By Electric Moccasins


2019-02-05 21:11:33 8 Comments

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.

2 comments

@Mindlack 2019-02-05 21:21:53

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 $g(n) >> f(n)$.

@Robert Israel 2019-02-05 21:21:27

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 $M_k \le n < M_{k+1}$.

@orlp 2019-02-05 21:22:25

If we're being pedantic you need $f(n) = \dfrac{1}{n\log(n+1)^2}$ (or something similar) to prevent division by zero.

@Electric Moccasins 2019-02-06 00:54:19

Can you expand on how you know $\sum g(n)h(n)$ converges? Just trying to wrap my head around this

@Solomonoff's Secret 2019-02-06 01:45:56

@ElectricMoccasins Because $\sum g(n) h(n) \le \sum k 2^{-k}$, which converges.

@Electric Moccasins 2019-02-06 02:32:01

@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".

@Robert Israel 2019-02-06 05:23:34

$$\sum_{n=M_k}^{M_{k+1}-1} g(n) h(n) = k \sum_{n=M_k}^{M_{k+1}-1} g(n) < k 2^{-k}$$

@Electric Moccasins 2019-02-06 12:17:07

@RobertIsrael Awesome. Got it. Wow, what a proof to come up with in only a few minutes. I hope I can do something like that some day.

@Solomonoff's Secret 2019-02-06 12:26:31

@ElectricMoccasins You are a freshman in high school. Don't worry if a math professor can do proofs better than you!

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