# Prob. 16, Chap. 6, in Baby Rudin: Some Properties of Riemann Zeta Fucntion

Here is Prob. 16, Chap. 6, in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

For $1 < s < \infty$, define $$\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s}.$$ (This is Riemann's zeta function, of great importance in the study of the distribution of prime numbers.) Prove that

(a) $$\zeta(s) = s \int_1^\infty \frac{[x]}{x^{s+1} } \ \mathrm{d} x$$ and that

(b) $$\zeta(s) = \frac{s}{s-1} - s \int_1^\infty \frac{x-[x]}{ x^{s+1} } \ \mathrm{d} x.$$ where $[x]$ denotes the greatest integer $\leq x$.

Prove that the integral in (b) converges for all $s > 0$.

Hint: To prove (a), compute the difference between the integral over $[1, N]$ and the $N$th partial sum of the series that defines $\zeta(s)$.

My Attempt:

Here are the links to a couple of relevant posts of mine here on Math SE:

Prob. 8, Chap. 6, in Baby Rudin: The Integral Test for Convergence of Series

Theorem 6.12 (b) in Baby Rudin: If $f_1 \leq f_2$ on $[a, b]$, then $\int_a^b f_1 d\alpha \leq \int_a^b f_2 d\alpha$

Theorem 6.12 (c) in Baby Rudin: If $f\in\mathscr{R}(\alpha)$ on $[a, b]$ and $a<c<b$, then $f\in\mathscr{R}(\alpha)$ on $[a, c]$ and $[c, b]$

Theorem 6.12 (a) in Baby Rudin: If $f\in\mathscr{R}(\alpha)$ on $[a,b]$, then $cf\in\mathscr{R}(\alpha)$ for every constant $c$

Prob. 16 (a)

Let $b$ be a real number such that $b > 2$, and let $N$ be the integer such that $N \leq b < N+1$. Then $N \geq 2$ and we see that \begin{align} s \int_1^b \frac{ [x] }{ x^{s+1} } \ \mathrm{d} x &\geq s \int_1^N \frac{ [x] }{ x^{s+1} } \ \mathrm{d} x \\ &= s \sum_{k=1}^{N-1} \int_k^{k+1} \frac{ [x] }{ x^{s+1} } \ \mathrm{d} x \\ &\geq s \sum_{k=1}^{N-1} \int_k^{k+1} \frac{ k }{ x^{s+1} } \ \mathrm{d} x \\ &= s \sum_{k=1}^{N-1} k \int_k^{k+1} x^{-s-1} \ \mathrm{d} x \\ &= s \sum_{k=1}^{N-1} k \frac{ (k+1)^{-s} - k^{-s} }{ -s } \\ &= \sum_{k=1}^{N-1} k \left( \frac{ 1 }{ k^s } - \frac{ 1 }{ (k+1)^s } \right). \tag{1} \end{align} On the other hand, \begin{align} s \int_1^b \frac{ [x] }{ x^{s+1} } \ \mathrm{d} x &\leq s \int_1^{N+1} \frac{ [x] }{ x^{s+1} } \ \mathrm{d} x \\ &= s \sum_{k=1}^{N} \int_k^{k+1} \frac{ [x] }{ x^{s+1} } \ \mathrm{d} x \\ &\leq s \sum_{k=1}^{N} \int_k^{k+1} \frac{ k+1 }{ x^{s+1} } \ \mathrm{d} x \\ &= s \sum_{k=1}^{N} (k+1) \int_k^{k+1} x^{-s-1} \ \mathrm{d} x \\ &= s \sum_{k=1}^{N} (k+1) \frac{ (k+1)^{-s} - k^{-s} }{ -s } \\ &= \sum_{k=1}^{N} (k+1) \left( \frac{ 1 }{ k^s } - \frac{ 1 }{ (k+1)^s } \right) \\ &= \sum_{k=1}^{N} k \left( \frac{ 1 }{ k^s } - \frac{ 1 }{ (k+1)^s } \right) + \sum_{k=1}^{N} \left( \frac{ 1 }{ k^s } - \frac{ 1 }{ (k+1)^s } \right) \\ &= \sum_{k=1}^{N} k \left( \frac{ 1 }{ k^s } - \frac{ 1 }{ (k+1)^s } \right) + \left( 1- \frac{1}{ (N+1)^s } \right). \tag{2} \end{align}

By combining (1) and (2) we obtain $$\sum_{k=1}^{N-1} k \left( \frac{ 1 }{ k^s } - \frac{ 1 }{ (k+1)^s } \right) \leq s \int_1^b \frac{ [x] }{ x^{s+1} } \ \mathrm{d} x \leq \sum_{k=1}^{N} k \left( \frac{ 1 }{ k^s } - \frac{ 1 }{ (k+1)^s } \right) + \left( 1- \frac{1}{ (N+1)^s } \right) \tag{3}$$ for every real number $b > 2$ and for the natural number $N$ such that $N \leq b < N+1$, (i.e. $N = [b]$) and so $N \geq 2$.

Conversely, for every natural number $N > 2$, we can find a real number $b$ such that $N \leq b < N+1$, and so (3) holds.

Is what I've done so far correct? If so, then how to prove from here the identity asserted in (a)?

Prob. 16 (b)

We now assume that $$\zeta (s) = s \int_1^\infty \frac{ [ x ] }{ x^{s+1} } \ \mathrm{d} x \tag{4}$$ holds. Then \begin{align} \frac{s}{s-1} - s \int_1^\infty \frac{ x-[x] }{ x^{s+1} } \ \mathrm{d} x &= \frac{s}{s-1} - s \int_1^\infty \frac{ 1 }{ x^s } \ \mathrm{d} x + s \int_1^\infty \frac{ [x] }{ x^{s+1} } \ \mathrm{d} x \\ &= \frac{s}{s-1} - s \lim_{b \to \infty} \int_1^b \frac{ 1 }{ x^s } \ \mathrm{d} x \ + \ \zeta(s) \qquad \mbox{ [ using (4) ] } \\ &= \frac{s}{s-1} - s \lim_{b \to \infty} \frac{ b^{-s + 1} - 1 }{ -s+1 } + \zeta (s) \\ &= \frac{s}{s-1} - s \frac{ \lim_{b \to \infty} b^{-s + 1} \ - \ 1 }{ -s+1 } + \zeta (s) \\ &= \frac{s}{s-1} - s \frac{ 0 - 1 }{ -s+1 } + \zeta (s) \qquad \mbox{ [ because s > 1 ] } \\ &= \zeta (s), \end{align} as required.

Am I right?

Now as $0 \leq x - [x] < 1$ for all real numbers $x$ and as $x^{s+1} > 0$ for all $x \geq 1$, so $$0 \leq \frac{ x- [x] }{ x^{s+1 } } < \frac{ 1 }{ x^{s+1} },$$ and then we can conclude (from Theorem 6.12 (b) in Rudin ) that $$0 \leq \int_1^b \frac{ x- [x] }{ x^{s+1 } } \ \mathrm{d} x \leq \int_1^b \frac{ 1 }{ x^{s+1 } } \ \mathrm{d} x$$ for every real number $b \geq 1$. Therefore, \begin{align} 0 \leq \int_1^\infty \frac{ x-[x] }{x^{s+1} } \ \mathrm{d} x &= \lim_{b \to \infty} \int_1^b \frac{ x-[x] }{x^{s+1} } \ \mathrm{d} x \\ &\leq \lim_{b \to \infty} \int_1^b \frac{ 1 }{ x^{s+1 } } \ \mathrm{d} x \\ &= \int_1^\infty \frac{ 1 }{ x^{s+1 } } \ \mathrm{d} x. \tag{5} \end{align} Now the integral $\int_1^\infty \frac{ 1 }{ x^{s+1 } } \ \mathrm{d} x$ converges if and only if the series $\sum \frac{1}{n^{s+1} }$ converges (by Prob. 8, Chap. 6, in Rudin), since the function $f$ defined on $[1, \infty)$ by
$$f(x) \colon= \frac{1}{x^{s+1} }$$ is monotonically decreasing.

Now as $s > 1$, so $s+1 > 2$ and thus (by Theorem 3.28 in Rudin) the series $\sum \frac{1}{n^{s+1}}$ converges. So the last integral in (5) also converges, and then (5) implies the convergence of $\int_1^\infty \frac{ x-[x] }{x^{s+1} } \ \mathrm{d} x$, as required.

Is what I've asserted so far correct? Have I used the correct and rigorous enough logic in establishing whatever I have? Or, have I made any mistakes somewhere?

P.S.: After reading the answer and comments below, here is what I've thought of.

Theorem 3.41 in Rudin:

Given two sequences $\left\{ a_n \right\}$, $\left\{ b_n \right\}$, put $$A_n = \sum_{k=0}^n a_k$$ if $n \geq 0$; put $A_{-1} = 0$. Then, if $0 \leq p \leq q$, we have $$\sum_{n=p}^q a_n b_n = \sum_{n=p}^{q-1} A_n \left( b_n - b_{n+1} \right) + A_q b_q - A_{p-1} b_p. \tag{A}$$

So if we put $a_k \colon= 1 = k - (k-1)$ and $b_k \colon= \frac{1}{k^s}$ for $k \geq 1$ and $a_0 = 0$ in (A), then for any integer $r \geq 2$ we obtain \begin{align} \sum_{k=1}^r \frac{1}{k^s} &= \sum_{k=1}^{r-1} \frac{k - (k-1) }{ k^s } \\ &= \sum_{k=1}^{r-1} A_k \left( b_k - b_{k+1} \right) + A_r b_r - A_0 b_1 \\ &= \sum_{k=1}^{r-1} k \left( \frac{1}{k^s} - \frac{1}{ (k+1)^s } \right) + r \frac{1}{r^s} \\ &= \sum_{k=1}^{r-1} k \left( \frac{1}{k^s} - \frac{1}{ (k+1)^s } \right) + \frac{1}{r^{s-1} }, \end{align} and so $$\sum_{k=1}^{r+1} \frac{1}{k^s} = \sum_{k=1}^{r} k \left( \frac{1}{k^s} - \frac{1}{ (k+1)^s } \right) + \frac{1}{ (r+1)^{s-1} },$$ which implies that $$\sum_{k=1}^{r} k \left( \frac{1}{k^s} - \frac{1}{ (k+1)^s } \right) = \sum_{k=1}^{r+1} \frac{1}{k^s} \ - \ \frac{1}{ (r+1)^{s-1} }. \tag{6}$$

Then using (6) in (3) we obtain $$\sum_{k=1}^N \frac{1}{k^s} \ - \ \frac{1}{N^s} \leq s \int_1^b \frac{ [x] }{ x^{s+1} } \ \mathrm{d} x \leq \sum_{k=1}^{N+1} \frac{1}{k^s} + 1 - \frac{2}{ (N+1)^s }. \tag{7}$$ In (7) if we let $b \to \infty$, then $N \to \infty$ also, and then we obtain $$\sum_{k=1}^\infty \frac{1}{k^s} \leq s \int_1^\infty \frac{ [x] }{ x^{s+1} } \ \mathrm{d} x \leq 1 + \sum_{k=1}^\infty \frac{1}{k^s},$$
which is the same as $$\zeta(s) \leq s \int_1^\infty \frac{ [x] }{ x^{s+1} } \ \mathrm{d} x \leq 1 + \zeta(s).$$

Is what I've done so far the desired thing? If so, then what next?

• @SaaqibMahmuud..I've seen many of your posts and i believe that your determined to solve the life out of Baby Rudin..:P I praise that..good for you...+1 from me. – Marios Gretsas Aug 9 '17 at 12:32
• @Marios Gretsas thanks. What proportions of the problems are required to be solved in a typical course taught using Rudin at an American or European university, I wonder? – Saaqib Mahmood Aug 9 '17 at 12:46
• I don't know because Baby Rudin is a book recomended to my university in Greece, especially for introductory courses in advance calculus of the 3rdand 4rth semester.But this book is not used as a teaching materia by the professors in my university in such early semesters and ages,because its quite a difficult book. – Marios Gretsas Aug 9 '17 at 12:50
• Most of the people i know use this book for self study and for its challenging exercises to gain more mathematical maturity and a concrete picture of rigorus proofs – Marios Gretsas Aug 9 '17 at 12:52

## 1 Answer

$$\zeta(s) = \sum_{n=1}^\infty n^{-s} = \sum_{n=1}^\infty (n-(n-1)) n^{-s} = \sum_{n=1}^\infty n (n^{-s}-(n+1)^{-s}) \\= \sum_{n=1}^\infty n \int_n^{n+1} s x^{-s-1}dx = s\int_1^\infty \lfloor x \rfloor x^{-s-1}dx$$

The name of this is "summation by parts" and "Abel summation formula"

• how does the third equality in your answer occurs? Could you please elaborate? – Saaqib Mahmood Aug 9 '17 at 9:09
• @SimplyBeautifulArt I really meant $\sum_{n\ge 1} (n-(n-1))n^{-s}=\sum_{n\ge 1} n \, n^{-s}-\sum_{n\ge 0} n \, (n+1)^{-s}$ at least for $\Re(s) > 2$. It is not hard to adapt it to $\Re(s) > 1$. – reuns Aug 9 '17 at 11:03
• Oh, my bad. Summation by parts is the last step... (it could've been that other step too, but yeah, that's overcomplicating) – Simply Beautiful Art Aug 9 '17 at 11:08
• @SimplyBeautifulArt ? I don't follow you. Abel summation formula is $\sum_n n^{-s} = s \int_1^\infty \lfloor x \rfloor x^{-s-1}dx$, summation by parts is $\sum_n n^{-s} = \sum_n n(n^{-s}-(n+1)^{-s})$ – reuns Aug 9 '17 at 11:09
• ._. I shall go get my coffee and try waking up a bit more. – Simply Beautiful Art Aug 9 '17 at 11:14