New answers tagged sequences-and-series
0
votes
Power series where the number $e$ is a root
As $e=(1/e)^{(1/e)^{(1/e)^{...}}} $and $e^{ze^{ze^{ze^{\text{...}} }}}=\sum _{n=0}^{\infty } \frac{(n+1)^{n-1} z^n}{n!},
$
having $ {z}=$ $\frac{1}{e}$;
will make
$\sum _{n=0}^{\infty } \frac{(n+1)^{...
- 9
1
vote
Infinite summation formula for modified Bessel functions of first kind
Generated from Mathematica:
$$\int_0^{\frac{\pi }{2}} t I_0(2 \kappa \cos (t)) \, dt=\\\sum _{n=0}^{\infty } \frac{\, _2F_3\left(\frac{1}{2}+n,1+n;1,\frac{3}{2}+n,\frac{3}{2}+n;\kappa ^2\right)}{(1+2 ...
- 4,492
0
votes
Telescopic summation for AGP: $R_n=\sum_{k=1}^n k r^{k-1}$
First approach:
\begin{align}
\sum_{k=1}^n kr^{k-1} &= \sum_{k=1}^n \frac{d}{dr} r^k \\
&= \frac{d}{dr} \sum_{k=1}^n r^k \\
&= \frac{d}{dr} \left( r \sum_{k=1}^n r^{k-1} \right) \\
&= \...
- 149
1
vote
Accepted
Telescopic summation for AGP: $R_n=\sum_{k=1}^n k r^{k-1}$
It is easier to use the previous result for the geometric partial sum. We expand
$$kr^{k-1} = \frac{kr^{k-1} - kr^k}{1-r}$$
This is creates a telescopic behaviour, because consecutive terms cancel to ...
- 356
1
vote
Optimal strategy for uniform distribution probability game
The problem can be formulated as follows: Let $(X_t)_{t \geq 0}$ be a sequence of iid random variables, each is uniformly distributed on $[0,N]$. Adam is the guy who tries to maximize the expected ...
- 1,033
0
votes
IMO 2024 p-3,Sequence of Counts - Are Odd or Even Terms Eventually Periodic?
Solution
Interesting combinatorics in disguise of sequence. The key is to prove that sufficiently large numbers appear finitely many times in such sequence, and such number of appearance should be ...
- 59
8
votes
Accepted
Power series where the number $e$ is a root
You can use a series for $W(e^{1+z})$ where $W$ is the Lambert W function. See wikipedia (https://en.wikipedia.org/wiki/Lagrange_inversion_theorem):
$W(e^{1+z}) = 1 + \frac{z}{2} + \frac{z^2}{16} + \...
- 2,329
4
votes
Power series where the number $e$ is a root
Let $\alpha > 1$ be any real umber. Then by invoking the density of the rational numbers, we can choose $a_1, a_2, \ldots \in \mathbb{Q}_{>0}$ such that
$$ 1 - \frac{1}{n!} \leq \sum_{k=1}^{n} ...
- 170k
3
votes
Power series where the number $e$ is a root
Define a sequence $a_0,a_1,a_2,...$ of rational numbers recursively by
\begin{align*}
a_0&\,=\;1
\\[4pt]
a_{n+1}&\,=\;
\begin{cases}
-\frac{1}{3^{n+1}}&\text{if}\;s_n > 0\\[3pt]
\;\;\;\...
- 59.2k
5
votes
Power series where the number $e$ is a root
Assume $a_0=1/2$ and $a_k\neq 0,$ $0\le k\le n-1,$ are chosen so that $0<S_k<{1\over (k+1)!}$ for $0\le k\le n-1.$ There exists a nonzero negative rational $a_{n}$ satisfying $$0<S_{n}:=S_{n-...
- 33.9k
0
votes
Prove that lim$_{n \rightarrow \infty} (1 + \frac{z}{n})^{n} = e^z$ uniformly on all compact sets.
This problem appears in Ahlfors' book, where it is mentioned to use Taylor's Theorem as seen in Chapter 3. Therefore, I will try to give a proof based on this or at least how I think Ahlfors ...
- 195
3
votes
Accepted
Can the elements of this recursive sequence be calculated individually?
All numbers in these sequences are even, so let's divide them by $2$. Here's how you get $\frac12\cdot$ sequence $k$: List the numbers $1,\ldots,2^k-1$ in binary as a $k$-digit number (with leading ...
- 1,548
2
votes
Can the elements of this recursive sequence be calculated individually?
Your sequences may be 'extracted' from the sequence OEIS A030109 :
The trick for the sequence $m$ is to reverse the bits of $2^m-1$ consecutive numbers (subract $1$ and divide by $2$) starting at $2^{...
- 43.5k
0
votes
Optimal strategy for uniform distribution probability game
Let $X$ be a random variable representing Adam's final profit and $y = \mathbb{E}[X]$ denote the expected profit from playing this game.
Assume, at a given role, Adam has $v$ dollars. Then, if he re-...
- 540
1
vote
Example of a series where the ratio test $\limsup |a_{n+1}/a_n|$ can be applied, but $\lim |a_{n+1}/a_n|$ cannot
Consider the series $a_n$ obtained as follows
$a_1=1$,
$a_2=1$, $a_3=2$
$a_4=1$, $a_5=2$, $a_6=3$
$a_7=1$, $a_8=2$, $a_9=3$, $a_{10}=4$, etc.
That is, split $\mathbb{N}$ in blocks $A_k$, each of ...
- 40.8k
4
votes
Unexpected result, does $\Big\lfloor\frac{n-1}{2}\Big\rfloor=\sum_{i=1}^\infty\bigg\lfloor\frac{n+2^i-1}{2^{i+1}}\bigg\rfloor $
We shall instead prove the actual general form, that is \begin{align*} \sum_{k=0}^\infty \left\lfloor\frac{x+2^k}{2^{k+1}}\right\rfloor = \begin{cases} \lfloor x\rfloor &\text{if }\, x\geq 0 \\ \...
- 51
1
vote
Accepted
Limit of $\sqrt[n]{2-x_n}$
Define $f(x) = x^n(2-x)$ for $x\in (1,2)$ and $n \ge 1$. Then
$$f'(x)=-(n+1)x^{n-1}\left(x-\underbrace{\frac{2n}{n-1}}_{>2} \right) >0$$
So, $f(x)$ is increasing in function of $x$.
As you ...
- 16.8k
0
votes
Finding $\lim \frac{(2n^{\frac 1n}-1)^n}{n^2}$.
Here I want to present another solution to this old problem.
First observe that
$$a_n= \frac{(2n^{\frac 1n}-1)^n}{n^2}=\Big(\frac{2n^{1/n}-1}{n^{2/n}} \Big)^n=\Big(\frac{-(1-2n^{1/n}+n^{2/n})+n^{2/n}}{...
2
votes
$\displaystyle \lim_{n \rightarrow +\infty} \dfrac{1}{\log(n)}\sum_{k=1}^n \dfrac{1}{k}f\left(\dfrac{k}{n} \right) = f(0)$
Here I show that
For any bounded function function $f$ on $[0,1]$ that is continuous at $0$
$$\frac{1}{\log n}\sum^n_{k=1}\frac1k f(\frac{k}{n})\xrightarrow{n\rightarrow\infty}f(0)$$
Recall that ...
- 40.8k
1
vote
Interesting Weighted Sum over Even Fibonacci Numbers
Hint:
Like Prove that for every $ n \in \mathbb{N} \cup \{0\}$, $F_{n+5} + 2F_n$ is divisible by $5$.
and using
How can I evaluate $\sum_{n=0}^\infty(n+1)x^n$?,
$$S_\infty=-(0+1)F_0x^0+\sum_{n=0}^\...
- 276k
5
votes
Accepted
Interesting Weighted Sum over Even Fibonacci Numbers
Multiple approaches come to mind. The generating function for the Fibonnaci numbers under the definition
$$
F_0=0,\;\;\;\;F_1=1,\;\;\;\;F_{n+2}=F_{n+1}+F_n
$$
is
$$
\sum_{n=0}^\infty F_nx^n=\frac{x}{1-...
- 3,097
2
votes
Limit of $\sum_{k=1}^n \frac{1}{k+1/2} - \ln(n+1/2)$
Considering the hint provided by @openspace, I have got the correct result.
\begin{align*}
\lim_{n\to\infty}\sum_{k=1}^n \frac{1}{k+\frac{1}{2}} - \ln(n+\frac{1}{2}) &= \lim_{n\to\...
- 105
3
votes
Accepted
Limit of $\sum_{k=1}^n \frac{1}{k+1/2} - \ln(n+1/2)$
Let $\psi(z)=\Gamma'(z)/\Gamma(z)$. Then it is known that
$$
\psi(z+1)=-\gamma+\sum_{n=1}\left(\frac1n-{1\over n+z}\right),
$$
so there is
\begin{aligned}
\lim_{n\to+\infty}\left(\sum_{k=1}^n{1\over k+...
- 7,193
2
votes
$\displaystyle \lim_{n \rightarrow +\infty} \dfrac{1}{\log(n)}\sum_{k=1}^n \dfrac{1}{k}f\left(\dfrac{k}{n} \right) = f(0)$
Assume $f(0)=0$. Let $\epsilon >0$. Choose $j$ such that $|f(x)|<\epsilon$ for $ x <\frac 1j$. Now split $\dfrac{1}{\log(n)}\sum_{k=1}^n \dfrac{1}{k}f\left(\dfrac{k}{n} \right)$ into sum ...
- 39.3k
2
votes
Is $\sum _{n=0} ^{\infty} \frac{1}{(2n+1)^2(2n+3) \binom{n-1/2}{n}} = G - \frac{1}{2}$?
Going from binomila coeffcients to gamma function, if you consider
$$S(x)=\sum _{n=0} ^{\infty} \frac{x^{2n+1}}{(2n+1)^2(2n+3) \binom{n-1/2}{n}}=\frac{\sqrt{\pi }}{4}\sum _{n=0} ^{\infty}\frac{\Gamma (...
- 268k
1
vote
Computing the Limit $\lim_{n \to \infty} n\left[ \frac{a_{n+1}}{a_{n}} - \left(\frac{n}{n+1}\right)^{\frac{1}{3}} \right]$
Let $\;x_n=\dfrac n{n+1}\;$ for any $\;n\in\Bbb N\,.\quad(\implies\lim\limits_{n\to\infty}x_n=1\;)$
It results that, $\quad n(x_n-1)=-x_n\;\;$ for any $\;n\in\Bbb N\,.$
Moreover,
$\begin{align}b_n&...
- 12.4k
2
votes
Accepted
Is $\sum _{n=0} ^{\infty} \frac{1}{(2n+1)^2(2n+3) \binom{n-1/2}{n}} = G - \frac{1}{2}$?
Define:
$$f(x)=\sum _{n=0}^{\infty} \frac{x^n}{(2n+1)^2 \binom{n-1/2}{n}} = \sum_{n=0} ^{\infty} C_n$$
Using the definition for generalized hypergeometric functions, we can find the hypergeometric ...
- 31.7k
0
votes
Computing the Limit $\lim_{n \to \infty} n\left[ \frac{a_{n+1}}{a_{n}} - \left(\frac{n}{n+1}\right)^{\frac{1}{3}} \right]$
$$\frac{a_{n+1}}{a_n}=\frac{2n+1}{2n+2}$$
$$L:= \lim\limits_{ n \to \infty} n \left(\frac{2n+1}{2n+2} \left(\frac{n}{n+1}\right)^{\frac 1 3}\right)= \lim\limits_{ n \to \infty} n \left(\frac{2n+1}{2n+...
- 6,620
2
votes
Computing the Limit $\lim_{n \to \infty} n\left[ \frac{a_{n+1}}{a_{n}} - \left(\frac{n}{n+1}\right)^{\frac{1}{3}} \right]$
The expression can be represented as $${n\over 2(n+1)}-\left ({n\over n+1}\right )^{4/3}\,[-(n+1)]\left [\left (1-{1\over n+1}\right )^{2/3}-1\right ]$$ The limit of the last two factors is equal by ...
- 33.9k
1
vote
Accepted
Convergence of Sequences with respect to Ultrafilters on the Natural numbers in Compact Hausdorff Topological Spaces
For part $1$, assume $X$ is compact. Assume to the contrary that $\lim_\limits{\mathcal{F}} (x_n)$ is empty. Then by definition of convergence w.r.t. $\mathcal{F}$, we see that for any $x \in X$, ...
- 9,918
3
votes
Computing the Limit $\lim_{n \to \infty} n\left[ \frac{a_{n+1}}{a_{n}} - \left(\frac{n}{n+1}\right)^{\frac{1}{3}} \right]$
Let me call your while expression $A_n$. Notice, that there is the identity $\Gamma(1/2 + n) = \sqrt{\pi} \frac{(2n)!}{n! 4^n}$. Then you have $$a_n = \frac{\Gamma(n+1/2)}{\sqrt{\pi} n!}$$ and yields ...
- 121
3
votes
Computing the Limit $\lim_{n \to \infty} n\left[ \frac{a_{n+1}}{a_{n}} - \left(\frac{n}{n+1}\right)^{\frac{1}{3}} \right]$
Using $A^3-B^3 = (A-B)(A^2 +AB + B^2)$, we get
$$
\begin{split}
\frac{n+1/2}{n+1} - \sqrt[3]{\frac{n}{n+1}}
&= \frac{\frac{(n+1/2)^3}{(n+1)^3}- \frac{n}{n+1}} {\left(\frac{n+1/2}{n+1}\right)^2 +...
- 1,470
3
votes
Accepted
Computing the Limit $\lim_{n \to \infty} n\left[ \frac{a_{n+1}}{a_{n}} - \left(\frac{n}{n+1}\right)^{\frac{1}{3}} \right]$
Note that:
$$n\left[\frac{n+1/2}{n+1}-\left(\frac{n}{n+1}\right)^{\frac{1}{3}}\right]
=\frac{n}{n+1}\left[(n+1/2)-\sqrt[3]{n(n+1)^2}\right].$$
$$\lim_{n\to\infty}n\left[\frac{n+1/2}{n+1}-\left(\frac{n}...
- 8,455
4
votes
Sum of reciprocal Bernoulli numbers
I doubt a true closed form for either the sum or the power series exists, though it is not too difficult to derive the following alternative representation for the power series, valid for all $x\in\...
- 3,413
4
votes
Accepted
Finding the integer part of a sum
The function $n^{2/3}$, with an exponent less than $1$, is everwhere concave downwards for positive $n$:
$2n^{2/3}>(n+a)^{2/3}+(n-a)^{2/3},0<a<n$
We also get from AM-GM:
$n^{2/3}>[(n+a)(n-...
- 41.4k
-1
votes
The convergence of this series: $\sum\limits_{n=2}^\infty {1\over n^{\log n}}$
For any function $f$ such that $f(n)>\eta>1$ as of a certain $m$,
$$\sum_{n=1}^\infty\frac1{n^{f(n)}}<\zeta(\eta)+C_m$$
where $C_m$ is the finite constant corresponding to the replacement of ...
- 290
1
vote
Convergence of $\sum\limits_{n=1}^{\infty} \frac{(n!)^2}{(2n)!}x^n$
We have $$4^n=(1+1)^{2n}>{2n\choose n}={(2n)!\over ( n!)^2}$$ Hence for $ x=\pm 4$ the absolute value of the general term is greater than $1,$ which excludes the convergence.
Remark 1 The radius ...
- 33.9k
1
vote
The convergence of this series: $\sum\limits_{n=2}^\infty {1\over n^{\log n}}$
Just for your curioisity.
It does converge by the integral test
$$\int \frac {dx}{x^{\log(x)}}=\int e^{-t^2+t}\,dt=e ^{1/4}\int e^{-\left(t-\frac{1}{2}\right)^2}\,dt$$
$$\int_2^\infty \frac {dx}{x^{\...
- 268k
5
votes
Accepted
What is the limit of the alternating series $F(z)=\sum_{n=1}^\infty(-1)^n z^{T_n}$ as $z\to1$ for a sequence $T_n\sim cn$?
The answer is no. For example,
$$
\sum_{n=1}^\infty (-1)^n z^{4n+(-1)^n} = - \sum_{m=1}^\infty z^{4(2m-1)-1} + \sum_{m=1}^\infty z^{4(2m)+1} = -\frac{z^3}{1-z^8} + \frac{z^9}{1-z^8} = -\frac{z^3+z^5+z^...
- 82.1k
6
votes
Convergence of $\sum\limits_{n=1}^{\infty} \frac{(n!)^2}{(2n)!}x^n$
This is perhaps overkill but I think the maths is cute and it doesn't require any appeals to Stirling's approximation etc:
For $x = 4$,
$\displaystyle\sum\limits_{n=1}^{\infty} \frac{(n!)^2}{(2n)!}2^{...
- 365
3
votes
Convergence of $\sum\limits_{n=1}^{\infty} \frac{(n!)^2}{(2n)!}x^n$
Stirling's approximation (good for intuition, even if it's not acceptable for an explanation) gives you
$$
\frac{4^n(n!)^2}{(2n)!} \approx \frac{4^n(\sqrt{2\pi n} n^n e^{-n})^2}
{\sqrt{2\pi(2n)} (...
- 27k
3
votes
G.P. geometry relation
Progressions with a common ratio govern some nested polygon structures. Below is such a structure involving triangles.
Start with $\triangle ABC$. Along ray $\overset{\rightarrow}{AB}$ identify point ...
- 41.4k
3
votes
Is there a closed form for the quadratic Euler Mascheroni Constant?
Using this answer, it's not hard to show that the limit equals
$$
L=\color{blue}{\pi\big(2\gamma-\ln\pi+4\ln\Gamma(3/4)\big)}\approx 2.584981759579253217+
$$
Indeed, we have $L=\lim\limits_{N\to\infty}...
- 40.1k
1
vote
Accepted
Proving $\lim_{y\to 0} y^2 \ln|yx^2|=0$ using sequences
There is no difference between the proofs for the particular sequence $(b_n)$ and a general one: $b_n^{2}\ln (|b_n|x^{2})=b_n^{2}\ln |b_n|+b_n^{2}\ln (x^{2})$ and each of the two terms tends to $0$. [...
- 39.3k
0
votes
Does the value of a convergent series equal the sum of its non-negative terms minus the sum of the absolute values of its negative terms?
Yes it is true, and more elementary than the two theorems you are quoting:
Assuming $\sum a_n$ is absolutely convergent, simply write
$$\sum_{n\le N}a_n=\sum_{n\le N\atop a_n\ge0}a_n-\sum_{n\le N\atop ...
- 39.7k
2
votes
I have a series and I cannot find a way to calculate its value
Hint
Using Catalan numbers $$a_k= \frac{(4k-1)!}{(2k+1)!\,(2k-1)!}=\frac{1}{2}\,C_{2 k}$$ which have the generating function
$$\frac{1}{2}+\frac{1}{\sqrt{2} \sqrt{1+\sqrt{1-16x}}}$$ This would give ...
- 268k
0
votes
Continuum among converging sum of reciprocals of natural numbers sandwiched between consecutive terms
My intuition is that a careful greedy-ish algorithm will suffice, for a much broader class of problems. I've made this answer Community wiki in case anyone wants to flesh out this idea (assuming it's ...
1
vote
General form of Jacobi Theta Transformation $\sum_{n \in \mathbb{Z}} e^{n^2 \pi / x} = \sqrt{x} \sum_{n \in \mathbb{Z}} e^{n^2 \pi x} $
The functional equation listed comes from the Poisson summation formula; the main idea is that $e^{-\pi t^2}$ is, up to a constant, its own Fourier transform. You can experiment with different ...
- 25.6k
1
vote
Closed formula for probability of n-digit numbers containing three consecutive sixes
We use a generating function approach to derive a formula for the wanted sequence of probabilities. In order to do so we reformulate the problem a bit. We consider an alphabet $\mathcal{V}=\{0,1,2,3,4,...
- 109k
1
vote
Accepted
Does there exist a positive sequence with these two properties?
While the answer by @John B is of course correct, I wanted to give a simple explicit construction of such a sequence.
Ignoring the $(-1)^n$ for a moment, you essentially want multiples of $\sqrt2$ ...
- 1,548
Top 50 recent answers are included