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Take the module of the ratio and pass to the limit you find $$(2|z|)^2\frac{2k(2k-1)}{(2k+2)(2k+1)}\xrightarrow{k\to\infty}4|z|^2<1\iff |z|<\frac12$$ so the radius of convergence is $R=\frac12$. We verify easily that the domain of convergence is $\overline D(0,\frac12)$: the closed disc of radius $\frac12$.

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You can solve $b_n$ for general form and find a suitble sequnece to compare whor it. $$x^k=2x^{k-1}+x^{k-2}$$ So $$x^2=2x+1$$ And $$x_1= \frac{2+\sqrt 8}{2} , x_2= \frac{2-\sqrt 8}{2}$$ Then $$b_n =(\frac{1}{2}-2\sqrt 2)(1+\sqrt 2)^n +(\frac{1}{2}+2\sqrt 2)(1-\sqrt 2)^n$$ If $a_n= 3^n (\frac{1}{2}-2\sqrt 2)+1$ then $b_n >a_n$ and since $\sum ... 1 Some thoughts: I am not fond of negative numbers, so let us keep the recurrence and let$a_0=-1$and$a_1=7$. That just switches all the signs. It may now be a good idea to write down the first few terms. Because of the$2a_{k-1}$part, the terms (except at the beginning) keep more than doubling, so we can compare the$\frac{1}{a_k}$with$\frac{1}{2^k}$. ... 3 Yes, the sequence$(x_n)$converges to$x/2$. Fix$\epsilon>0$. Let$N \in \mathbb{N}$be an index such that$|x-2x_n| < \epsilon$for all$n \geq N$. Then$|x/2 - x_n| < \epsilon/2 < \epsilon$. So$(x_n)$converges to$x/2$. 2 If$|x-2x_n| <\epsilon $, then$|x/2-x_n| <\epsilon/2 $so$x_n$converges to$x/2$. 1 $$n \cdot n^{1/n} = n e^{\log{n}/n} = n \left [ 1+ \frac{\log{n}}{n} + O\left ( \frac{\log^2{n}}{n^2} \right ) \right ] = n + \log{n}+ O\left ( \frac{\log^2{n}}{n} \right )$$ Now,for sufficiently large$n$(i.e.,$n \gt 4$),$n+\log{n} \lt n \log{n} $, so the sum is larger than $$\sum_{n=2}^{\infty} \frac1{n \log{n}}$$ which diverges. 3$\require{cancel}$Using the limit comparison test with the harmonic series we see $$\lim_{n\to\infty} {\cancel{(1/n)}\cdot 1\over \cancel{(1/n)}\sqrt[n]{n}}=1$$ hence the series diverges since the harmonic series does. 8 Since$\lim_{n\to\infty}\sqrt[n]n=1$then for sufficiently large$n$we have $$\frac1{n\sqrt[n]n}\ge \frac1{2n}$$ Can you take it from here? 1 You could use the fact that for a series of positive terms,$\sum_{n=1}^\infty a_n$converges if and only if$\prod_{n=1}^\infty (1+a_n)$converges. Applying this result to the given problem: The given series converges if and only if the infinite product$\prod_{n=1}^\infty 2^{\frac1n}$For this infinite product, the partial products are of the form ... 0 Since$2^x=1+x\ln 2+O(x^2)$as$x\to 0$then $$\sum_{n\ge 1}\left(2^{1/n}-1\right)\asymp \sum_{n\ge 1}\frac{1}{n},$$ which diverges. 0 First the radius of convergence is defined for a power series but not for whatever series. Then the power series converge inside the disk of convergence, the one centered in zero and having for radius the radius of convergence. And it diverges outside this disk. 1 I suppose you mean strictly increasing. Then the answer is no: Take$a_{(m, n)} = 1$if$m < n$, and$\ldots = 2$if$m \geq n$. Then$L = 1$. Suppose$m : \mathbb{N} \to \mathbb{N}$is strictly increasing; then$m(n) \geq n$, so that$\lim_{n \to \infty} a_{(m(n), n)} = 2$. 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$, and we will use ... 0 This is known as the Law of Large Numbers. (Well, almost. The LLN states that the sum in part 1 converges almost surely/in probability to the expectation$m$. The sum in part 2 is merely a subsequence of the sum in part 1.) Given that we have bounded variance, it is not hard to show the sum converges in probability; perhaps try this for yourself using ... 0 "$\Rightarrow$": Suppose that$X_n \to X$in probability. For any$k \in \mathbb{N}$there exists$N=N(k)$such that $$\mathbb{P} \left( |X_n-X|> \frac{1}{2^k} \right) < \frac{1}{2^k} \qquad \text{for all} \, \, n \geq N(k).$$ Without loss of generality,$N(1)< N(2)<\dots$If we set$n_k := N(k)$, then $$\sum_{k \geq 0} \mathbb{P} \left( ... 2 Long story short,$$ 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}$$that diverges due to Cauchy condensation test (applied twice). 2 Recall again the prime number theorem, which gives that$$ 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 ... 1 Hints: For the first you can use the ratio test and for the second use the alternating series test. 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 ... 3 It's not true. Consider the functions$f_n(x) = x/n$on$(0,\infty)$. 0 I can't reach the contradiction. I get the characteristic function of the partial sums to be$\phi_{n}(\theta) = \prod_{i=1}^{n} \left(\frac{1}{1-i\frac{\theta}{\lambda_{i}}} \right)$. So$|\phi_{n}(\theta)| = \prod_{i=1}^{n} \frac{1}{\sqrt{1 + (\frac{\theta}{\lambda_{i}})^2}} \leq 1$. What am I doing wrong? 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 ... 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. ... 0 given$\varepsilon \in (0,1)$. Note that$P(|X_{n} -1|\geq \varepsilon) = (1-\varepsilon)^n$.$\sum_{n=1}^{\infty}(1-\varepsilon)^n < \infty$thus$P(|X_{n} -1| \geq \varepsilon \ i.o.)=0$by the Borel-Cantelli lemma. That is$P(|X_{n} -1| < \varepsilon \ evt.) = 1$. Hence$X_{n} \to 1$almost surely. 0 As I said in my comment, the answer to the first part of your question is "yes". This is known as the Lebesgue differentiation theorem, see e.g. http://en.wikipedia.org/wiki/Lebesgue_differentiation_theorem. The answer to the second part of your question is also "yes". Seen in the right way, this becomes a question about approximation by convolution (with ... 0 Actually I used the geometric series as noted in this website : Pauls Online Notes Thank you guys anyway for such a fast response! 1 Just simplify to$(\frac{3}{8})^n$and it becomes much easier. 0 It seems the following. It is easy to see that$\zeta(k_1,k_2,\dots,k_n)$converges for all natural$k_1>1$,$k_2$, ...,$k_n$iff$\zeta(2,1,\dots,1)$converges. But $$\zeta(2,1,\dots,1)=\sum_{m_1>m_2>\cdots>m_n>0}\frac{1}{m_1^{2}m_2\cdots m_n}\le$$ $$\sum_{m_1\ge m_2, m_3, \cdots,m_n\ge 1}\frac{1}{m_1^{2}m_2\cdots m_n}=$$ $$... 2 Hint: Since$$\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.$$1 For a fixed t, taking logarithm gives \log \phi_{Y_n}(t) = \sum_{k=1}^n \log \cos(kt \sqrt{\dfrac{3}{n^3}}) Using$$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})$$and ... 0 Hint: now you want to look at how close$$\cos\left(k t \sqrt{\frac{3}{n^3}}\right) \exp\left(\frac{3 k^2 t^2}{2 n^3}\right)$$is to 1. 2 Consider the probability space ((0,1),\mathcal{B}(0,1)) endowed with the Lebesgue measure \lambda and the random variables$$X(\omega) := 1_{(0,1/2)}(\omega) \qquad \qquad Y(\omega) := 1_{(1/2,1)}(\omega), \qquad \omega \in (0,1).$$Then X \sim Y. Set X_n(\omega) := Y(\omega) for all n \in \mathbb{N}, \omega \in (0,1). X_n \to X in ... 4 By definition, \displaystyle\sum_{n=-\infty}^\infty r^{n^2}=\theta_3(0,r). See Jacobi elliptic \theta function. 0 Here's an approach that doesn't rely on logarithms. If a_1,\ldots,a_n are numbers in [0,1) with sum less than 1 then the following inequality holds:$$\prod_{k=1}^n(1+a_k)\leq\frac{1}{1-\sum_{k=1}^na_k}.$$This follows from$$1+a\leq\frac{1}{1-a}$$for a\in[0,1) and induction. In your case this shows that ... 3 There is something which is sometimes called the subsequence principle. It states that x_n \to x holds if and only if every subsequence (x_{n_k})_k admits a further subsequence (x_{n_{k_l}})_l, which converges to x. Note that the limit x has to be fixed. This holds as soon as the notion of convergence is induced by a topology. This should help ... 2 For p\geqslant 1, we have \lVert X_n\rVert_p^p=2/n, hence X_n\to 0 in \mathbb L^p. Since (in general) convergence in \mathbb L^1 implies convergences in probability, which implies in turn convergence in distribution, we can answer questions 2), 3) and 4). Note that for these question, we did not use independence. Question 1) can be solved using ... -1 The first answer is correct because the distribution function of your random variable converges to a function describing a non (Kolmogorovian) probability measure: uniform over the real line. The second answer is wrong. The average converges to 1 with probability 1. That is the distribution functions converges to the distribution function of a constant ... 0 Fine. So you already have the series$$\sum_{n=1}^\infty\log\left(1+\frac1{2^n}\right)$$Define for \;x\ge0\;:$$f(x)=\log(1+x)- x\implies f'(x)=\frac1{1+x}-1=\frac{-x}{x+1}<0\implies\;f(x)$$is monotone descending, and thus$$\forall x\ge 0\;,\;\;f(x)\le f(0)=0$$and we're done. Finally: use the comparison test for infinite positive series 6 We will prove a more general statement: If a_n is a positive sequence then \prod (1+a_n) converges iff \sum a_n converges. Proof: We have$$\log\left(\prod (1+a_i)\right) = \sum \log(1+a_i) \leq \sum a_i$$where we have used \log(1+x)\leq x which is valid for all x > -1. The reversed implication is proven here. Now take a_n = ... 1 An easy way to see that \sum \frac{1}{n^{2}} converges is to compare with the larger telescoping series$$ 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. $$This ad hoc trick may be unsatisfying for a couple of reasons: How does one come up with similar estimates for other series? ... 2 Think about Zeno's paradox in reverse.$$ \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \dots $$You want to walk a mile. First you walk half of it. Then take a bit of rest. Then walk half of the remaining, then take a bit of rest. And so on. Will you ever go beyond the mile? You have an infinite number of numbers (of the form \frac{1}{2^n}), but you ... 1 The note says that if |z| \leq 1, then we have absolute convergence, since the sum of the probabilities is bounded above by 1. Therefore, the radius of convergence is at least 1, hence r_X \geq 1. We do not have enough information to conclude how much bigger (if at all) the radius of convergence is. 2 The point of contention is x{}={}2 and what happens there. Being careful about how we approach this point of discontinuity for F(x), using a limiting sequence in terms of the F_n(x) and the right continuity of F(x), note that for all k>0,$$ F_n\left(2+\dfrac{1}{k}\right){}={}1\,,\ \ \forall\ n>k\,. $$So, for all k>0,$$ ... 2 You are given$\displaystyle \lim_{n \to \infty} P(\{|X_n - 1| \ge \epsilon\}) = 0$for every$\epsilon > 0$. If$0 < \epsilon < 1then $$\left| \frac{1}{X_n} - 1 \right| \ge \epsilon \iff -\epsilon \le \frac{1}{X_n} - 1 \le \epsilon \iff \frac{1}{1 + \epsilon} \le X_n \le \frac{1}{1 - \epsilon}$$ and $$\frac{1}{1 + \epsilon} \le X_n \le ... 2 We have to show that$$\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 ... 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 ... 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 $$\sup_{0 \le t \le 1} |f(t)| \le \sup_{0 \le ... 0 From your computation it is easy to see that$$ \lim_{n\to\infty}S_n=\begin{cases}0 & \text{if }a=0,\\ -\infty & \text{if }a>0,\\ +\infty & \text{if }a<0.\end{cases} $$3 I will address both parts of your question: Part 1: Proving that a_n>0. From your final form of a_n$$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 ...

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As PhoemueX in the comments said, the Dirichlet function is a counterexample which is in Baire class 2, but not in Baire class 1 because the function is nowhere continuous and class 1 functions can only be discontinuous on a meagre set.

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