New answers tagged

2

First you write the numbers in base $3$ $0,1,2,10,11,12,20,21,22,100,101,102,110,111,112,120,\dots$ Then you add up the (base 3) digits in each number (presumably in base $10$) $0,1,2,1,2,3,2,3,4,1,2,3,2,3,4,3,\dots$ Since we start from zero, we express $2011$ in base $3$, getting $2202111_3$, then add the digits to get $9$


0

You can use this equation to get all kinds of power series expansions for Ln(x+y+1). The equation adapts itself because you may chose to hold x or y constant and integrate it. Study the equation and play with it. It yields all kinds of neat results, Including some very important results. Hint let x=X^2. It provides insight into series. $$\sum_{n=0}^\...


0

We can write the series as $\displaystyle\frac{\sqrt{3}}{2}\sum_{n\in S}(-1)^{r(n)}\tan\left(\frac{\pi}{n+2}\right)$, where $S=\{n: n\ge1, n\not\equiv0{\pmod 3}\}$ and $r(n)$ is the remainder when $n$ is divided by 3, so the series converges by the Alternating Series Test since $\tan\left(\frac{\pi}{n+2}\right)$ is decreasing and converges to 0.


4

You may write $(-1)^n$ as $\cos(\pi n)$, so the sequence given by: $$ a_n = (-1)^n\sin\left(\frac{\pi}{3}n\right) = \frac{1}{2}\left[\sin\left(\frac{4\pi}{3}n\right)-\sin\left(\frac{2\pi}{3}n\right)\right]$$ has partial sums bounded by $\frac{\sqrt{3}}{2}$ in absolute value and $$ \sum_{n\geq 1} a_n \tan\left(\frac{\pi}{n+2}\right) $$ is conditionally ...


7

The function $\;\tan\frac\pi{n+2}\;$ is monotone descending and positive, and also $\;\lim\limits_{n\to\infty}\tan\frac\pi{n+2}=0\;$ , and since $$\sin\frac{n\pi}3=\begin{cases}\pm\frac{\sqrt3}2\;,&n=1,2\pmod 3\\{}\\0\,,&n=0\pmod 3\end{cases}$$ and its series is bounded: $$\left|\sum_{n=1}^N\sin\frac{n\pi}3\right|=\left|\frac{\sqrt3}2+\frac{\...


2

Directly using Cauchy-Hadamard formula with the $\;n\,$ th root: $$\sqrt[k]{\frac{k^2}{3^k}}=\frac{\sqrt[k]{k^2}}3\xrightarrow[k\to\infty]{}\frac13$$ and thus the radius of convergence is $\;R=3\;$...so you were right! (Though your use of equality signs isn't very careful...). But there is only one for each power series, not several.


2

Let $b_n = \ln( a_n )$, then according to $a_n = \sqrt{3a_{n-1}}$, we have: $b_n = \frac{1}{2}[ \ln(3)+b_{n-1} ]$ with $b_0=0$ which is a classical problem. We can easily find its solution: $b_n = \ln(3)[ 1 - (\frac{1}{2})^n]$. It is trivial to convert $b_n$ to $a_n = 3^{1-(\frac{1}{2})^n}$


3

Assuming that there are two rearrangements leading to different sums, that means that the original series is not absolutely convergent (otherwise, any rearrangement would lead to the same value). Any series that is conditionally converging but not absolutely convergent can be rearranged in order to have sum $r$ for any $r\in\mathbb{R}$ by the Riemann(-Dini) ...


3

As @Dr.MV said, we can use the comparison test to solve this problem. $\lvert n \rvert < 1 \rightarrow$ In this case, we have $\lvert n^{k^2} \rvert < \lvert n^k \rvert$, so since $\sum_{k=0}^\infty n^k$ converges, so does $\sum_{k=0}^\infty n^{k^2}$. $\lvert n \rvert \geq 1 \rightarrow$ In this case, we have $\lvert n^{k^2} \rvert \geq \lvert n^k \...


1

The series diverges and does not seem to be a power series, so the question in my opinion is badly worded. To see why it diverges, rewrite your series as: $$1+\sum_{n=1}^\infty\frac{1}{n^{-k^2}}.$$ Now using the $p-$series test, for convergence we need $-k^2>1$, which is impossible for all real $k$.


2

We have: $$ E=\sum_{n\geq 1}\frac{1}{2^n-1}=\sum_{n\geq 1}(2^{-n}+2^{-2n}+\ldots)=\sum_{n\geq 1}\frac{d(n)}{2^n} $$ and: $$ 1\left(1-\frac{1}{2}\right)+2\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}\right)+\ldots+n\left(\frac{1}{2^n-1}+\ldots-\frac{1}{2^{n+1}-2}\right) $$ equals: $$ \sum_{k=1}^n k\left(A_{2^{k+1}-2}-A_{2^k-2}\right),\qquad A_k = \...


15

If we start with $a_0$, what about $a_n=3^{\left(1-\frac{1}{2^{n+1}}\right)}$? Note that the terms are $3^{1/2}$, $3^{3/4}$, $3^{7/8}$, and so on.


0

Divergent series diverge and can not be treated as sums. Therefore $1 + 2 + 3 + ..... = -1/12$ is not true and makes no sense. [Yes, I know $\zeta(-1) = -1/12$ but $\zeta(-1) \ne 1 + 2 + 3 + 4 + 5+ ...$. Yes I know that IF $1 + 2 + 3 + 4 + ... $ converged it would equal $\zeta(-1)$ but but it DOESN'T converge and so it DOESN'T equal $\zeta(-1)$.] ...


0

Have a look at this question. We have that: $$ e^{A\sin x}= I_0(A) + 2\sum_{n\geq 0}(-1)^n I_{2n+1}(A)\sin((2n+1) x)+2\sum_{n\geq 1}(-1)^n I_{2n}(A)\cos(2nx) $$ hence by considering the even part and replacing $x$ with $Bx$ we get: $$ \cosh(A\sin(Bx))= I_0(A) + 2\sum_{n\geq 1}(-1)^n I_{2n}(A)\cos(2nBx) $$ and: $$ I_0(A)+2\sum_{n\geq 1}(-1)^{n}I_{2n}(A)\frac{...


6

By your definition of $a_n$, there is a permutation $σ$ such that $∑_{n=1}^∞a_{σ(n)}=∞$. Now take $b_n=a_{σ(n)}$, then $b_n$ satisfies your conditions (as permutations form a group) and $∑_{n=1}^∞b_n=∞$.


4

The converse does not hold. For start with the usual series $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots$. This can be rearranged so that the rearranged series does not converge. Let $a_1+a_2+a_3+\cdots$ be such a rearrangement. Then $a_1+a_2+a_3+\cdots$ does not converge, but the terms can be rearranged to give any desired sum.


3

Directly ratio test: $$\left|\frac{a_{n+1}}{a_n}\right|=\frac{|\alpha-1|(n+1)!}{(n+2)!-(n+1)!+1}\frac{(n+1)!-n!+1}{n!}=$$ $$=|\alpha-1|\frac{\left(n\cdot n!+1\right)(n+1)}{(n+1)(n+1)!+1}=|\alpha-1|\frac{n+\frac1{n!}}{(n+1)+\frac1{n!}}\xrightarrow[n\to\infty]{}|\alpha-1|$$ and the series converges exactly when $\;|\alpha-1|<1\;$


1

Hint. One may write, as $n \to \infty$, $$ \frac{n!}{\left(n+1\right)!-n!+1} =\frac1{n+1/n!} \sim \frac1n $$ and the initial series is convergent for $|\alpha-1|<1$. When $|\alpha-1|=1$ we obtain a conditionally convergent series.


0

Denote by $d(x,y):=\sup_{0\leq t\leq T}|x(t)-y(t)|$ the "usual" distance in $X:=C([0,T])$. Then $$\rho(x,y)\leq d(x,y),\quad d(x,y)\leq e^{LT}\rho(x,y)\qquad\forall x,\>y\in X\ .$$ This shows the Cauchy sequences in $(X,\rho)$ and $(X,d)$ are the same. Since $(X,d)$ is known to be complete (see below) we can at once infer that $(X,\rho)$ is complete as ...


4

If $\lim na_n=0$ then clearly $a_n\sim \dfrac{a_n}{1+na_n}$ and the two series $\sum a_n$ and $\sum \dfrac{a_n}{1+na_n}$ have the same nature. Thus $\sum \dfrac{a_n}{1+na_n}$ is divergent. If $\lim na_n=\ell$ with $\ell>0$ or $\ell=+\infty$ then $\frac{a_n}{1+na_n}\sim\frac{k}{n}$ where $k=1$ if $\ell=+\infty$ and $k=\ell/(1+\ell)$ otherwise. But $\sum\...


1

We have $$(-1)^m n \int_0^1 x^{n-1} \ln^m(1-x)dx=\lim_{z\to 0} \frac{d^m}{dz^m} \,n \int_0^1 x^{n-1} (1-x)^{-z}dx\\ =\lim_{z\to 0} \frac{d^m}{dz^m} \, \frac{n!\, \Gamma(1-z)}{\Gamma(n-z+1)}=\lim_{z\to 0} \frac{d^m}{dz^m} \, \prod_{k=1}^n \frac{1}{1-z/k}. \tag{1}$$ But, in view of the generating function of the Complete homogeneous symmetric polynomials, we ...


1

I won't insert all the details into a hint, but consider this oft-used technique: $|x(t_2)-x(t_1)|<|x(t_2)-x_m(t_2)|+|x_m(t_2)-x_m(t_1)|+|x_m(t_1)-x(t_1)|$, the RHS of which is arbitrarily small. The existence of the function $x(t)$, the 'limit' of the Cauchy sequence, or more formally that function for which $\sup_{0\leq t\leq T}|x(t)-x_m(t)|$ is ...


2

Extending what Zach466920 said: $$\partial_t[k(t,n)]=F(n+1)\cdot k(t,n+1)\tag1$$ We can develop a power series solution for this. Let $$k(t,n)=\sum_{m=0}^\infty \kappa(n, m)\ t^m$$ We equation $(1)$ as: $$\sum_{m=1}^\infty m\ \kappa(n, m)\ t^{m-1} =\sum_{m=0}^\infty(m+1)\ \kappa (n, m) \ t^m =F(n+1)\cdot\sum_{m=0}^\infty \kappa(n+1, m)\ t^m$$ Which gives ...


2

Change variables $x=1-t$: $$ I=\int_0^1 x^n \log^m (1-x) \, {\rm d}x=\int_0^1dt (1-t)^n\log^m t=\sum_{k=0}^n {n\choose k}(-1)^{n-k}\int_0^1 dt \ t^{n-k}\log^m t\ . $$ Now change variable $t=\exp(z)$ and get $$ I=\sum_{k=0}^n {n\choose k}(-1)^{n-k}\int_{-\infty}^0 dz\ e^{(n-k+1)z}z^m= \boxed{\Gamma (m+1)\sum_{k=0}^n {n\choose k}\frac{(-1)^{n-k+m} }{ (n+1-k)^{...


0

The function $\Psi:\mathbb{Z}^+\to\mathbb{Z}$, defined by, $$\Psi(n)=\left\lfloor\dfrac{\langle n\pmod 6\rangle}{3.5}\right\rfloor$$generates the sequence, $$0,0,0,1,1,1,0,0,0,\ldots$$ where $\langle n\pmod 6\rangle$ denotes the least possible positive integer $r$ such that $n\equiv r\pmod 6$.


1

If one knows the following Taylor series expansion, as $u \to 0$, $$ \log(1+u)=u+O(u^2) $$ then one may write, for any fixed real number $t$, as $n \to \infty$, $$ \left( 1+\frac{t}{n} \right)^{n}=e^{\large n\log\left(1+\frac{t}{n}\right)}=e^{\large n\left(\frac{t}{n}+O\left(\frac{1}{n^2}\right)\right)}=e^{t+O\left(\frac{1}{n}\right)} $$ which gives $$ \...


1

Let $X_n=n^{1/\sqrt{logn}}\mathbb{1}\{[\sum_{i=1}^{n-1}1/i,\sum_{i=1}^{n}1/i)\;mod\; 1\}$ for $n\ge 2$ on the unit interval with lebesgue measure. Then $|X_n|_r^r=n^{r/\sqrt{logn}}/n\to0$ for any $r>0$. On the other hand, $n^{1/\sqrt{logn}}\to\infty$, and each $\omega\in [0,1)$ is in infinitely many of the $X_n$.


3

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'...


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} \...


0

The very first paragraph of the paper you link describes Oblath's problem: find all perfect powers $n^m$ whose decimal representation consist of repetitions of the same digit. The paper then poses the question: find all squares having all equal digits but one (intentionally phrased by me using the language from the title of your research). Unpacking your ...


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^...


4

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 ...


12

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 ...


6

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 ...


2

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \...


0

For a decreasing function $f$ on $[0,\infty)$, we have $$ f(0)+f(1)+\dots+f(m-1)\ge\int_0^m f(t)\,dt $$ Since $f(t)=\frac{1}{t^2+1}$ has $$ f'(t)=\frac{-2t}{(t^2+1)} $$ which is negative for $t>0$, we can conclude $$ \sum_{n=0}^{m-1}\frac{1}{n^2+1} \ge \int_0^{m} \frac{1}{t^2+1}\,dt $$ and, passing to the limit $$ \sum_{n=0}^{\infty}\frac{1}{n^2+1}\ge \...


0

To compute this series, let us look at $$ \sum_{k=1}^{\infty} \frac{k^{k-1}}{k!} (- z e^z)^k = \sum_{k=1}^{\infty} \frac{(-1)^k k^{k-1}}{k!} z^k e^{kz} $$ as a formal power series. This series can be rewritten in the form $$ \sum_{m=1}^{\infty} \alpha_m z^m. $$ where \begin{align*} \alpha_m &= \sum_{k=1}^m \frac{(-1)^k \, k^{k-1}}{k!} (\text{coef. by }...


1

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \...


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=\...


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: $$...


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

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


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

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)}{...


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 ...


0

We follow the method of this paper, using the Fourier transform. Following his normalizations, $$ \hat f (\omega) = \frac{1}{2\pi} \int_{\mathbb R} f(x) e^{-i \omega x} \, dx, \qquad f(x) = \int_{\mathbb R} \hat f(\omega) e^{i \omega x} \, d\omega, $$ the key facts to be used are: The identity $\cos(bx) = \frac{e^{i b x} + e^{-i b x}}{2}$ If $f(x) = e^{...


0

Let define the following function: $$f:x\mapsto\frac{2x}{7+x}.$$ Some observations $f([0,+\infty[)\subseteq[0,+\infty[$ and $f$ is an increasing function on $[0,+\infty[$. Therefore, $(a_n)_n$ is an increasing sequence. Therefore, $(a_n)_n$ is convergent if and only if $(a_n)_n$ is bounded. Notice that if $(a_n)_n$ is convergent to $l$, $l$ can only be a ...


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 ...


0

Like continued fractions, if it's not periodic (or some sort of nice pattern), you have to do the innermost first. See my trial using Mathematica below:


0

Let $\displaystyle P=\prod_{i=1}^n\cos\left(\frac x{2^i}\right)$. $\displaystyle P=\cos\left(\frac x{2}\right)\cos\left(\frac x{4}\right)\cos\left(\frac x{8}\right)\cdots\cos\left(\frac x{2^n}\right)$ $\displaystyle P\sin\left(\frac x{2^n}\right)=\cos\left(\frac x{2}\right)\cos\left(\frac x{4}\right)\cos\left(\frac x{8}\right)\cdots\cos\left(\frac x{2^n}\...



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