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11

The way to do this without generating functions is to start with the ansatz that $$ a_n = x^n $$ satisfies the recursion but possibly not the starting points at $n=0$ and $1$. If we have that solution then any $a_nkx^n$ satisfies the recursion as well. And if we have two such solutions $x^n$ and $y^n$ then any linear combination $a_n=kx^n+my^n$ will ...


8

$$0,1,\frac 1 2, \frac 3 4, \frac 5 8, \frac{11}{16}, ...$$ Now, the trick is that if you take the common differences: $$1, -\frac 1 2, \frac 1 4, -\frac 1 8, \frac 1 {16}$$ It's a geometric series of first term $1$ and ratio $-\frac 1 2$. Thus, this sequence is just a sequence of partial sums of this geometric sequence, so we just use the partial sum ...


5

This is called a linear homogeneous recurrence relation. If we look at the recursive case, we find that the coefficient of $a_{n-1}$ is $10$ and the coefficient of $a_{n-2}$ is $-21$. This means the "characteristic polynomial," which is basically the polynomial which tells us what the bases of the explicit formula will be, is like this: $$x^2-10x+21$$ Notice ...


5

Interpreted as linear recurrence with constant coefficients $$ x_n = \frac{1}{2} x_{n-1} + \frac{1}{2} x_{n-2} $$ with characteristic polynomial $$ p(t) = t^2 - \frac{1}{2} t - \frac{1}{2} $$ where we deduce the roots from $$ 0 = \left( t - \frac{1}{4} \right)^2 - \frac{1}{2} - \frac{1}{16} \iff \\ t = \frac{1 \pm 3}{4} $$ so we have the solutions $$ x_n = ...


3

This is a homogeneous linear recurrence relation with constant coefficients. From $$ a_n = 10 a_{n-1} -21 a_{n-2} $$ you can infer the order $d=2$ and the characteristic polynomial: $$ p(t) = t^2 - 10 t + 21 $$ Calculating the roots: $$ 0 = p(t) = (t - 5)^2 - 25 + 21 \iff t = 5 \pm 2 $$ this gives the general solution $$ a_n = k_1 3^n + k_2 7^n $$ The two ...


3

Where did you find that equation? It's quite different from what I got, which I shall explain now. First, a common power series is $$ \frac{1}{1-x} = \sum_{i\geq0} x^{i}. $$ Using the substitution $x=-t^{n}$, $$ \frac{1}{1+t^{n}} = \sum_{i\geq0} (-t^{n})^{i} = \sum_{i\geq0} (-1)^{i}t^{in}. $$ Then, $$ \int_{0}^{x} \frac{1}{1+t^{n}} dt = \int_{0}^{x} \...


3

If a sequence is convergent then we know that $\lim_{n \to \infty} a_{n+1} = \lim_{n \to \infty} a_n = L$. take the limit of both sides and you will get: $$\lim_{n \to \infty} a_{n+1} = \lim_{n \to \infty} \frac{2a_n}{7+a_n} = \frac{2 \cdot \lim_{n \to \infty} a_{n}}{7 + \lim_{n \to \infty} a_n} \implies L = \frac{2L}{7+L}$$ This should be an easy ...


3

The "standard" example is the sequence $\{g_n\}$ defined by $$ g_1=1_{[0,1]},\; g_2=1_{[0,\frac{1}{2}]},\;g_3=1_{[\frac{1}{2},1]},\; g_4=1_{[0,\frac{1}{4}]},\;g_5=1_{[\frac{1}{4},\frac{1}{2}]},\dots$$ where $1_A$ is the indicator function of the set $A$, i.e. $1_A(x)=1$ if $x\in A$ and $1_A(x)=0$ otherwise. Note that $$\lim_{n\to\infty}\int_0^1g_n(x)\;dx=...


2

Since you say you can evaluate the remaining infinite sum in the $d=2$ case, I gather that reducing the multiple sum to a single infinite sum would already be sufficient progress. Then you want to count the number $\pi_d(n)$ of ways in which a positive integer $n$ can be written as an ordered product of positive integers. Find the prime factorisation, $n=\...


2

(Just for the cauchy part) For some integers $m,n$ where $m>n>0$, $|x_n-x_{n+1}|=|x_n-\frac12(x_n-x_{n-1})|=\frac12|x_{n-1}-x_n|=...=\frac1{2^{n-1}}|x_1-x_2|=\frac1{2^{n-1}}\\ |x_n-x_m|=|(x_n-x_{n+1})+(x_{n+1}-x_{n+2})+...+(x_{m-1}-x_m)|\\ \leq \frac1{2^{n-1}}+\frac1{2^n}+...\\ =\frac1{2^n}\to 0\text{ as }n\to\infty$


2

If we start from $$ \cos(x) = \prod_{n\geq 0}\left(1-\frac{4x^2}{(2n+1)^2 \pi^2}\right) \tag{1}$$ we have: $$ \log\cos(x) = -\sum_{n\geq 0}\sum_{m\geq 1}\frac{4^m x^{2m}}{m(2n+1)^{2m} \pi^{2m}}=-\sum_{m\geq 1}\frac{(4^m-1)\zeta(2m)\,x^{2m}}{m\pi^{2m}}\tag{2} $$ hence: $$ \sum_{l\geq 1}\log\cos\left(\frac{x}{3^l}\right)=-\sum_{m\geq 1}\frac{(4^m-1)\zeta(2m)}{...


2

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2

Let $$ f(t)=e^{t^2}\int_0^t e^{-x^2}\,\mathrm{d}x\tag{1} $$ Then, the product rule says $$ f'(t)=1+2tf(t)\tag{2} $$ Since $f$ is the product of an analytic function and the integral of an analytic function, it is analytic. Therefore, we can write $$ f(t)=\sum_{k=0}^\infty a_kt^k\tag{3} $$ Then $(2)$ becomes $$ \overbrace{a_1+\sum_{k=2}^\infty ka_kt^{k-1}}^{f'...


1

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1

We may just find the explicit form of $a_n$. By setting $a_n=\frac{p_n}{q_n}$ we have: $$\begin{pmatrix} p_{n+1} \\ q_{n+1}\end{pmatrix}=\begin{pmatrix} 2 & 0 \\ 1 & 7\end{pmatrix}\begin{pmatrix} p_{n} \\ q_{n}\end{pmatrix}$$ where the characteristic polynomial of the involved matrix is $(x-2)(x-7)$. By the Cayley-Hamilton theorem it follows that: $$...


1

by using Fourier series of $f(x)=1, x\in[0,1]$ $$1=\sum_{n=1}^\infty\frac{4}{(2n-1)\pi}\sin (2n-1)\pi x$$ integrate both sides when integration limits are $x=0 \rightarrow 1$ $$\int_{0}^{1}1.dx=\int_{0}^{1} \sum_{n=1}^\infty\frac{4}{(2n-1)\pi}\sin (2n-1)\pi x dx$$ $$1=\sum_{n=1}^\infty\frac{8}{(2n-1)^2\pi^2}$$ $$\sum_{n=1}^\infty\frac{1}{(2n-1)^2}=\frac{\pi^...



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