# Proof verification: $\lim\limits_{s\to\infty}\zeta(s)=1$

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 $$N\in\mathbb{N}$$,

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

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 $$L-\varepsilon<\zeta(n) for every $$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) 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.

• 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 $p$. Commented Jul 16, 2021 at 4:47
• 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. Commented Jul 16, 2021 at 4:47
• Beautiful identity. Commented Jul 16, 2021 at 4:55
• @stochasticboy321 you're right! I can't believe I overlooked that! Commented Jul 16, 2021 at 5:56
• I added a correction at the bottom of the post. Commented Jul 16, 2021 at 5:56

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.
• 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. Commented Jul 16, 2021 at 13:46