)\ln_k(n)}$

still diverges (cf. proof that the sum of the reciprocals of the primes diverges for k = 1) but

$\sum_{n=N_k}^\infty\frac1{n\ln(n)\ln_2(n)\cdots\ln_{k-1}(n)(\ln_k(n))^{1+\varepsilon}}$

converges for every ε > 0. Here ln_k denotes the k-fold composition of the natural logarithm defined recursively by

$\ln_k(x)= \begin{cases} \ln(x)&\text{for }k=1,\\ \ln(\ln_{k-1}(x))&\text{for }k\ge2. \end{cases}$

Furthermore, N_k denotes the smallest natural number such that the k-fold composition is well-defined and ln_k N_k ≥ 1, i.e.

$N_k\ge \underbrace{e^{e^{\cdot^{\cdot^{e}}}}}_{k\ e'\text{s}}=e \uparrow\uparrow k$

using tetration or Knuth's up-arrow notation.

To see the divergence of the first series using the integral test, note that by repeated application of the chain rule

$\frac{d}{dx}\ln_{k+1}(x) =\frac{d}{dx}\ln(\ln_k(x)) =\frac1{\ln_k(x)}\frac{d}{dx}\ln_k(x) =\cdots =\frac1{x\ln(x)\cdots\ln_k(x)},$

hence

$\int_{N_k}^\infty\frac{dx}{x\ln(x)\cdots\ln_k(x)} =\ln_{k+1}(x)\bigr|_{N_k}^\infty=\infty.$

To see the convergence of the second series, note that by the power rule, the chain rule and the above result

$-\frac{d}{dx}\frac1{\varepsilon(\ln_k(x))^\varepsilon} =\frac1{(\ln_k(x))^{1+\varepsilon}}\frac{d}{dx}\ln_k(x) =\cdots =\frac{1}{x\ln(x)\cdots\ln_{k-1}(x)(\ln_k(x))^{1+\varepsilon}},$

hence

$\int_{N_k}^\infty\frac{dx}{x\ln(x)\cdots\ln_{k-1}(x)(\ln_k(x))^{1+\varepsilon}} =-\frac1{\varepsilon(\ln_k(x))^\varepsilon}\biggr|_{N_k}^\infty<\infty.$

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