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6

We will prove a more general statement: If $a_n$ is a positive sequence then $\prod (1+a_n)$ converges iff $\sum a_n$ converges. Proof: We have $$\log\left(\prod (1+a_i)\right) = \sum \log(1+a_i) \leq \sum a_i$$ where we have used $\log(1+x)\leq x$ which is valid for all $x > -1$. The reversed implication is proven here. Now take $a_n = ...


4

By definition, $\displaystyle\sum_{n=-\infty}^\infty r^{n^2}=\theta_3(0,r)$. See Jacobi elliptic $\theta$ function.


3

It's not true. Consider the functions $f_n(x) = x/n$ on $(0,\infty)$.


3

I will address both parts of your question: Part 1: Proving that $a_n>0$. From your final form of $a_n$ $$a_n=\frac{2n-2\sqrt{(n-1)(n+1)}}{2\sqrt{n}+\sqrt{n-1}+\sqrt{n+1}}$$ we can notice that the denominator is always positive (when $n\ge1$), so it remains to show that $2n>2\sqrt{(n-1)(n+1)}=2\sqrt{n^2-1}$. This is quite easy (thanks to Steven ...


3

There is something which is sometimes called the subsequence principle. It states that $x_n \to x$ holds if and only if every subsequence $(x_{n_k})_k$ admits a further subsequence $(x_{n_{k_l}})_l$, which converges to $x$. Note that the limit $x$ has to be fixed. This holds as soon as the notion of convergence is induced by a topology. This should help ...


2

For $p\geqslant 1$, we have $\lVert X_n\rVert_p^p=2/n$, hence $X_n\to 0$ in $\mathbb L^p$. Since (in general) convergence in $\mathbb L^1$ implies convergences in probability, which implies in turn convergence in distribution, we can answer questions 2), 3) and 4). Note that for these question, we did not use independence. Question 1) can be solved using ...


2

We have to show that $$\mathbb{P} \left( \left| \frac{1}{X_n} - 1 \right|> \varepsilon \right) \to 0 \qquad \text{as} \, \, n \to \infty$$ for any $\varepsilon>0$. To this end, we note that $$\left| \frac{1}{X_n} - 1 \right|> \varepsilon \iff |X_n-1|> \varepsilon \cdot |X_n|.$$ This implies $$\begin{align*} \left\{ \left| \frac{1}{X_n} - 1 ...


2

You are given $\displaystyle \lim_{n \to \infty} P(\{|X_n - 1| \ge \epsilon\}) = 0$ for every $\epsilon > 0$. If $0 < \epsilon < 1$ then $$\left| \frac{1}{X_n} - 1 \right| \ge \epsilon \iff -\epsilon \le \frac{1}{X_n} - 1 \le \epsilon \iff \frac{1}{1 + \epsilon} \le X_n \le \frac{1}{1 - \epsilon}$$ and $$ \frac{1}{1 + \epsilon} \le X_n \le ...


2

Consider the probability space $((0,1),\mathcal{B}(0,1))$ endowed with the Lebesgue measure $\lambda$ and the random variables $$X(\omega) := 1_{(0,1/2)}(\omega) \qquad \qquad Y(\omega) := 1_{(1/2,1)}(\omega), \qquad \omega \in (0,1).$$ Then $X \sim Y$. Set $X_n(\omega) := Y(\omega)$ for all $n \in \mathbb{N}, \omega \in (0,1)$. $X_n \to X$ in ...


2

You need to fix $x>0$. Then there is $K$ such that $\frac{1}{2^k}<x$, for $k>K$. Then $f_k(x)=0$, for $k>K$. Then $f(x)=0$. For $x\leq0$, $f_k(x)=0$, for all $k$. Then $f(x)=0$. Now, for $L^1$ we can integrate. ...


2

Think about Zeno's paradox in reverse. $$ \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \dots $$ You want to walk a mile. First you walk half of it. Then take a bit of rest. Then walk half of the remaining, then take a bit of rest. And so on. Will you ever go beyond the mile? You have an infinite number of numbers (of the form $\frac{1}{2^n}$), but you ...


2

If $g$ and $h$ are real functions defined on a domain $D$, then $$\sup_{x \in D} (g(x) + h(x)) \le \sup_{x \in D} g(x) + \sup_{x \in D} h(x).$$ This is pretty simple to verify using the definition. Thus $$\sup_{0 \le t \le 1} |f_n(t)| \le \sup_{0 \le t \le 1} |f_n(t)| + \sup_{0 \le t \le 1} |f_n(t) - f(t)|$$ and $$\sup_{0 \le t \le 1} |f(t)| \le \sup_{0 \le ...


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 ...


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

Hint: Since $$\sum_{n \geq 1} \mathbb{P}(X_n \neq -1) < \infty$$ it follows from the Borel-Cantelli lemma that for almost all $\omega \in \Omega$ there exists $N = N(\omega)$ such that $$X_n(\omega)=-1 \qquad \text{for all} \, n \geq N.$$


2

The point of contention is $x{}={}2$ and what happens there. Being careful about how we approach this point of discontinuity for $F(x)$, using a limiting sequence in terms of the $F_n(x)$ and the right continuity of $F(x)$, note that for all $k>0$, $$ F_n\left(2+\dfrac{1}{k}\right){}={}1\,,\ \ \forall\ n>k\,. $$ So, for all $k>0$, $$ ...


1

We only need the fact that $X_n$ are uncorrelated, independence is not necessary. $E\left(\sum_{i=1}^n \dfrac{X_i}{n} - m\right)^2 = E\left(\sum_{i=1}^n \dfrac{X_i-m}{n}\right)^2 = \sum_{i=1}^n E\left(\dfrac{X_i-m}{n}\right)^2 \leq\sum_{i=1}^n \dfrac{K}{n^2} = \dfrac{K}{n} $ So we get $\sum_{i=1}^n \dfrac{X_i}{n}$ converges to $m$ in $L^2$, and we will use ...


1

An easy way to see that $\sum \frac{1}{n^{2}}$ converges is to compare with the larger telescoping series $$ 1 + \sum_{n=2}^{\infty} \frac{1}{n(n-1)} = 1 + \sum_{n=2}^{\infty} \left[\frac{1}{n - 1} - \frac{1}{n}\right] = 2. $$ This ad hoc trick may be unsatisfying for a couple of reasons: How does one come up with similar estimates for other series? ...


1

For a fixed $t$, taking logarithm gives $\log \phi_{Y_n}(t) = \sum_{k=1}^n \log \cos(kt \sqrt{\dfrac{3}{n^3}})$ Using $$1-\cos(x) = \dfrac{x^2}{2} - O(x^4), x\to 0$$ $$\log(1-x) = -x + O(x^2), x\to 0$$ we have $$1- \cos(kt \sqrt{\dfrac{3}{n^3}}) = \dfrac{3k^2t^2}{2n^3} + O(\dfrac{9k^4t^4}{n^6}) = \dfrac{3k^2t^2}{2n^3} + O(\dfrac{k^4}{n^6})$$ and ...


1

Let $\varepsilon > 0$. Since $(f_n)$ is decreasing to $0$ pointwise on $[0,1]$, we have $$[0,1] \subseteq \bigcup_{n = 1}^\infty \bigcap_{n\ge N} E_n,$$ where $E_n := \{x\in [0,1] : f_n(x) < \varepsilon\}$. In fact, since $(f_n)$ is decreasing, $(E_n)$ is increasing. Hence, $\cap_{n \ge N} E_n = E_N$ for all $N$. Furthermore, since $f_n$ is ...


1

Your sequence is going to zero pointwise. On the one hand, $f_k(0) \equiv 0$. On the other hand, if $x>0$ and $k>-\log_2(x)$, then $f_k(x)=0$. Additionally, your sequence is dominated by $f(x) = \frac{1}{\sqrt{x}} \chi_{[0,1]}(x)$, which is in $L^p$ for $p \in [1,2)$. So it is definitely going to zero in $L^p$ for $p \in [1,2)$. It does not go to zero ...


1

Just simplify to $(\frac{3}{8})^n$ and it becomes much easier.


1

I suppose you mean strictly increasing. Then the answer is no: Take $a_{(m, n)} = 1$ if $m < n$, and zero if $m \geq n$. Then $L = 1$. Suppose $m : \mathbb{N} \to \mathbb{N}$ is strictly increasing; then $m(n) \geq n$, so that $\lim_{n \to \infty} a_{(m(n), n)} = 0$.


1

Hints: For the first you can use the ratio test and for the second use the alternating series test.


1

The note says that if $|z| \leq 1$, then we have absolute convergence, since the sum of the probabilities is bounded above by 1. Therefore, the radius of convergence is at least 1, hence $r_X \geq 1$. We do not have enough information to conclude how much bigger (if at all) the radius of convergence is.


1

You should note that the Cauchy principlal value of $\zeta(1)$ is $\gamma$ (Euler-Mascheroni constant): $$\lim_{h\to0}\frac{\zeta(1+h)+\zeta(1-h)}2=\gamma$$ The same value can be obtained using the Ramanujan's summation of harmonic series. So to directly answer your question, yes, we can assign a value to the sum of harmonic series, and at least the two ...



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