I ask this because, according to Euler's product formula, Riemann's zeta function =(1/something), so how could that be zero?
Also, how could one find zeros that are on the negative side and find a zero in the "critical strip", that is, when the real part of the input is between 0 and 1?
[Math] Why does the Riemann zeta function have zeros in the complex plane? How is it possible to find them
analytic-number-theorycomplex-analysisriemann-zeta
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Your Google Books link doesn't work for me; here's one that does.
I think there's nothing mysterious going on here, and no assumption about the multiplicities of the zeros of $\zeta$ is being made. The sum is written as a sum over the zeros, but it's implied that these contribute according to their multiplicity. The residue isn't computed in any special way (unless you count forming the residue of the logarithmic derivative as a special way); the residue from $\zeta'/\zeta$ is the multiplicity, and this gets multiplied by $x^\rho/\rho$.
Hardy proved in 1914 that an infinity of zeros were on the critical line ("Sur les zéros de la fonction $\zeta(s)$ de Riemann" Comptes rendus hebdomadaires des séances de l'Académie des sciences. 1914).
Of course other zeros could exist elsewhere in the critical strip.
Let's exhibit the main idea starting with the Xi function defined by : $$\Xi(t):=\xi\left(\frac 12+it\right)=-\frac 12\left(t^2+\frac 14\right)\,\pi^{-\frac 14-\frac{it}2}\,\Gamma\left(\frac 14+\frac{it}2\right)\,\zeta\left(\frac 12+it\right)$$ $\Xi(t)$ is an even integral function of $t$, real for real $t$ because of the functional equation (applied to $s=\frac 12+it$) : $$\xi(s)=\frac 12s(s-1)\pi^{-\frac s2}\,\Gamma\left(\frac s2\right)\,\zeta(s)=\frac 12s(s-1)\pi^{\frac {s-1}2}\,\Gamma\left(\frac {1-s}2\right)\,\zeta(1-s)=\xi(1-s)$$ We observe that a zero of $\zeta$ on the critical line will give a real zero of $\,\Xi(t)$.
Now it can be proved (using Ramanujan's $(2.16.2)$ reproduced at the end) that : $$\int_0^\infty\frac{\Xi(t)}{t^2+\frac 14}\cos(x t)\,dt=\frac{\pi}2\left(e^{\frac x2}-2e^{-\frac x2}\psi\left(e^{-2x}\right)\right)$$ where $\,\displaystyle \psi(s):=\sum_{n=1}^\infty e^{-n^2\pi s}\ $ is the theta function used by Riemann
Setting $\ x:=-i\alpha\ $ and after $2n$ derivations relatively to $\alpha$ we get (see Titchmarsh's first proof $10.2$, alternative proofs follow in the book...) : $$\lim_{\alpha\to\frac{\pi}4}\,\int_0^\infty\frac{\Xi(t)}{t^2+\frac 14}t^{2n}\cosh(\alpha t)\,dt=\frac{(-1)^n\,\pi\,\cos\bigl(\frac{\pi}8\bigr)}{4^n}$$ Let's suppose that $\Xi(t)$ doesn't change sign for $\,t\ge T\,$ then the integral will be uniformly convergent with respect to $\alpha$ for $0\le\alpha\le\frac{\pi}4$ so that, for every $n$, we will have (at the limit) : $$\int_0^\infty\frac{\Xi(t)}{t^2+\frac 14}t^{2n}\cosh\left(\frac {\pi t}4\right)\,dt=\frac{(-1)^n\,\pi\,\cos\bigl(\frac{\pi}8\bigr)}{4^n}$$
But this is not possible since, from our hypothesis, the left-hand side has the same sign for sufficiently large values of $n$ (c.f. Titchmarsh) while the right part has alternating signs.
This proves that $\Xi(t)$ must change sign infinitely often and that $\zeta\left(\frac 12+it\right)$ has an infinity of real solutions $t$.
Probably not as simple as you hoped but a stronger result! $$-$$
From Titchmarsh's book "The Theory of the Riemann Zeta-function" p. $35-36\;$ and $\;255-258$ :
Best Answer
Wherever it converges, the Riemann zeta function $\zeta(s)$ is equal to the $p$-series - and it converges only when the real part of $s$ is greater than one. The Euler product analogously only converges when $\mathrm{Re}(s)>1$, so the reasoning that $\zeta=1/\text{something}\ne0$ does not apply where the product is an invalid representation of zeta.
Moreover, the idea is flawed all by itself: the "$\text{something}$" is an infinite product, so if that product diverges to infinity in the limit, we have $\text{something}^{-1}\to0$. For example, formally we have
$$0\le\frac{1}{(1+1)(1+1/2)(1+1/3)\cdots}\le\frac{1}{1+1/2+1/3+\cdots}=\frac{1}{\infty}=0.$$
In general, properties (e.g. inequalities) satisfied by expressions with a finite number of terms do not necessarily carry over to an infinite number of terms (but we can still apply logic to the partial sums if need be).
So if neither the original series nor the product converges outside the right side of $1$, how can it be then that $\zeta(s)$ can be computed on or to the left of this? The answer is analytic continuation, which I gave a brief introductory explanation to just recently. But then precisely how can it be explicitly extended left of $1$? The simplest, but limited, way is through the Dirichlet eta function:
$$\eta(s)=1-\frac{1}{2^s}+\frac{1}{3^s}-\frac{1}{4^s}-\cdots=\left(1+\frac{1}{2^s}+\frac{1}{3^s}+\cdots\right)-2\left(\frac{1}{2^s}+\frac{1}{4^s}+\cdots\right)$$ $$=\zeta(s)-\frac{2}{2^s}\zeta(s)=(1-2^{1-s})\zeta(s).$$ The eta function converges for $\mathrm{Re}(s)>0$, so if we write $\zeta(s)=(1-2^{1-s})^{-1}\eta(s)$ we have a way to evaluate zeta to the left of where we originally could (by $1$ at most). Analytic continuation beyond this requires more sophisticated machinery - a functional equation was first established by Bernhard Riemann in the paper "On the Number of Primes Less Than a Given Magnitude,"
$$\zeta(s)=2^s\pi^{s-1}\sin\left(\frac{\pi s}{2}\right)\Gamma(1-s)\zeta(1-s).$$ This means that what $\zeta$ evaluates to on the left of $s=1/2$ is determined by its evaluation on the same point reflected across $s=1/2$. With this and the $\eta$ continuation we are given the ability to compute $\zeta$ anywhere we like. Note that with the presence of the $\sin$ above, it is trivially true that $s=-2n$ are zeros of $\zeta$ for $n=1,2,3,\dots$ Moreover, since $\zeta(s)$ has no zeros right of $\mathrm{Re}(s)=1$, the functional equation predicts it has no other nontrivial zeros to the left of $\mathrm{Re}(s)=0$: this means all of the nontrivial zeros lie in the critical strip.
The Riemann Hypothesis is that all nontrivial zeros have real part $1/2$, but it has not been proven - we can, however, calculate the zeros to arbitrary accuracy and prove when particular zeros have real part exactly half, see this question linked by J.M.
As a sidenote, while the Euler product doesn't converge globally, I believe the Hadamard product (via Weierstrass factorization) does globally converge - see my comment above.