New answers tagged convergence-divergence
0
votes
Does sequence produced by Newton's method converge
Note that, for all $u, v\in \mathbb{R}$,
$$\begin{pmatrix}
u\\
v
\end{pmatrix} - (\nabla^2 f(u,v))^{-1}\nabla f(u, v)
= \frac{u^2(4u + 3)}{6u^2 + 6u + 1} \begin{pmatrix}
1\\
1
\end{pmatrix}.$$
...
- 42.7k
0
votes
Find $x_0$ such that the sequence $ 2^n (x_n - \sqrt{3})$ is convergent
Not an answer, it's a remark on the behavior of $x_n$.
As mentioned before, since $x_0>0$, the only possible limit is $\ell = \sqrt{3}$. It was also mentioned that an easy choice would be to take $...
- 22.3k
1
vote
Accepted
Cardinality of two sets where some series diverges .
Hint: the series
$$\sum_{n=1}^{\infty} \frac{1}{n^{1+\frac{1}{n}}}$$
is divergent, that's everything you need to know.
- 12.5k
3
votes
Find $x_0$ such that the sequence $ 2^n (x_n - \sqrt{3})$ is convergent
If you set $f(x) = \frac{1+\sqrt{1+x^2}}{x}$, you have $x_{n+1} =f(x_n)$, thus if you find a fixed point, it is enough.
Now things are well made since $f(\sqrt{3}) = \frac{1+\sqrt{1+3}}{\sqrt{3}} = \...
- 5,008
4
votes
Accepted
How is $x_n$ bounded above by 3?
If $x_n\leq2$, then
$$
x_{n+1}=\sqrt{x_n+2}\leq\sqrt{2+2}=2.
$$
This allows you to show by induction that $x_n\leq2$ for all $n$.
- 213k
2
votes
When is $X_n \sim N(\mu_n, \sigma_n^2)$ uniformly integrable?
Here's the proof of @geetha290krm's comment:
The collection $\{X_n: n \ge 1\}$ where $X_n \sim \mathcal{N}(\mu_n, \sigma_n^2)$ is UI iff $(\mu_n), (\sigma_n)$ are bounded sequences.
"$\...
- 3,023
0
votes
A problem on Lp spaces I found on internet
Assume the sequences $\|f_n\|_p$ and $\|f_n\|_q$ are bounded and $\|f_n-f\|_p\to 0.$ Then $f_n$ contains a subsequence $f_{n_k}$ convergent to $f$ almost everywhere. Thus by the Fatou's lemma
$$\int\...
- 39.5k
0
votes
Convergence in $L^1$ for partial sums of random variables
As stated in LNT's comment, the most straightforward approach is to deal with $L^2$ convergence first. Then our calculations work out as:
$$\mathbb{E}\Bigg(\bigg(\sum_{j=n+1}^\infty X_j/j\bigg)^2\Bigg)...
- 333
0
votes
How to find the limit of $\frac{\ln(n+1)}{\sqrt{n}}$ as $n\to\infty$?
You can use the Cesaro-Stolz lemma(which is the discrete version of the l'Hospital rule; actually the proof of l'Hospital's rule is based on Heine's theorem, the Cesaro-Stolz lemma and Cauchy's ...
- 1
3
votes
Kolmogorov's three series theorem - a valid reference
Perhaps we could add to MSE a complete proof of the 3-series Theorem.
Let $X_i$ be a sequence of real valued independent random variables. $\sum X_i$ converges almost everywhere iff for every $A>0$...
- 2,701
0
votes
If $X_n$ is a sequence with $\lim(X_n)=+\infty$, can we conclude that series generated by this sequence is divergent?
If $X_n \to \infty$ then there exists $M>0$, $ \forall n>n_0, \ X_n>M$ . So that :
$$ \sum_{k=n_0}^n X_k > (n-n_0+1)M \to \infty $$
Id est a properly divergent resulting serie.
- 2,292
3
votes
Accepted
Is there a characterization of convergence in Hausdorff metric in terms of that in the underlying metric?
There is indeed such a characterization. It is given in Section 17.6.3 of the book "Point Sets" by E. Cech. It says roughly that, when $E$ is compact, a necessary and sufficient condition ...
- 590
1
vote
Convergence in measure is not given by a seminorm
Suppose such a seminorm exists, obviously the seminorm of the constant function $1$ is positive because it does not converges to $0$ in measure and the seminorm is neccessarily nonnegative.
Let $\{g_{...
- 11
-1
votes
Prove whether $x_n=\frac{2^n+n^{2^n}}{(-3)^{3n}-3}$ converges or diverges
A quick answer can be seen as this :
$$ x_n=\dfrac{2^nn^{2n}}{(-3)^{3n}-3}$$
So it is equivalent to so that :
$$ |x_n| \sim\left |\dfrac{2^nn^{2^n}}{(-3)^{3n}}\right| $$
And clearly $ (-3)^{3n} = o(n^{...
- 2,292
2
votes
How to show this bounded sequence must be constant
Firstly we rewrite the recurrence as
\begin{equation}
(n^2 + 2n)a_{n+1} = (n+1)^2a_n^2 - 1
\end{equation}
which gives us
\begin{equation}
a_{n+1} = a_n^2 + \frac{a_n^2 - 1}{(n^2 + 2n)}
\end{equation}
...
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2
votes
Why does an individual random walk continue to behave randomly as time goes on?
Both the proportion of one of the types of balls in Polya's urn and the simple symmetric random walk (i.e. $p=1/2$) are martingales. However, a fundamental difference between the two is that the ...
- 17.1k
1
vote
Why isn't the moment generator function for the beta distribution divergent?
Beside what is given here, assuming that $\alpha$ and $\beta$ are positive,
$$M_X(t) = E[e^{tX}] = \int_0^1 e^{tx}\, \frac{x^{\alpha-1}\,(1-x)^{\beta-1}}{B(\alpha,\beta)}\, dx=\, _1F_1(\alpha ;\alpha +...
- 274k
1
vote
Why isn't the moment generator function for the beta distribution divergent?
The MGF is never divergent for a bounded random variable.
Proof:
If $X$ is bounded then $\exp(t X)$ is bounded (for fixed $t$).
If $Y = \exp(t X)$ is bounded then $E(Y)$ is bounded too.
QED.
NB: ...
0
votes
Accepted
If $f$ is continuous at $x_0$, then the left and right limits equal $f(x_0)$
Per the comments, I decided to write an alternative proof using the epsilon-delta method.
Since $f$ is continuous at $x_0$, we have that for any $\varepsilon > 0$, there is a $\delta > 0$ such ...
- 1,716
1
vote
Accepted
continuous functional calculus on C$^*$-algebra
You don’t have to choose those specific functions. By Stone-Weierstrass, given any closed interval $[a, b]$ where $0 < a < b$, you always have a sequence of polynomials $f_n(x)$ which converges ...
- 18.6k
0
votes
Difference between Almost sure convergence and Convergence in probability
We consider a probability space $(\Omega, \mathcal A, \text{Pr})$, and for concision we define the preimage
$$S_n(\varepsilon) := \{ \omega \in \Omega : \lvert X_n(\omega) -
X(\omega) \rvert \geq \...
- 747
0
votes
Convergence of block coordinate descent over nonconvex sets
We can reformulate each of your subproblems as,
$$\underset{\mathbf{x}_{n} \, : \, \| \mathbf{x}_{n} \| = 1}{\mathrm{minimize}} \ \frac{\mathbf{x}_{n}^{\mathrm{T}} \mathbf{A} (\{\mathbf{x}_{i}\}_{i \...
- 31
3
votes
Accepted
Asymptotic limit of a recursive sequence with quartic convergence behaviour
This answer is heavily inspired by the first half of metamorphy's answer to $\{\log n-n(1-na_n)\}$ is convergent for $a_{n+1}=a_n(1-a_n)$ and $0<a_1<1$ (without using asymptotic analysis), which ...
- 6,606
5
votes
Does the sequence $\{\sin^n(n)\}$ converge?
By request of bounty, I'll present a more detailed and clear proof to this question. Please notice that this exact question, both about the sequence and about the summation, have a large number of ...
- 4,210
0
votes
Convergence of $\sum_{n\geq1}\frac{\sin^n(n)}{n}$
This series converges, as do a variety of similar series. This follows from a mild generalization of a paper by Ravi B. Boppana which was about a similar problem. Just a few months ago Ravi uploaded a ...
- 4,210
1
vote
Accepted
Does $\sum_{k=1}^\infty\frac{(-1)^k}{\sqrt{2k-1}}$ converge or diverge?
For $k\geq 1$, consider the sequence defined as $$a_k=\frac{1}{\sqrt{2k-1}}$$
So, $(a_k)_k$ is a positive decreasing sequence:
$$a_k=\frac{1}{\sqrt{2k-1}}>\frac{1}{\sqrt{2(k+1)-1}}=\frac{1}{\sqrt{...
- 1,497
1
vote
Accepted
Proof of a result concerning limits of subsequences
The idea behind your proof is correct. However, not all such $n_k$ as you described form a valid subsequence. What we need is not the condition that $n_k \to \infty$ as $k\to \infty$, but instead that ...
- 159
0
votes
Accepted
Prove that $\lim_{n\to \infty}\int_0^{n^{-\beta}}nf(x)dx=0$, for all $f\in L^q[0,1]$
In fact it is true in the case $\beta=q/(q-1)$. Indeed, applying Hölder's inequality with the exponents $p=q/(q-1)$ and $q$, and $g=n\mathbf{1}_{(0,n^{-\beta})}$, $h=\mathbf{1}_{(0,n^{-\beta})}f$, we ...
- 178k
1
vote
Accepted
Why convergence in probability?
The intuition for your question is that (omitting $\text{log}$ factors which are a technical complication but no essential difficulty, and also omitting taking real parts of the eigenvalues):
$X(t)$ ...
- 446
0
votes
Rate of convergence in probability
Recall that we usually say for a (real) deterministic sequence $(u_n)_{n\ge 1}$ that it converges to a limit $u$ with a rate of convergence $(r_n)_{n\ge 1}$ whenever
$$|u_n-u| = O(r_n), \tag1$$
where $...
2
votes
Does the infinite power tower $\left(e^{1/e} + \frac{1}{1^s}\right)^{\left(e^{1/e} + \frac{1}{2^s}\right)^{\cdots}}$converge?
It converges if and only if $s>2$. Gennady Bachman analyzed this type of question quite thoroughly in his 1995 paper Convergence of Infinite Expoenentials. If $e^{1/e} \le a_n < e^{1/e}(1+\frac{...
- 6,933
2
votes
Prove that the sequence $(s_n)$ is bounded above by 3 and increasing
As you correctly wrote
$$
\left(1+\frac{1}{n}\right)^n =\sum_{k=0}^{n} {n\choose k} \frac{1}{n^k}
=\sum_{k=0}^{n} a_{k,n}
$$
where
$$
a_{k,n}=\frac1{k!} \left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\...
- 5,059
3
votes
Accepted
If a sum $\sum_{k<x} a_k$ is asymptotically less than $\sum_{k<x} b_k$ , then $\sum_{k<x} (a_k)^n < \sum_{k<x} (b_k)^n $ if the second sum converges.
In general, $\sum_{k\geq 1} (a_k)^n < \sum_{k\geq 1} (b_k)^n$ will not hold. For example, let $b_k = 1/k$ and define
$$ a_1 = 10, \quad \quad a_k = \frac{1}{k\log k} \quad k \geq 2,$$
then we have ...
- 803
3
votes
$a_n$ increasing, $\sum\frac{1}{a_n}$ diverges $\implies\exists\ n_1<n_2<n_3,$ such that $\vert (a_{n_3}-a_{n_2})-(a_{n_2}-a_{n_1})\vert<\varepsilon$
The proposition is true. Indeed, if all members of the sequence $(a_n)_{n\in\mathbb N}$ are negative then it converges, which implies the required conclusion. Otherwise, replacing $(a_n)_{n\in\mathbb ...
- 103k
9
votes
Accepted
Topology for the complex numbers endowed with different notions of convergence
To answer what I think is the real question here: There is no topology on $\mathbb{C}$ such that $\lim_{n\to \infty} a_n = x$ in this topology if and only if $\lim_{n\to \infty} \frac{a_1 + \cdots + ...
- 22.1k
1
vote
If $X_n \overset{p}{\to} X$ and $Y_n \overset{p}{\to} Y$, then ${X_n\choose Y_n} \overset{p}{\to}{ X\choose Y} $.
Building upon geetha290krm's comment, the inequality is geometrically evident.
2
votes
Study the convergence/divergence of this series with only upper/lower bounds
Note that \begin{align*}
\sum\frac{(-1)^k}{\sqrt k} &= -1+\sum_{n=1}^\infty \big(\frac1{\sqrt{2n}}-\frac1{\sqrt{2n+1}}\big) = -1+\sum_{n=1}^\infty \frac{\sqrt{2n+1}-\sqrt{2n}}{\sqrt{2n(2n+1)}}\\ &...
- 121k
1
vote
Determining if the function converges uniformly
How is $\bigg|\frac{nx}{\sqrt{1+n^2x^2}}-1\bigg| = \frac{nx-\sqrt{1+n^2x^2}}{\sqrt{1+n^2x^2}}$? What if $x$ is negative?
So, it is quite difficult to accept what you did. Firstly, notice that if
$$f(...
- 2,891
1
vote
Determining if the function converges uniformly
When we find the piecewise limit we get for $x\neq 0$
$$\lim_{n \to \infty} f_n(x) = \lim_{n\to \infty} \frac{nx} {\sqrt{1+n^2x^2}} = \frac{x}{|x|} = \text{sgn} (x) $$
while for $x=0$
$$\lim_{n \to \...
- 2,076
1
vote
Accepted
Infinite limsup and bounded
$\lim\sup_t \theta_t = \infty$ is already sufficient to conclude that the sequence is unbounded. With regards to the variance decaying, this does not really effect the boundedness of the sequence (...
- 803
2
votes
Accepted
Convergence of infinite series $\frac{(-1)^k\ln(\ln k)}{\sqrt{\ln k}}$
You can apply the Dirichlet test since $\frac{\ln(\ln{k})}{\sqrt{\ln{k}}}$ is eventually decreasing, and the first finitely many terms do not affect convergence. Indeed, for $x > 0$,
$$(\frac{\ln{x}...
- 18.6k
2
votes
$\sum_{n=2}^{\infty }\big( \sum_{i=1}^n \frac 1i \big) ^2\frac 1{n^2} $ and a related series.
The harmonic number $\sum_{i=1}^n\frac1i$ is asymptotically equivalent to $\ln n$, and the Bertrand series $\sum_{n\ge2}\frac1{n^\alpha\,(\ln n)^\beta}$ converges as soon as $\alpha>1$, hence so ...
- 44.2k
0
votes
the power series converges in compact convergence topology
Adding an explicit formula for $f_n$
$$F_n(x)=\sum\limits_{k=0}^n x^k=\dfrac{1-x^{n+1}}{1-x}$$
$$F_{n}^{'}(x)=\sum\limits_{k=1}^n kx^{k-1}=\frac{1-(n+1)x^{n}+nx^{n+1}}{(1-x)^2}$$
$$f_n(x)=x F_{n}^{'}(...
- 759
0
votes
$\sum_{n=2}^{\infty }\big( \sum_{i=1}^n \frac 1i \big) ^2\frac 1{n^2} $ and a related series.
For $S_1$, use the fact that for (some) $0<\delta<1$, there exists an integer $N$
$$\frac{(1+\frac{1}{2}+\cdots+\frac{1}{n})^2}{n^2}\leq \frac{1}{n^{1+\delta}}$$
for any $n\geq N$.
Hint. The ...
- 1,271
2
votes
Continuation of convergence in $L^1$
Use Dominated Convergence Theorem. Since $X_n\to X$ in $L^1$, you get that $X_n\to X$ in probability. Thus, $X_nY\to XY$ in probability. Now $|X_nY|\leq M|Y|$ which is integrable, so by DCT, $X_nY\to ...
- 13.3k
2
votes
Determine all $x$ such that the series $\sum_{n=0}^\infty \frac{x^nn^2}{(2n)\cdots (6)(4)(2)}$ converges using the ratio test.
Indeed "the limit of the final expression is $0$, meaning that this will converge for all $x$", but L'Hopital is useless here: as $n\to\infty$,
$$\frac{a_{n+1}}{a_n}=\frac x2\frac{n+1}{n^2}=\...
3
votes
Determine all $x$ such that the series $\sum_{n=0}^\infty \frac{x^nn^2}{(2n)\cdots (6)(4)(2)}$ converges using the ratio test.
Hint: you don't need anything complicated to evaluate the limit, just carry on with some more algebraic simplification to separate concerns and reduce to limits you should know:
$$\frac{x(n+1)}{2n^2} =...
- 50.3k
0
votes
Resnick - Probability Path - Exercise 7.19
Let $\sum_n \mathbb{E}\big[\frac{X_n}{1+X_n}\big] < \infty.$ Then, by M.C.T., $$\mathbb{E}\big[\sum_n \frac{X_n}{1+X_n}\big] < \infty \hspace{1cm} (\because X_n\ge0).$$
So, $\sum_n \frac{X_n}{1+...
- 11
1
vote
there exists $n_0\in\mathbb{N}$ s.t. $x_n\neq 0$ for all $n \geq n_0$, and $\lim_{n\to\infty}\left(\frac{1}{x_n}\right)_{n \geq n_0}=\frac{1}{L}$.
First note that $(x_n)_{n = 1}^{\infty}$ converges to $L$ and since $L \neq 0$, and so $|L| \neq 0$, so then there exists an $N_1 \in \mathbb{N}$ such that $|x_n - L| < \frac{|L|}{2}$ for all $n \...
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