2

Is the limit comparison test the best method of determining the divergence of an almost harmonic series?

Let $$a_n=\frac{1}{n},$$ $$b_n=\frac{1}{n^{1+f(n)}}$$ where $f(x):[1,\infty) \to [0,\infty)$ is a continuous function. What I'm looking for is the biggest $f(n)$ in order to have each term $b_n$ as small as possible and the series $\sum_{n=k>1}^\infty b_n$ to still diverge. I know that for $f(n)=c, c\gt0$ where $c$ is a constant no matter how small, the series converge. Yet there are functions $f(n)$ that are not constant but the series still diverges. Such a function would be $f(x)= \frac 1x$.

Using LCT we have $$\lim_{n\to\infty}\frac{b_n}{a_n}=\lim_{n\to\infty}\frac{\frac1n*n^{-\frac1n}}{\frac1n}=e^{\lim_{n\to\infty}{-\frac1n}*\ln n}=e^0=1$$ therefore $\sum_{n=k>1}^\infty b_n$ diverges. From $L=e^{\lim_{n\to\infty}{-f(n)}*\ln n}$ we notice that $f(x)$ must be decreasing and $\lim_{x\to\infty}f(x)=0$ in order to offset the $\ln n$ term in order for $L$ to be fininte such that the series over $b_n$ to diverge. From the requirement that $b_n$ terms to be as big as possible we notice that $f(x)$ must be the slowest decreasing possible function. The one that I found is $f(x)=\frac{c}{\ln x},c\gt0$.

My questions: Is there an even slower decreasing $f(x)$ such when we compute the LCT, $L=0$ or $\infty$ therefore the the LCT is inconclusive but the series over $b_n$ still diverges? Is there a stronger test other than LCT that we can use in order to find such a function or prove that one doesn't exist and $f(x)=\frac{c}{\ln x},c\gt0$ is the best we can get?

Note: I stumbled upon this problem when I learned that the harmonic series is the threshold of divergence for $\sum_{n=k>1}^\infty c_n= \frac{1}{n^p}$. I found it counterintuitive that for the smallest number $p \gt1$ the series converges yet there are exponents $p \gt1$ that depend on $n$ and the series diverges. I figured that because $f(x)$ is decreasing, $\exists k\gt0$ such that $b_n \ge c_n,\forall p\searrow 1$. That's why the series over $b_n$ diverges even if the first $k$ terms are smaller than $\frac {1}{n^p}$

aku jack
  • 129
  • 14

1 Answers1

1

This is not a general solution but a long comment trying to explain that no general apprach may exists for the type of series the OP is discussing.

If for example $f(x)>c>0$ for all $x$ large enough, then the situation is rather simple: comparison with the series $\sum^\infty_{n=1}\frac{1}{n^{1+c}}$ will do.

However, the problem may have sharp teeth if $f(x)$ oscillates and gets closed to $0$ very often. Consider the series $$\sum^\infty_{n=1}\frac{1}{n^{2-\cos n}}=\sum^\infty_{n=1}\frac{1}{n^{1+f(n)}}\tag{1}\label{one}$$ where $f(n)=1-\cos n$. There is an interesting posting by Andrés E. Caicedo here.

The article Enrico Laeng, Vittorino Pata. A convergence–divergence test for series of nonnegative terms, Expositiones Mathematicae 29 (4), (2011) 420–424. MR2861768 (2012m:40002) presents a rather general, albeit not elementary, approach to deal with some series as in \eqref{one}.

Edit: Another interesting posting on this matter is here. One series they consider there is

$$\sum^\infty_{n=1}\frac{1}{n^{1+|\cos n|}}$$

Mittens
  • 46,352