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

Question

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?

Answer 1

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

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