Tag Info

New answers tagged

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 ...


4

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


3

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 $$\log n \le \sum_{k=1}^{n-1} \frac{1}{k} < \sum_{k=1}^n \frac{1}{k}.$$



Top 50 recent answers are included