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5

Let me write $z = s-1$ for a little more convenient notation in the expansions. From the pole of $\zeta$ we have the Laurent expansion $$\zeta(1+z) = \frac{1}{z} + \gamma + O(z).$$ Taylor expansion of $\zeta$ and $\zeta'$ around $\rho_n$ yields \begin{align} \frac{\zeta'(z+\rho_n)}{\zeta(z+\rho_n)} &= \frac{\zeta'(\rho_n) + \zeta''(\rho_n)z + ... 1 The formula used for the spiral is r=a\cdot e^{b\cdot\phi}. The arc below and above a selected arc have radii r_\pm=a\cdot e^{b\cdot(\phi\pm2\pi)}. The radius of a circle on the arc should be proportional to r, say c⋅r, with the condition that circles on neighboring arcs do not overlap, r_-+c⋅r_-\le r-c⋅r\land r+c⋅r\le r_+-c⋅r_+ $$where both of ... 2 The power series$$ \sum_{n\ge0}\frac{x^n}{n!} $$defines a function on the whole real line, let's call it “exp”. Since power series can be differentiated term by term, we see that \exp'=\exp and also that \exp0=1. Consider now the function$$ f(x)=\exp(a-x)\exp x. $$We have$$ f'(x)=-\exp(a-x)\exp x+\exp(a-x)\exp x=0 $$so that f'(x)=0. Therefore f ... 3 I'm not giving you the full formal proof - you can look that up in just about any calculus textbook - but the basic idea is to use$$ (a + b)^n = \sum_{k=0}^n \binom{n}{k} a^k b^{n-k} \text{ where } \binom{n}{k} = \frac{n!}{k!(n-k)!} $$to get$$ \left(1 + \frac{x}{n}\right)^n = \sum_{k=0}^n \frac{n!}{n^kk!(n-k)!}x^k = \sum_{k=0}^n ...

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HINT : Use the General Binomial Theorem on the expression on the right hand side.

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Hint: Use the series definition of the zeta function, $\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}$, to rewrite your sum as a double sum, and then change the order of summation. $$\sum_{m=2}^{\infty}(-1)^{m}\frac{\zeta (m)}{m}=\sum_{m=2}^{\infty}\frac{(-1)^{m}}{m}\sum_{n=1}^{\infty}\frac{1}{n^m}\\ ... 0 A simple comparison suffices: consider the functions$$f(x) = \frac{1}{x}, \quad g(x) = \frac{1}{\lfloor x \rfloor}.$$Clearly, g(x) \ge f(x) for all x \ge 1. Integrating both on [1, n] gives$$\int_{x=1}^n f(x) \, dx \le \int_{x=1}^n g(x) \, dx,$$or$$\log n \le \sum_{k=1}^{n-1} \frac{1}{k} < \sum_{k=1}^n \frac{1}{k}.

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