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17

Using the series $$(1-4x)^{-1/2}=\sum_{n=0}^\infty\binom{2n}{n}x^n\tag{1}$$ we get \begin{align} f(x) &=\sum_{n=0}^\infty\frac{(-1)^{n+1}(2n+1)!}{(n+2)!\,n!}x^{n+2}\\ &=\frac12\sum_{n=1}^\infty(-1)^n\binom{2n}{n}\frac{x^{n+1}}{n+1}\\ &=\frac12\int_0^x\left[(1+4t)^{-1/2}-1\right]\,\mathrm{d}t\\ &=\frac14(1+4x)^{1/2}-\frac ... 9 A few easy manipulations yield the equivalent series\frac{13}{8}-\frac1{64}\sum_{n=0}^\infty \frac{(2n+1)!}{n!(n+2)!}\left(-\frac1{16}\right)^n$$The problem, then, is to evaluate$$f(z)=\sum_{n=0}^\infty \frac{(2n+1)!}{n!(n+2)!}z^n$$Using the duplication formula for the factorial, we have$$\begin{align*} f(z)&=\sum_{n=0}^\infty ...

7

some straightforward manipulations of the sum brings it to the form $$S=\frac{1}{2}\sum_{n=0}^{\infty}\frac{(-1)^{n+1}(2(n+1))!}{(n+2)!(n+1)! 4^{2n+3}}$$ Using the defintion of the catalan numbers this nicely rewrites as $$S=\frac{1}{2}\sum_{n=0}^{\infty}\frac{(-1)^{n+1}C_{n+1}}{ ... 6 The only problem is, for the past 150 years \zeta(-3) = 1/120, but now all of a sudden I've made \zeta(-3) = 1/144 and as we all know, 120≠144? I'm gonna tell you right now the same thing I told another guy a year or two back: If you can accept that \infty=\dfrac1{120} despite the fact that \infty\neq\dfrac1{120}, then you can also accept ... 4 HINT your RHS is a generating function of a certain power series. You have to simplify that generating function and try to understand what counting process results in having this generating function as a result. Note in particular that$$1-x-x^{17}+x^{18} = (1-x) - x^{17}(1-x) = (1-x)\left(1-x^{17}\right),$$so your GF is a product of two. What series ... 4 If you don’t see the factorization, you can still solve the problem. Let g(x)=\sum_{n\ge 0}c_nx^n. Then$$\left(1-x-x^{17}+x^{18}\right)g(x)=1\;,$$so$$\begin{align*} \sum_{n\ge 0}c_nx^n&=g(x)\\ &=xg(x)+x^{17}g(x)-x^{18}g(x)+1\\\\ &=x\sum_{n\ge 0}c_nx^n+x^{17}\sum_{n\ge 0}c_nx^n-x^{18}\sum_{n\ge 0}c_nx^n+1\\\\ &=\sum_{n\ge ...

4

Leibniz formula tells us: $$\sum_{k=0}^{\infty} \frac{(-1)^k}{2k+1}=\frac{\pi}4=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}\dots$$ Group the terms by two, the difference is positive, so there is an obvious increasing sequence of rationals that converges to $\pi$: $$4\left(1-\frac{1}{3}\right)+4\left(\frac{1}{5}-\frac{1}{7}\right)+\dots$$ Or more formally ...

3

To build on my comment, there are probably a few ways to define that sequence. One "lazy" way to do it is to recognize that $\pi$ as an irrational has the property that for any $n \in \mathbb{N}$, we can find the $n^{th}$ decimal in $\pi$'s decimal expansion. So for $b_i \in \{0,1,\ldots,8,9\}$ we can write $$\pi = b_0.b_1b_2b_3b_4\ldots ... 3 In french it is called the conjugué. Multiply each fraction by it ot remove the square root in the denominator. $$\dfrac{1}{\sqrt{0}+\sqrt{1}}=\dfrac{1}{\sqrt{0}+\sqrt{1}}\times \dfrac{\sqrt{1}-\sqrt{0}}{\sqrt{1}-\sqrt{0}}= \dfrac{\sqrt{1}-\sqrt{0}}{1}$$ Apply the same thing to every fraction and see what you'll have. 2 By using induction we can prove that |a_{k+1}-a_k|\le\frac{1}{2^{k-1}}|a_2-a_1|, then, if n and m are integers with n>m>0 we have, from triangle inequality$$|a_n-a_m|\le\sum_{j=1}^{n-m}|a_{n-j+1}-a_{n-j}|\le \sum_{j=1}^{n-m}\frac{1}{2^j}|a_{m+1}-a_m|\le\sum_{j=1}^{n-m}\frac{1}{2^{m+j-1}}|a_{2}-a_1|\tag{1}$$Now, since ... 2 Let n_0:=\lfloor\pi\rfloor and define inductively$$n_{k+1}:=\left\lfloor10^{k+1}\left(\pi-\sum_{i=0}^k\dfrac{n_i}{10^i}\right)\right\rfloor.$$Then$$\lim_{m\to\infty}\sum_{k=0}^m\dfrac{n_k}{10^k}=\sum_{k=0}^\infty\dfrac{n_k}{10^k}=\pi.$$2 The sequence in graydad's answer can be written in terms of the floor function as$$n \mapsto 10^{-n} \lfloor 10^n \pi \rfloor.$$This suggests some cute generalizations. Instead of the decimal expansion for \pi, we can use the expansion in any base q:$$n \mapsto q^{-n} \lfloor q^n \pi \rfloor$$converges to \pi for any whole number q \ge 2. In ... 2 Your argument is not correct. Every sequence in C has a convergent subsequence because K is compact. However, the subsequential limit is in K. You need to show that every subsequential limit is in C. Let \{x_n\} be a sequence in C. Because, K is compact, \{x_n\} has a convergent subsequence \{x_{n_k}\} which converges to a point in ... 1 Well, n is either even or odd. If n is even, n=2k and$$\lfloor\frac {n-1}2\rfloor=\lfloor k-\frac 12 \rfloor=k-1$$So$$a_{2k}=k-1+2k=3k-1$$If n is odd, n=2k+1 and$$\lfloor\frac {n-1}2\rfloor=\lfloor k \rfloor=k$$So$$a_{2k+1}=k+2k+1=3k+1$$Either way a_n can't be divisible by 3. 1 There are at least two ways to do this. Firstly, because you are looking at divisibility, you can generally split the proof into cases. The second option you have is to use induction since it seems you are trying to prove this for n\ge 1 (or some other integer). The first way is more general in that you can show that a_n is not divisible by 3 for any ... 1 By now, I've found a closed-form by doing some integral evaluation, a lot of hypergeometric, polylogarithm and polygamma manipulation.$$ S = \sqrt{\pi}\left(\frac{\pi}{12}\zeta(3)+\frac{1}{192\sqrt3}\psi^{(3)}\left(\tfrac13\right)-\frac{\pi^4}{72\sqrt3}-1\right). $$1 Here, we have$$\begin{align} a_n-b_n&=\frac{1}{n^2}\sum_{k=1}^{n}\left((k+1/2)\log (k+1/2)-k\log k\right)\\\\ &=\frac{1}{n^2}\sum_{k=1}^{n}\left(\frac12 \log k+\left(k+\frac12\right)\log\left(1+\frac{1}{2k}\right)\right)\\\\ &=\frac{1}{n^2}\sum_{k=1}^{n}\left(\frac12 \log k+O\left(1\right)\right)\\\\ &=\frac{1}{n^2}O\left(n\log n\right)\\\\ ...

1

Let's restate your first product in terms of logarithms: $$a_k \downarrow 0: \lim_{n\to \infty}\sum_{k=1}^n \beta^k\log a_k \to \log c \in \mathbb{R}$$ Intuitively, $\beta^k \downarrow 0$ so that the diverging $\log a_k$ values are kept in check; therefore $0\leq \beta<1$. Now, lets look at the modified version in your post: $$\lim_{n\to ... 1 Start with the original Newton's method recurrence:$$x_{n+1}=x_n - \frac{f(x_n)}{f'(x_n)} $$and write x_n=r+e_n, where r is the root f(r)=0 and we assume e_n is small in the sense that we may perform Taylor expansions, viz.$$\begin{align}e_{n+1} &= e_n - \frac{f(r)+e_n f'(r) + \frac12 e_n^2 f''(r)+\cdots}{f'(r)+e_n f''(r)+\frac12 e_n^2 ...

1

You can check that $$\left| \frac{z-\alpha}{1-\bar \alpha z} \right| < 1$$ precisely when $|z|<1$. (See for example this, but there are many many others on this site as well.) Hence the series converges for $|z| < 1$ (and diverges for $|z| > 1$ where $\left| \frac{z-\alpha}{1-\bar \alpha z} \right| > 1$). Finally, by Dirichlet's test, ...

1

$$\frac{1}{\sqrt{n} +\sqrt{n+1}} = \frac{\sqrt{n} -\sqrt{n+1}}{(\sqrt{n} +\sqrt{n+1})(\sqrt{n} -\sqrt{n+1})}$$ The denominator becomes $$(\sqrt{n} +\sqrt{n+1})(\sqrt{n} -\sqrt{n+1}) = n -n -1 = -1$$ Thus the sum becomes $$-\sum_{n=0}^k \sqrt{n}-\sqrt{n+1}$$ Which is a telescope sum.

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