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0

Let be the partial sum $H_n = \frac11 + \frac12 + \frac13 + \cdots + \frac1n$. Using Cesàro-Stolz: $$\lim_{n\to\infty}\frac{H_n}{\log n} = \lim_{n\to\infty}\frac{H_{n+1}-H_n}{\log(n+1)-\log n} = \lim_{n\to\infty}\frac{\frac1{n+1}}{\log(1+1/n)} = \lim_{n\to\infty}\frac{\frac1{n+1}}{\frac1n} = 1$$ and $$\sum_{n=1}^\infty\frac1n = \lim_{n\to\infty}H_n = ... 0 Not just for some closed interval (say [0,1]), but also for any interval which is a subset of [-|M|,|M|] for some M, your statement (in title) is true. According to the Sequential Compactness Theorem, every sequence has a monotone subsequence, and since every subsequence of a bounded sequence is also bounded and since a monotone bounded sequence is ... 0 I think one of the most important ways in which series are useful are in approximating real-world phenomena. A simple example is the equation for a pendulum: \ddot{\theta} = -\frac{g}{l} \sin \theta. This ODE doesn't have a closed-form solution but by taking the Taylor series approximation of \sin \theta \approx \theta we get a very standard, ... 0 For me, I have always been amazed at the relationship between infinite series and trigonometric functions. For example:$$ \sin x = x - \frac{x^3}{3!}+\frac{x^5}{5!}+\cdots = \sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{(2n+1)!} $$and$$ \cos x = 1 - \frac{x^2}{2!}+\frac{x^4}{4!}-\cdots = \sum_{n=0}^\infty \frac{(-1)^nx^{2n}}{(2n)!} $$and many others. Of ... 0 Here are some properties of the radius of convergence R: If |z|<R then the series \sum |a_n z^n| is convergent ( that is, \sum a_n z^n is absolutely convergent) If |z| >R then the series \sum a_n z^n is not convergent. Equivalently 1' If \sum_n |a_n z^n| is divergent then |z|\ge R 2' If \sum_n a_n z^n  convergent then |z|\le ... 2 A real sequence is actually an injection f:\mathbb{N}\to\mathbb{R}, while a subsequence of a sequence \{f(n)\}_n is f\circ\varphi:\mathbb{N}\to\mathbb{R}, where \varphi:\mathbb{N}\to\mathbb{N} is a strictly increasing function, so that to form a subsequence we pick infinitely many elements of a sequence without disturbing their original order. You ... 0 This is really a question of language and logic. The set of peaks for a given sequence may be finite (including empty) or it may not be. Either the statement it is finite is true or it is false. If it is false then you are in case 2: there are infinitely many peaks. If it is true then you are in case 1. 0$$ 4\log_2\log_2(k+1) = 4\log_2\log_2(\sqrt{k+1})^2 = 4\log_2(2\log_2\sqrt{k+1}) =  = 4 + 4\log_2\log_2\sqrt{k+1} \gt 1 + 4\log_2\log_2\lfloor\sqrt{k+1}\rfloor \ge 1 + a_{\lfloor\sqrt{k+1}\rfloor} = a_{k+1} $$4 Notice that f^{2}(x) = \frac{x-1}{x} and that f^{3}(x) = x. Then reduce 653 mod 3. This gives (653)\equiv 2 ~mod ~3, so f^{653}(x) =f^{2}(x). 0 New one on me. Notice that the same sort of thing happens if we are adding something really tiny, for example$$ y_{n+1} = y_n + \frac{1}{e^{y_n}}; $$it just takes much longer to pass any given point. So I wanted to emphasize that, we say the sequence is unbounded, how long does it take to get to specified points? This is the original example,$$ x_{n+1} ...

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I've been intrested in products aswell (see my question) the methode i used was this one, i hope it helps. $$\prod_{i=b}^c 1+a_i$$= $$1+\sum_{i=b}^{c} (a_i)+$$ $$1/2!((\sum_{i=b}^c (a_i))^2-\sum_{i=b}^c (a_i)^2$$ $$1/3!((\sum_{i=b}^c (a_i))^3-3\sum_{i=b}^c (a_i)^2\sum_{i=b}^c (a_i)+2\sum_{i=b}^c (a_i)^3)$$ $$1/4!((\sum_{i=b}^c (a_i))^4-6(\sum_{i=b}^c ... 0 From what you write, it seems you are defining the following sequence of sets: (A_n)_{n\ge 1} such that A_1=A_2=\{1\} and A_{n+1} is the set of all possible non-empty sums with values in A_1,\ldots,A_n and such that they do not belong to A_1 \cup \cdots \cup A_n. Well, at this point, it is not difficult to prove by induction that$$\forall n\ge 4, ...

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It's straightforward to show by induction that the sequence is strictly positive and strictly increasing. Therefore the sequence converges if and only if it is bounded above. Suppose there exists $M$ such that $x_n<M$ for all $n$. Then $\frac1{x_{n-1}}>\frac1M$ so $\frac1{x_{n-1}^2}>\frac1{M^2}$. But this implies that ...

1

Suppose $\{x_n\}$ is a bounded sequence, that is $0<x_n \leq M$ for some $M$. Then: $x_n = (x_n-x_{n-1})+(x_{n-1}-x_{n-2})+\cdots +(x_2-x_1)+x_1 = \dfrac{1}{x_{n-1}^2}+\dfrac{1}{x_{n-2}^2}+ \cdots +\dfrac{1}{x_1^2}>\dfrac{1}{M^2}+\dfrac{1}{M^2}+\cdots +\dfrac{1}{M^2}=\dfrac{n-1}{M^2}$. If we select $n$ be such that $n > M^3+1$, then $x_n > M$, ...

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Another option would be the ratio test. $$\lim_{n\to\infty}\Bigl\lvert\frac{a_{n+1}}{a_n}\Bigr\rvert < 1$$ Then \begin{align} \lim_{n\to\infty}\biggl\lvert\frac{2^{n+1}(z^{2n}+1)}{(z^{2(n+1)}+1)2^n}\biggr\rvert & = \lim_{n\to\infty}2\biggl\lvert\frac{z^{2n}+1}{z^{2(n+1)}+1}\biggr\rvert\\ &= ...

2

Note that, for $|z|> \sqrt{2}$, you have $|z^{2n}+1|> |z|^{2n}-1>2^n-1$. Moreover (suppose $|z|> \sqrt{2} + \varepsilon$) you can show that $$\left| \frac{2^n}{z^{2n}+1} \right| < \rho^n$$ with suitable $\rho <1$ depending on $\varepsilon$, so the series converges (at least) in the unlimited domain $\{ |z| > \sqrt{2} \}$.

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Hint #1: The sequence is monotone increasing. If it were bounded, it would have a limit. What property would that limit satisfy? Hint #2: That limit would satisfy $$L=L+\frac{1}{L^2}$$ However no $L$ satisfies this equation.

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Since $a_1 > 0$ and $b_1 > 0$, show by induction that $\alpha_n > 0$ and $\beta_n > 0$ for all $n\in \Bbb N$. By the AM-GM inequality, $$\alpha_{n+1}\beta_{n+1} = \alpha_{n+1}\sqrt{\alpha_n\beta_n} \le \alpha_{n+1}\frac{\alpha_n + \beta_n}{2} = \alpha_n \beta_n$$ for all $n\in \Bbb N$. Hence, the sequence $(\alpha_n\beta_n)_{n\in \Bbb N}$ is a ...

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The problem is well discussed here! One may write functions in Mathematica that work with OEIS data and some of the answers include functionality that one may build upon.

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It's a cycling sequence, with period 8. \text{Let }k = (\sqrt{2} + 1) \\ \text{then }k^2 = (2+2\sqrt{2} + 1) \\ \text{and }k^2-2k = (2-1)=1 \\ \begin{align} \hspace{2in}x_2 &= \frac{kx_1-1}{k+x_1} \\ x_3 &= \frac{k\frac{kx_1-1}{k+x_1}-1}{k+\frac{kx_1-1}{k+x_1}}\\ &= \frac{k(kx_1-1)-(k+x_1)}{k(k+x_1)+(kx_1-1)}\\ ... 0 Combine the first two approaches. Start witha_0\in A$and an upper bound$b_0$. If$a_0=b_0$, you’re done; otherwise, let$x_0=\frac12(a_0+b_0)$. If$[x_0,b_0)\cap A\ne\varnothing$, pick any$a_1\in[x_0,b_0)\cap A$and let$b_1=b_0$. If$[x_0,b_0)\cap A=\varnothing$, let$a_1=a_0$and$b_1=x_0$. Continue in this fashion. At each stage you’ll have$a_n\in ...

2

Just note this for $|z|\neq 1$ $$\sum_{n}\frac{2^n}{|1+z^{2n}|} < \sum_{n}\frac{2^n}{|z^{2n}|}\implies \frac{2}{|z|^2}<1 \implies \dots.$$

0

The sequence $s_n:=\sup\{x_m:m\geq n\}$ is non-increasing. To see this just notice that if $n\le k$, then $\{x_m \colon m\geq n\}\supseteq\{x_m \colon m\geq k\}$ and thus $s_n=\sup \{x_m \colon m\geq n\} \ge x_k = \{x_m \colon m\geq k\}$. Every bounded monotone sequence is convergent. See, for example, Wikipedia. If the sequence is not bounded below, ...

1

I can't comment because apperently just guys with super high level can help you. About this approach with the mean valued integral theorem: I've tried it.It won't work simply because $\xi$ depends on $n$ if it was a constant for any $n$ I would agree with you, this limit is 1. But the sequence $a_n$ you created is not necessarly monotonic. Because $\xi$ is a ...

1

In $h(p)$ it is defined as a sum that runs from 0 to x. $k{n \choose k } = k\frac{n!}{k! (n-k)!} = \frac{n!}{(k-1)! (n-k)!} = (n-(k-1))\frac{n!}{(k-1)! (n-(k-1))!} = (n-k+1){n \choose k-1}$ Note $(n-k+1) = (n-(k-1))$, as k runs from 1 to n in the sum, k-1 runs from 0 to n-1. So we can replace the (k-1) term by k and change the range our summand runs ...

1

That's a bit too many question, this is discouraged on MSE. Nevertheless: This is CDF, i.e. the expression is $P( Y \leq x)$, which, by the way, doesn't exist in closed form (partial sum of rows of Pascal's triangle), You take the derivative of the product of two functions of $p: p^k$ and $(1-p)^{n-k}$ Do the algebra. You may try logging the expression, ...

0

To the get the linear equation of type: y=kx+b In which, y is the number of total balls and x is the sequence element's number. If you have access to 2 elemnts and its corresponding number of balls simply solve this system of equations. For example, y(1)=6 and y(2)=10, then | 6=k*1+b <=> |k=4 | 10=k*2+b |b=2 So the final equation would ...

0

Note that the lattice problem in your question is equivalent to going from $(0,0)$ to $(2n,0)$ using steps $(1,1)$ and $(1,-1)$ and staying in the region $y\geq 0$. You just rotate your problem by angle $-\frac\pi 4$ around origin, reflect using $x$ axis as mirror and scale it by a factor of $\sqrt 2$ to get this one. Coming back to the original question, ...

0

$$\sum^\infty_{n=1}\frac{sin(nx)}{n^3}= ?$$ We know that $$\sum^\infty_{n=1}\frac{sin(nx)}{n}= \frac{x-\pi}{2}$$. $$(\frac{sin(nx)}{n^3})''= (\frac{cos(nx)}{n^2})'=-\frac{sin(nx)}{n}$$ $$-\sum^\infty_{n=1}\frac{sin(nx)}{n}= -(\frac{x-\pi}{2})$$ $$- \int \frac{x-\pi}{2}= -\int \frac{x}{2}-\frac{\pi}{2} = \frac{x^2}{4}+\frac{\pi x}{2}+c$$ ...

0

We have $$\sum_{k=1}^21/r^{k²}=( r³+1)/r⁴$$ $$\sum_{k=1}^31/r^{k²}= (r⁸+r⁵+1)/r⁹$$ $$\sum_{k=1}^41/r^{k²}= (r¹⁵+r¹²+r⁷+1)/r¹⁶$$ $$\sum_{k=1}^51/r^{k²}= (r²⁴+r²¹+r¹⁶+r⁹+1)/r²⁵$$ So, for a fixed $n$ we have $$\sum_{k=1}^n1/r^{k²}= w_{n-1}(r)/r^{n²}$$ where $w_{n-1}(r)$ is a polynomial of degree $n²-1$ in $r$. But obtaining the formula when $n$ goes to ...

1

write \begin{align*}6^k &= (6^k)(3-2) \\&= 3^{k+1}2^k - 3^k2^{k+1}\\ &= 3^{k+1}2^k - 3^k2^{k+1} + 3^{2k+1} - 3^{2k+1}\end{align*} take (first and fourth) and (second and third) term and factorise it will split into the req form

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You can apply the root test, yet it is best to ignore the specifics of the coefficients and rather use very crude bounds for them (from above and below). Then you can study the "upper" and "lower" bound, and luckily both give the same radius. A reason this works is that when you have sequences $(a_n)_n,(b_n)_n, (c_n)_n$ such that $|b_n| \le |a_n| \le ... 2 Hint;$1\leq a_n\leq n^{50}$. Can you apply the root test to both sides? 1 Where they write $$a_1 + (a_1 + 1) + (a_1 + 1) + \cdots + (a_1 + 1) = ka_1 + k - 1$$ is that what you're wondering about? Because$a_1$is written$k$times, and$1$is written$k-1$times (after each of the$k$instances of$a_1$, except the first one), so that's what it becomes. Clearer ways of writing the right-hand side in that respect include$k(a_1 + ...

3

See if this converts to a Riemann sum which converges to an integral: $$g_n(u)=\sqrt[n]{\frac{1}{nx^{n-1}}} \cdot x^{\frac{n+1}{2}} \cdot e^{\frac 1n\sum_{k=1}^n \ln(1 - x^k)}=\sqrt[n]{\frac{1}{nx^{1-1/n}}} \cdot u^{\frac{n+1}{2n}} \cdot e^{\frac 1n\sum_{k=1}^n \ln(1 - x^k)}\\\sim \sqrt{u}\exp\int_0^1\ln(1-u^t)dt=\sqrt u\exp\frac{\zeta(2)-{\rm Li}_2(u)}{\ln ... 1 As$$\sqrt{x^k(1-x^k)}\le\frac{x^k+(1-x^k)}2=\frac12\iff x^k(1-x^k)\le\frac14$$So:$$\lim_{n \to \infty } \sqrt[n]{\int_0 ^1 x^{\frac{n(n+1)}{2}}(1-x)(1-x^2)\cdots(1-x^n)d x}=\lim_{n \to \infty } \sqrt[n]{\int_0 ^1 \prod_{k=1}^nx^k(1-x^k)}\le\lim_{n\to\infty}\sqrt[n]{\int_0^1\left(\frac14\right)^ndx}=\frac14$$So the limit is \displaystyle\le\frac14 Note ... 1 You've just made a simple error. Since b_n\leq m, 0\leq b_1r+\ldots+b_nr^n\leq Mr+\ldots+Mr^n from which you can take out a common factor. Can you finish it off? 2 Hint. You may just write$$ \frac{te^{(1-x)t}}{e^t-1}=\frac{t\:e^{t}e^{x(-t)}}{e^t(1-e^{-t})}=\frac{t\:e^{x(-t)}}{1-e^{-t}}=\frac{(-t)e^{x(-t)}}{e^{(-t)}-1} $$and use the definition. 3 Because you are looking for 3's in the last two terms of a sequence cycle of 7, the function defined below will work:$$a_n=\begin{cases} 3 &if \,\,\,\, n \mod 7=6,0\\ 2 &otherwise\\ \end{cases}$$4 The easiest way is to look up arithmetic modulo 7 as suggested by @Fundamental. Here is another answer: let$$g(n)=\frac{\sin\dfrac{2\pi(n+1)}{7}}{\sin\dfrac{2\pi}{7}} \frac{\sin\dfrac{2\pi(n+2)}{7}}{\sin\dfrac{4\pi}{7}}\cdots \frac{\sin\dfrac{2\pi(n+6)}{7}}{\sin\dfrac{12\pi}{7}}\ ;$$then take$$f(n)=2+g(n)+g(n+1)\ ;$$then for ... 0 For example$$ a_n=2+\left\lfloor \frac{n \mod 7}{5} \right\rfloor. $$0 Try finding a number c such that  \lfloor\frac {\sin \frac {(n-3)\pi}{7} +2.5 +c}{2} \rfloor satisfies. It'll work for some number c, since this number will fluctuate between 2+c and 3+c, and is rounded down by the floor function over a period of 7. You just need to make sure that the function is above 3 for two of the seven parts of the cycle. 1 The difference between \frac 43(2-x)^{3/2} and -\frac43i(x-2)^{3/2} is: An overall minus sign. A factor of i. The thing raised to the (3/2)th power has been negated. Of these (3) is what causes (2), since (-1)^{3/2}=\pm i. As for (1), the overall sign of the square root of something is not well defined when we're working with complex numbers. ... 2 Long story short,$$ p_n \leq C n \log n $$by Chebyshev's theorem (\pi(n)\geq D \frac{n}{\log n}), hence the series is lower bounded by a multiple of$$ \sum_{n\geq 3}\frac{1}{n\log n \log\log n}$$that diverges due to Cauchy condensation test (applied twice). 2 Recall again the prime number theorem, which gives that$$ p_n \approx n\log n.$$Then the sum we want can be compared to$$ \sum_p \frac{1}{p \log \log p} \sim \sum_n \frac{1}{n\log n \log \log n}.$$I claim this latter sum diverges based on the integral test for convergence. Claim: The following series diverges:$$\sum_n \frac{1}{n \log n \log \log ...

4

We have: $$\begin{eqnarray*} I_n &=& \int_{0}^{1}x^{n+1}(1-x)(1-x^2)\cdot\ldots\cdot(1-x^n)\,dx \\&\leq& \int_{0}^{1}x^{n+1}(1-x^n)^n\,dx = \frac{\Gamma(n)\,\Gamma\left(1+\frac{2}{n}\right)}{\Gamma\left(2+n+\frac{2}{n}\right)}\leq\frac{1}{n^2},\tag{1}\end{eqnarray*}$$ while on the other hand, if we set: $$A(n,m)\triangleq\int_{0}^{1} x^n ... 2 Firstly,$$\int_0^1 f_n(x)dx=n\int_0^1 xe^{-nx^2}dx=n\left[-\frac{1}{2n}e^{-nx^2}\right]_0^1=n\left(\frac{1}{2n}-\frac{1}{2n}e^{-n}\right)=\frac{1}{2}-\frac{1}{2}e^{-n}$$and thus$$\lim_{n\to\infty }\int_0^1 f_n(x)dx=\frac{1}{2}.$$Secondly,$$\lim_{n\to\infty }nxe^{-nx^2}=0$$and thus$$\int_0^1\lim_{n\to\infty }f_n(x)dx=0.$$We conclude that ... 1 The radius of convergence is the distance to the nearest singularity For (b) you can advance as (based on your calculations) using partial fraction$$P(x)= \frac{x}{(1-2x-x^2)}= \frac{A}{x-a} + \frac{B}{x-b} $$where a,b are the roots of 1-2x-x^2 and A and B need to be determined. The calculations gives:$$\frac{-1-\sqrt{2}}{2 \sqrt{2} ...

0

The terms in your sum are $p_nx^n$. If you solve the recurrence relation, you can find that $p_n$ grows as $r^n$ for some $r$, the larger root of the characteristic equation. You need the terms to decease faster than $\frac 1n$, so need $|x| \lt \frac 1r$

1

As pointed in the other answer, the series is pointwise convergent on $\left(0,\frac{\pi}{2}\right)$ by Dirichlet's test, since: $$\sum_{n=1}^{N}\sin(x)^n \leq \frac{\sin x}{1-\sin x}$$ and $\frac{1}{n}$ decreases towards zero. However, in a left neighbourhood of $x=\frac{\pi}{2}$ the original series behaves like the harmonic series:  ...

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