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We just hit convergence tests in calculus, and learned that $\sum_{n=1}^{\infty} \frac{1}{n^p}$ converges for all $p \gt 1$. I thought that this was sort of a "barrier" between what converges and what diverges. Specifically, that setting $a_n=\frac{1}{n^{1+\epsilon}}$ is sort of the "greatest function" (I'll make this precise later) for which $\sum a_n$ converges.

But, I did realize that there are functions that dominate $\frac{1}{n^{1+\epsilon}}$ but not $\frac1n$, such as $\frac{1}{n\log(n)}$. Now, the sum of that specific example diverges, but it got me wondering about whether $\frac{1}{n}$ is truly the "boundary". So, this leads me to two questions.

1) Is there a function $f$ that dominates $\frac{1}{n^p}$ for all $p>1$, meaning: $$\lim_{x\to\infty} \frac{f(x)}{\frac{1}{x^p}}=\infty$$ Such that: $$\sum_{n=1}^\infty f(n)$$ converges?

2) If so, up to a constant is there a function $g$ such that $\sum_{n=1}^\infty g(n)$ converges, such that $g$ dominates $f$ for all other functions $f$ such that $\sum_{n=1}^\infty f(n)$ converges?

I'm just a freshman in high school so I apologize if this is a stupid question.

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    $\begingroup$ I tried to format the question such that it excludes all the "trivial" answers, i.e. "no, because you can just take this function and multiply by any constant to get another function that dominates it and still converges". The constant thing is just excluded by the fact that the limit has to diverge, rather than be equal to some constant, but something similar could still possibly happen. I tried to include a sort of "parameter" on $f$ to patch this, but I couldn't get it formal. But the sort of "philosophy" of the question is about the form of the function, rather than a "cop-out" like that. $\endgroup$ – Electric Moccasins Feb 5 at 21:16
  • $\begingroup$ How about $f(x) = \frac{1}{x\ln^2(x)}$ for your first question $\endgroup$ – Jakobian Feb 5 at 21:19
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    $\begingroup$ It's a good question; part of a good answer will surely be to identify the most fruitful way to make the vague intuition behind it more precise. $\endgroup$ – Henning Makholm Feb 5 at 21:21
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    $\begingroup$ From math.stackexchange.com/a/452074/42969: Let $\sum_{n=1}^{\infty} c_n$ be any convergent series with positive terms. Then, there exists a convergent series $\sum_{n=1}^{\infty} C_n$ with much bigger terms in the sense that $\lim_{n\rightarrow\infty} C_n/c_n = \infty$. $\endgroup$ – Martin R Feb 5 at 21:22
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    $\begingroup$ There are lots of other great answers in that linked thread "Is there a slowest rate of divergence of a series?", which I'm sure the OP will find very interesting. $\endgroup$ – Rahul Feb 6 at 11:36
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1) Yes, $f(n)=\frac{1}{n(\ln{n})^2}$.

2) No. Assume $f \geq 0$ and $\sum_{n \geq 1}{f(n)} < \infty$.

Then there exists an increasing sequence $N_n$ and some constant $C > 0$ such that $\sum_{N_n+1}^{N_{n+1}}{f(k)} \leq C2^{-n}$.

Then set $g(n)=(p+1)f(n)$, where $N_p < n \leq N_{p+1}$.

Then $\sum_{N_n+1}^{N_{n+1}}{g(k)} \leq C(n+1)2^{-n}$, thus $\sum_{n \geq 1}{g(n)}$ is finite and $g(n) >> f(n)$.

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1) $$f(n) = \frac{1}{n \log(n)^2}$$

2) No. Given any $g > 0$ such that $\sum_n g(n)$ converges, there is an increasing sequence $M_k$ such that $$\sum_{n \ge M_k} g(n) < 2^{-k}$$
Then $ \sum_n g(n) h(n)$ converges, where $h(n) = k$ for $M_k \le n < M_{k+1}$.

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    $\begingroup$ If we're being pedantic you need $f(n) = \dfrac{1}{n\log(n+1)^2}$ (or something similar) to prevent division by zero. $\endgroup$ – orlp Feb 5 at 21:22
  • $\begingroup$ Can you expand on how you know $\sum g(n)h(n)$ converges? Just trying to wrap my head around this $\endgroup$ – Electric Moccasins Feb 6 at 0:54
  • $\begingroup$ @ElectricMoccasins Because $\sum g(n) h(n) \le \sum k 2^{-k}$, which converges. $\endgroup$ – Solomonoff's Secret Feb 6 at 1:45
  • $\begingroup$ @Solomonoff'sSecret Ok, I'm really not sure how you're getting that inequality. I get the inequality in Prof. Isreal's answer, but not how it translates to the one you said. Sorry, this is my first ever brush with "hard math". $\endgroup$ – Electric Moccasins Feb 6 at 2:32
  • $\begingroup$ $$\sum_{n=M_k}^{M_{k+1}-1} g(n) h(n) = k \sum_{n=M_k}^{M_{k+1}-1} g(n) < k 2^{-k}$$ $\endgroup$ – Robert Israel Feb 6 at 5:23

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