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$\zeta (s)={\sum }_{n\ge 1}{n}^{-s}$ for $\Re s>1$; its Euler product is ${\prod }_p(1-p^{-s}{)}^{-1}$.

For $c=1+1/\log x$, $1\le y\le x$ and $2\le T\le x$, apply the truncated Perron formula with Von Mangoldt function coefficients at both endpoints on the same vertical line. The difference kernel is $((x+y{)}^s-x^s)/s={\int }_x^{x+y}{u}^{s-1}du$, of modulus $O(y)$. The usual near-integer error bound contributes $O(x{\log }^{2}x/T)$. The logarithmic derivative on this line is finite because of the Euler product.

Under Riemann hypothesis, integrate a smoothed explicit formula from $1/2+\delta$ to two, taking $x=\log |t|$. The prime contribution is $O_{\delta }(x^{1-2\delta })$, while the local zero count bounds the zero contribution by $O_{\delta }(x^{-\delta }\log |t|/\log x)$. Both are $o(\log |t|)$ for fixed $0<\delta <1/4$, yielding the displayed bound; other fixed positive shifts follow by standard strip bounds and the Euler product.

The Euler product and Mertens second theorem give ${\sum }_{n\le D}{3}^{\omega (n)}/n\ll {\log }^{3}D$. The log-weighted convolution bound for three to the prime omega, with the Chebyshev estimate $\psi (y)\ll y$, then gives ${\sum }_{\sqrt{D}\le n\le D}{3}^{\omega (n)}\ll D{\log }^{2}D$. A direct bound for the full range follows from $3^{\omega (n)}\le {\tau }_3(n)$, where ${\tau }_3(n)=\#\{(a,b,c):abc=n\}$:

For $F=-{\zeta }^{\prime }/\zeta$ and $\sigma >1$, positivity ofterm by term in the Euler product givesComparing this with the pole of $F$ at one and the local poles of ${\zeta }^{\prime }/\zeta$ at zeros proves the classical zero-free region.

For a finite set of primes choose $0<g(p)<1$, put $k(p)=g(p)/(1-g(p))$, and let $W={\prod }_p(1+k(p))$. Give each squarefree integer $d$ on these primes probability $k(d)/W$. Inclusion of each prime is represented by a Bernoulli random variable with parameter $g(p)$; these are independent random variables, so $\mathbb{E}[\log d]={\sum }_{p}g(p)\log p$. If this is at most $\theta \log L$ with $0<\theta <1$, the Markov inequality givesThis turns a full Euler product into a lower bound for the truncated normalizing sum of a Selberg sieve.

Suppose a function has least period $r\mid N$ and distinct values on distinct period cosets. Two independent Fourier samples have the form $c_j=s_{j}N/r$ with uniform $s_j\in {\mathbb{Z}}_r$. The displayed candidate is $r/\gcd (r,s_1,s_2)$, so exact recovery occurs when no prime divisor of $r$ divides both residues. The Chinese remainder theorem gives probability ${\prod }_{\ell \mid r}(1-{\ell }^{-2})\ge 1/\zeta (2)=6/{\pi }^2$, using the Euler product. Unlike a single-sample coprimality probability, this bound is uniformly greater than one half. A candidate-period test makes successful recovery heralded when such a test is available.

The logarithm of the Euler product and $3+4\cos \theta +\cos 2\theta \ge 0$ give $\zeta (\sigma {)}^3|\zeta (\sigma +it){|}^4|\zeta (\sigma +2it)|\ge 1$. At $\sigma =1+a/{\log }^9|t|$, upper bounds $\zeta (\sigma )\ll {\log }^9|t|/a$ and $|\zeta (\sigma +2it)|\ll \log |t|$ imply the lower bound $|\zeta |\gg a^{3/4}{\log }^{-7}|t|$. The Cauchy estimate for derivatives gives $|{\zeta }^{\prime }|\ll {\log }^2|t|$ nearby, allowing a sufficiently small leftward displacement of order ${\log }^{-9}|t|$. This elementary region is weaker than the classical logarithmic region but gives an explicit reciprocal bound by simple estimates.