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Note: I'm aware that there are much simpler proofs for this result. I decided to go with this approach because $(1)$ it was a nice challenge, and $(2)$ it makes use of a nice identity shown below.

I've recently been spending time evaluating integrals involving the floor function. One of them was

$$\int_0^1\frac{x^p}{\left\lfloor\frac{1}{x}\right\rfloor}dx$$

for integers $p\geq 0$. By splitting the integral

$$\int_\frac{1}{k}^1\frac{x^p}{\left\lfloor\frac{1}{x}\right\rfloor}dx$$

into a sum of integrals indexed by the intervals $[1/(i+1),1/i]$ for $i=1,2,3,...,k-1$, evaluating each of the integrals, and manipulating the resulting sum, I found that

$$\int_\frac{1}{k}^1\frac{x^p}{\left\lfloor\frac{1}{x}\right\rfloor}dx=\frac{1}{(p+1)k}-1+\frac{1}{p+1}\left(\sum_{n=1}^{k-1}\frac{1}{n^{p+2}}+\sum_{m=2}^{p+1}\sum_{n=1}^{k}\frac{1}{n^m}\right)$$

which, after letting $k\to\infty$, yields

\begin{align} \int_0^1\frac{x^p}{\left\lfloor\frac{1}{x}\right\rfloor}dx &= -1+\frac{1}{p+1}\left(\zeta(p+2)+\sum_{m=2}^{p+1}\zeta(m)\right)\\ &= -1+\frac{1}{p+1}\sum_{m=2}^{p+2}\zeta(m) \end{align}

Pretty neat! I believe I can use this equation to give an overkill proof of $\lim_{s\to\infty}\zeta(s)=1$ ($s$ is a real parameter). Here is my attempt:

Beginning

We first note that the sequence $\{\zeta(n)\}_{n=2}^\infty$ certainly has a limit, since $\zeta$ is strictly decreasing and bounded below by $1$ (both of these facts easily follow from the series $\sum_{n=1}^\infty 1/n^s$), so

\begin{align} \lim_{n\to\infty}\zeta(n) &= L & (1) \end{align}

for some real number $L\geq 1$. Fixing an arbitrary $\varepsilon>0$, we infer that for some $N\in\mathbb{N}$,

$$L-\varepsilon<\zeta(n)<L+\varepsilon\text{ for every }n>N$$ $$\implies \sum_{n=2}^{p+2}(L-\varepsilon)<\sum_{n=2}^{p+2}\zeta(n)<\sum_{n=2}^{p+2}(L+\varepsilon)\text{ for every }p>N$$ $$\implies (p+2-1)(L-\varepsilon)<\sum_{n=2}^{p+2}\zeta(n)<(p+2-1)(L+\varepsilon)\text{ for every }p>N$$ $$\implies (p+1)(L-\varepsilon)<\sum_{n=2}^{p+2}\zeta(n)<(p+1)(L+\varepsilon)\text{ for every }p>N$$ $$\implies L-\varepsilon<\frac{1}{p+1}\sum_{n=2}^{p+2}\zeta(n)<L+\varepsilon\text{ for every }p>N$$

Since $\varepsilon>0$ was fixed arbitrarily, we can apply the prior sequence of implications to any positive real number. This shows that

\begin{align} \lim_{p\to\infty}\frac{1}{p+1}\sum_{n=2}^{p+2}\zeta(n) &= L & (2) \end{align}

Now consider the identity

$$\int_0^1\frac{x^p}{\left\lfloor\frac{1}{x}\right\rfloor}dx=-1+\frac{1}{p+1}\sum_{m=2}^{p+2}\zeta(m)$$

We can write

\begin{align} 0<\int_0^1\frac{x^p}{\left\lfloor\frac{1}{x}\right\rfloor}dx &= \int_0^\frac{1}{2} \frac{x^p}{\left\lfloor\frac{1}{x}\right\rfloor}dx+\int_\frac{1}{2}^1\frac{x^p}{\left\lfloor\frac{1}{x}\right\rfloor}dx\\ &\leq \int_0^\frac{1}{2}\frac{\left(\frac{1}{2}\right)^p}{\left\lfloor\frac{1}{x}\right\rfloor}dx+\int_\frac{1}{2}^1 \frac{x^p}{1}dx\\ &= \left(\frac{1}{2}\right)^p\int_0^\frac{1}{2}\frac{1}{\left\lfloor\frac{1}{x}\right\rfloor}dx+\frac{1}{p+1}-\frac{\left(\frac{1}{2}\right)^{p+1}}{p+1} \end{align}

and thus

$$0<-1+\frac{1}{p+1}\sum_{m=2}^{p+2}\zeta(m)\leq\frac{1}{2^p}\int_0^\frac{1}{2}\frac{1}{\left\lfloor\frac{1}{x}\right\rfloor}dx+\frac{1}{p+1}-\frac{1}{(p+1)\cdot 2^{p+1}}$$

Since

$$\lim_{p\to\infty}\left(\frac{1}{2^p}\int_0^\frac{1}{2}\frac{1}{\left\lfloor\frac{1}{x}\right\rfloor}dx+\frac{1}{p+1}-\frac{1}{(p+1)\cdot 2^{p+1}}\right)=0$$

the Squeeze Theorem gives

$$\lim_{p\to\infty}\left(-1+\frac{1}{p+1}\sum_{m=2}^{p+2}\zeta(m)\right)=0$$

which is equivalent to $\lim_{p\to\infty}\frac{1}{p+1}\sum_{m=2}^{p+2}\zeta(m)=1$. Combining this with $(2)$, we see that we must have $L=1$. We deduce from $(1)$ that

$$\lim_{n\to\infty}\zeta(n)=1$$

We now do the final push and get $\lim_{s\to\infty}\zeta(s)=1$. Fix an arbitrary $\varepsilon>0$. Since the sequence $\{\zeta(n)\}$ converges to $1$ and $\zeta(t)>1$ for every real $t>1$, there is an $N\in\mathbb{N}$ such that

$$0<\zeta(n)-1<\varepsilon\text{ for every }n>N$$

We know that $\zeta(s)<\zeta(n)$ for every real $s>n$, so

$$0<\zeta(s)-1<\zeta(n)-1<\varepsilon\text{ for every real }s>n>N$$

Since $\varepsilon$ was fixed arbitrarily, we can apply the preceding argument to every positive real number, so we are done. $\blacksquare$

Let me know what you think! If you identify any errors or optimizations, feel free to share them with me.

Edit: as @stochasticboy321 kindly pointed out, my proof has a small error. You see, I can't deduce the inequality

$$\sum_{n=2}^{p+2}(L-\varepsilon)<\sum_{n=2}^{p+2}\zeta(n)<\sum_{n=2}^{p+2}(L+\varepsilon)$$

from the fact that $L-\varepsilon<\zeta(n)<L+\varepsilon$ for every $n>N$, since this assumes that the latter inequality is also true for $2,3,...,N$. A correct approach would $(1)$ use the integral to prove that

$$\lim_{p\to\infty}\frac{1}{p+1}\sum_{n=2}^{p+2}\zeta(n)=1$$

$(2)$ write

$$\sum_{n=2}^{p+2}\zeta(n)=\sum_{n=2}^{N}\zeta(n)+\sum_{n=N+1}^{p+2}\zeta(n)$$

for an arbitrary $p\geq N-1$, $(3)$ apply the inequality $L-\varepsilon<\zeta(n)<L+\varepsilon$ to get

$$(L-\varepsilon)(p+2-N)+\sum_{n=2}^{N}\zeta(n)<\sum_{n=2}^{p+2}\zeta(n)<(L+\varepsilon)(p+2-N)+\sum_{n=2}^{N}\zeta(n)$$ $$\implies (L-\varepsilon)\left(\frac{p+2}{p+1}-\frac{N}{p+1}\right)+\frac{1}{p+1}\sum_{n=2}^{N}\zeta(n)<\frac{1}{p+1}\sum_{n=2}^{p+2}\zeta(n)$$ $$\text{ and }$$ $$\frac{1}{p+1}\sum_{n=2}^{p+2}\zeta(n)<(L+\varepsilon)\left(\frac{p+2}{p+1}-\frac{N}{p+1}\right)+\frac{1}{p+1}\sum_{n=2}^{N}\zeta(n)$$

and $(4)$ let $p\to\infty$ to yield

$$L-\varepsilon\leq\lim_{p\to\infty}\frac{1}{p+1}\sum_{n=2}^{p+2}\zeta(n)=1\leq L+\varepsilon$$

from which $L=1$ follows because $\varepsilon$ was fixed arbitrarily, so the inequality $|L-1|\leq\varepsilon$ will hold for any $\varepsilon>0$.

For the sake of honesty, I won't edit the original proof.

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    $\begingroup$ There's a minor error - if $\zeta(n) \le L + \epsilon$ for $n >N$, then this does not imply that $\sum_{2}^{p+2} \zeta(n) \le \sum_2^{p+2} (L+\epsilon)$ for large $p$, since the $\zeta(n)$ for $n \le N$ may be bigger than $L+\epsilon$. What you do have is that if you define $\sum_{n = 2}^{N} \zeta(n) = C_N,$ then $\sum_{n = 2}^{p+2} \zeta(n) \le C_N + (p+2 - N) (L+\epsilon),$ and a similar lower bound. Of course, this doesn't change the conclusion, since $C_N/p$ will go to zero with $p$. $\endgroup$ Commented Jul 16, 2021 at 4:47
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    $\begingroup$ BTW the argument that if $a_n$ converges then the ergodic average $\sum_{n \le N} a_n/N$ converges to the same limit follows directly by Stolz-Cesaro, which is a nice tool to know. The rest of your argument seems fine to me. $\endgroup$ Commented Jul 16, 2021 at 4:47
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    $\begingroup$ Beautiful identity. $\endgroup$ Commented Jul 16, 2021 at 4:55
  • $\begingroup$ @stochasticboy321 you're right! I can't believe I overlooked that! $\endgroup$ Commented Jul 16, 2021 at 5:56
  • $\begingroup$ I added a correction at the bottom of the post. $\endgroup$ Commented Jul 16, 2021 at 5:56

1 Answer 1

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Over the real line, $\lim_{s\to +\infty}\zeta(s)=1$ is a straightforward consequence of the dominated convergence theorem applied to the series defining $\zeta(s)$. Through concrete inequalities,

$$ 1 < \zeta(s)=1+\sum_{n\geq 2}\frac{1}{n^s}\leq 1+\int_{1}^{+\infty}\frac{dx}{x^s}=\frac{s}{s-1} $$ and the claim follows by squeezing.

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    $\begingroup$ I see that this is your $3$rd answer since $2020$, welcome back Jack! I love so many of your posts, like the ones that show how to solve really tough integrals, you're an inspiration for me. Thanks so much for them. $\endgroup$ Commented Jul 16, 2021 at 13:46
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    $\begingroup$ I agree with @A-Level Student, it's great to have you back Jack. $\endgroup$ Commented Jul 16, 2021 at 14:41

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