#### [SOLVED] Is there a "greatest function" that converges?

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!