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11

$(1+\frac12)+(1+\frac14)+(1+\frac18)+\cdots$ certainly diverges; but if you subtract $1$ from each term you get the convergent series $\sum \frac{1}{2^n}$

10

Consider the series $\sum a_n$. If the sequence $a_n$ does not converge, then regardless of what $L$ is, $a_n-L$ does not go to zero, so $\sum (a_n-L)$ still diverges. Suppose the sequence $a_n$ converges. Let $L$ be the limit. Then $a_n-L$ converges to zero. So the series $\sum (a_n-L)$ has a chance of convergence. But the example $\sum\frac{1}{n}$ ...

7

Note that $$\left|\sum_{k=1}^n \sin k\right|= \frac{|\sin(n/2)\sin[(n+1)/2]|}{\sin(1/2)}\leqslant \frac{1}{\sin(1/2)}.$$ Derivation: \begin{align} 2 \sin(1/2)\sum_{k=1}^n\sin k &= \sum_{k=1}^n2 \sin(1/2)\sin k \\ &= \sum_{k=1}^n2 \sin(1/2)\cos (k +\pi/2) \\ &= \sum_{k=1}^n[\sin(k + 1/2 + \pi/2)-\sin (k -1/2 + \pi/2)] \\ &= \sin(n + 1/2 + ... 5a_n > 0a_{n+2}=\frac{1}{1+a_{n+1}}=\frac{1}{1+\frac{1}{1+a_n}}=\frac{1+a_n}{2+a_n}=1-\frac{1}{2+a_n}\therefore \frac{1}{2}\leq a_{n}\leq 1\quad$$4 First of all, you have to "feel" how does this sequence evolve. Computing the first terms, we have:$$1, 0.5, 0.66666666, 0.6, 0.625, 0.615, 0.619, ....$$We see that this sequence go up and down in a decreasing interval and that the values are all in [0.5, 1]. We have a feeling about what's happening. So now, you only have to prove (by recurrence) that ... 4 Because of the weak convergence, there is a probability space (\Omega,\mathcal F,\Bbb P) and random variables X_1,X_2,\ldots,X defined thereon, with values in S, such that (i) P_n is the distribution of X_n (i.e., \Bbb P(X_n\in B)=P_n(B) for all Borel B\subset S), (ii) P is the distribution of X, and (iii) \lim_{n\to\infty} ... 4 The answer is sometimes, but only in very specific circumstances. For example, it certainly works when the series is eventually constant, that is there exists an N and a c such that a_n=c for all n>N. In this case, it should be clear that subtracting the constant c will result in a finite, and hence convergent, sum. In general, subtracting a constant will ... 4 Hint: consider the sequence of functions \chi_{[0,1]},\chi_{[0,1/2]},\chi_{[1/2,1]},\chi_{[0,1/3]},\chi_{[1/3,2/3]}\dots 4 Since$$\lim\limits_{n\rightarrow\infty}\left(1-\frac{1}{n}\right)^n=\frac{1}{e}$$the sum cannot converge 3 One has:$$\lim_{n\to+\infty}\left(1-\frac{1}{n}\right)^n=\frac{1}{e}\neq 0.$$Therefore, your series does not converge. 3 HINT: Notice,$$\int_{1}^{\infty}\frac{dt}{t^4+t^2+1}=\lim_{z\to \infty}\int_{1}^{z}\frac{dt}{t^4+t^2+1}=\lim_{z\to \infty}\int_{1}^{z}\frac{\left(\frac{1}{t^2}\right)dt}{t^2+\frac{1}{t^2}+1}=\frac{1}{2}\lim_{z\to \infty}\int_{1}^{z}\frac{\left(1+\frac{1}{t^2}\right)-\left(1-\frac{1}{t^2}\right)}{t^2+\frac{1}{t^2}+1}\ dt=\frac{1}{2}\lim_{z\to ...

3

Arguing a little sloppily, where "$\sim$" means "within a factor of", $\begin{array}\\ \frac{1}{n^2}\sum_{k=1}^n \frac{k^2}{(k+1)\log(k+1)} &\sim \frac{1}{n^2}\sum_{k=2}^n \frac{k^2}{k\log(k)} \qquad\text{since we can ignore }k=1\\ &\sim \frac{1}{n^2}\sum_{k=2}^n \frac{k}{\log(k)}\\ &\sim \frac{1}{n^2\log(n)}\sum_{k=2}^n k \qquad\text{since ... 3 Since$(b_n)$is bounded, there exists$b_-$and$b_+$such that $$\liminf_{n\to\infty} b_n = b_- \qquad \text{and}\qquad \limsup_{n\to\infty} b_n = b_+.$$ Now suppose by contradiction that$b_-<a$, then there exists$N$such that $$|b_N-b_-|< \frac{b_--a}{2} \implies b_N<\frac{b_-+a}{2}<a \implies \text{contradiction}$$ Similarly, if ... 3 Assuming the$X_nare independent, then it follows from the second Borel-Cantelli lemma that $$\mathbb P\left(\limsup_{n\to\infty} \{X_n=1\}\right)=1.$$ (See for example here for a proof of the Borel-Canelli lemmas.) However, $$\mathbb P\left(\liminf_{n\to\infty}\{X_n=1\} \right) = \mathbb P\left(\bigcup_{n=1}^\infty\bigcap_{k=n}^\infty ... 3 The sequence (a_n)_n is Cauchy: If m>n>N then$$|a_m-a_n|\le\sum_{k=n}^{m-1}|a_{k+1}-a_k|\le \sum_{k=N}^\infty|a_k-a_{k+1}|$$and the latter is <\epsilon for N large enough for this expresses exactly that \sum_{k=1}^\infty|a_k-a_{k+1}| converges. Hence with a:=\lim_{k\to\infty} a_k, we have \lim_{k\to\infty}|a_1-a_k|=|a_1-a|. 3 Question 1: Suppose \sum_{k} a_{k} is a divergent series; does there exist a real number c such that \sum_{k}(a_{k} - c) converges? As Luke Hamblin's and Gerald Edgar's answers note, the question boils down to a minor variant of "does a (general) infinite series converge?" An obvious necessary condition is (a_{k}) \to c; the terms ... 3 It follows directly from Fatou's lemma that \|f\|_p \leq 1; in particular, f \in L^p. To prove the second claim, we fix \epsilon>0 and write$$\begin{align*} \|f_n-f\|_r^r &= \int_{|f_n-f| \leq \epsilon^{-1}} |f_n-f|^r \,dx + \int_{|f_n-f|>\epsilon^{-1}} |f_n-f|^r \, dx \\ &=: I_1+I_2\end{align*}$$We estimate the terms separately. ... 3 Hint:$$\sum_{n=1}^\infty \frac{(-2)^n+n^2}{n\cdot2^n} = \sum_{n=1}^\infty \frac{(-1)^n}{n} + \sum_{n=1}^\infty \frac{n}{2^n}$$Think about the convergence of these two terms. 2 \sin\left(2\pi n\left(1+\frac{1}{2m+2}\right)\right) =\sin\left(\frac{2\pi n}{2m+2}\right) =\sin\left(\frac{\pi n}{m+1}\right) so \frac1{n}\sin\left(2\pi n\left(1+\frac{1}{2m+2}\right)\right) =\frac1{n}\sin\left(\frac{\pi n}{m+1}\right) =\frac{m+1}{\pi n}\sin\left(\frac{\pi n}{m+1}\right) =\frac{\pi}{m+1}\frac{m+1}{\pi n}\sin\left(\frac{\pi ... 2 Not always. Consider 1+2+3+4+\dotsb; this diverges no matter what constant you subtract from the terms. That is:$$(1-c)+(2-c)+(3-c)+(4-c)+\dotsb$$always diverges. 2 There are multiple distinct definitions of convergence, usually by specifying a topology on the set of functions one is considering. For example, the space of all functions from a set X to a topological space Y can be equipped with the product topology, also knows as "pointwise convergence", i.e., a sequence of functions f_n:X\to Y converges to f if and ... 2 I do not know if I am off-topic. If this is the case, please forgive me. May be, we could compute the antiderivative by parts$$u=\log(1+\frac {a^2}{x^2}) \implies du=-\frac{2 a^2}{a^2 x+x^3}dxdv=dx \implies v=x\int \log(1+\frac{a^2}{x^2})\,dx=x \log \left(1+\frac{a^2}{x^2}\right)+\int\frac{2 a^2}{a^2+x^2}\,dx$$The second integral is quite ... 2 As Daniel fischer noted this argument only works for the special case of non negative terms b_i . Hint: if the set \{n: b_n \geq 1\} is infinite then you should know what to do. However if that is finite,then try to make use of the following fact: for every n such that b_n<1 we have$$\frac {b_n}{1+b_n} > \frac {b_n}{2} $$2$$|(a_1-a_r)-(a_1-a_s)|=|a_r-a_s|= |\sum_{k=r}^{s-1} a_{k+1}-a_k | \le \sum_{k=r}^{s-1} |a_{k+1}-a_k |$$This is convergent. 2 As you have defined:$$ p_nf = \sum_{k=1}^{n}\left[n\int_{\frac{k-1}{n}}^{\frac{k}{n}}f(y)dy\right]\chi_{[\frac{k-1}{n},\frac{k}{n}]}(x) $$The linear operator p_n is an orthogonal projection operator onto the linear span of the orthonormal set \left\{\sqrt{n}\chi_{[\frac{k-1}{n},\frac{k}{n}]}\right\}_{k=1}^{n}. Hence, p_n^2=p_n and \|p_nf\| \le ... 2 Obviously,$$|X_n Y_n-XY| \leq |Y| \cdot |X_n-X|+ |X_n| \cdot |Y_n-Y|.$$By the triangle inequality,$$\mathbb{P}(|X_n Y_n-XY| \geq \epsilon) \leq \mathbb{P}(|Y| \cdot |X_n-X| \geq \epsilon/2) + \mathbb{P}(|X_n| \cdot |Y_n-Y| \geq \epsilon/2).$$We estimate the terms separately. For the first one, note that$$\mathbb{P}(|Y| \cdot |X_n-X| \geq ... 2 Any continuous function is a uniform limit of polynomials, so pick your favourite non-differentiable continuous function, and that would work! In fact, iff$is such a function, say$f(x) := |x- 0.5|$, then take the sequence to be the sequence of Bernstein polynomials $$B_n(f)(x) := \sum_{k=1}^n {n\choose k} f\left(\frac{k}{n}\right) x^k (1-x)^{n-k}$$ ... 2 Let$l := \lim_{n}b_{n}$. If$a > l$, then$|b_{n}-l| < a-l$for large$n$, implying$b_{n} - l < a - l$for large$n$, implying$b_{n} < a$for large$n$, a contradiction; if$c < l$, then likewise we have$c < b_{n}$for large$n$, a contradiction again. 2 HINT: Both $$\sum_{n=1}^\infty \frac{1+1}{10^n}$$ and $$\sum_{n=1}^\infty \frac{1-1}{10^n}$$ converge. This is applicable because$-1\leqslant\sin x\leqslant1$2 Hints: Do decompose the fraction using the given factorization. You will get two fractions with a first degree numerator and second degree denominator. By adding a suitable constant to the numerator, you will turn it to the derivative of the denominator, making the fractions easy. Then you need to compensate with two integrands of the form$\$\frac ...

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