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7

Need to Use Monotonicity Since you did not use the monotonicity of $b_n$, you would need to know that the sum converges absolutely to make your argument work. Take for example $$ a_n=\frac{(-1)^{n-1}}n $$ and $$ b_n=1+(-1)^{n-1} $$ The sum of $a_n$ is a well known convergent series: $$ \sum_{n=1}^\infty a_n=\sum_{n=1}^\infty\frac{(-1)^{n-1}}n=\log(2) $$ ...


6

The parity of $\frac{n(n-1)}{2}$ is 4-periodic. Thus the sequence $(-1)^{\frac{n(n-1)}{2}}$ equals to: $$ 1, \, -1, \, -1, \, 1, \, 1, \, -1, \, -1, \, 1, \, 1, \, -1, \, -1, \, 1 , \cdots$$ The original series' partial sum truncated at $N$ equals to $$ \sum_{k=0}^{K} \left( \frac{1}{4k+1} - \frac{1}{4k+2} - \frac{1}{4k+3} + \frac{1}{4k+4}\right) + ...


5

Assuming at least that $\Bbb E|X_j|<\infty$: Clearly yes if $\sum|a_j|<\infty$; then Chebyschev says that $\sum P(|a_jX_j|>\epsilon)<\infty$ for every $\epsilon>0$. No in general. Assuming $X_j$ is not essentially bounded there exist $a_j\to0$ such that $\sum P(|a_jX_j|>1)=\infty$, so the less trivial half of Borel-Cantelli says that ...


5

Notice that $ \ln n < n$, so that $\sqrt[n]{\ln n} < n$, so that $$ \frac{1}{\sqrt[n]{\ln n}} > \frac{1}{n}.$$ Direct comparison of these two series now shows the first to diverge. $\diamondsuit$


5

Let $A_n$ be independent events with $\mathbb{P}(A_n)=1/n$, and define $X_n=1_{A_n}$. Then $X_n\to0$ in probability, but $X_n$ does not converge almost everywhere. Apply the second Borel-Cantelli lemma twice; once to the sequence $A_n$ and also to the sequence $A_n^c$, to conclude that $$P([X_n=1\mbox{ infinitely often}] \cap [X_n=0\mbox{ infinitely ...


5

Use the limit comparison test. Let $\displaystyle a_n=\frac{n^6+13n^5+n+1}{n^7+13n^4+9n+2}$ and $\displaystyle b_n=\frac{n^6}{n^7}=\frac{1}{n}$. Since $$\begin{align}\displaystyle\lim_{n\to\infty}\frac{a_n}{b_n}&=\lim_{n\to\infty}\frac{n^6+13n^5+n+1}{n^7+13n^4+9n+2}\cdot\frac{n}{1}\\ ...


4

For all $x \in [ \pi /4, \pi /2]$ you have $0 \le \cos x \le \frac{\sqrt{2}}{2}$. For all $n$ $$|f_n(x)| = \cos^n x (1- \cos^n x) = \cos^n x - \cos^{2n} x \le \cos^n x \le \left( \frac{\sqrt{2}}{2} \right)^n \to 0$$ so you have uniform convergence. The problem with your argument is that $$\cos^n x = \frac{1}{2} \Leftrightarrow x = \arccos 2^{-\frac{1}{n}}$$ ...


4

Some of the posted answers are using rearrangements or regrouping, but these are treacherous techniques to apply to series which diverge absolutely (as this one plainly does). Leibniz' Theorem as usually stated only applies to alternating series, but we can modify the proof of Leibniz' Theorem to cover this particular series. Leibniz Theorem: A series of ...


3

After $ \ n = 1 \ $ , the exponents of (-1) are binomial coefficients which are "double-alternating" between even and odd integers. So the series looks like $$ 1 \ - \frac{1}{2} \ - \ \frac{1}{3} \ + \ \frac{1}{4} \ + \ \frac{1}{5} \ - \ \frac{1}{6} \ - \ \frac{1}{7} \ + \ \ldots \ \ . $$ As lulu I think properly objected to my separation of terms ...


3

Since the integrand is continuous on $[1,+\infty)$, a potential problem is at $x \to +\infty$. As suggested by André Nicolas, using $$ \sinh x > x^3/6,\qquad x>0, $$ gives $$ \begin{align} \int_{1}^{\infty}\frac{\ln x}{\sinh x}dx&<6\int_{1}^{\infty}\frac{\ln x}{x^3}dx\\\\ &=\left.-\frac{3}{x^2} \ln ...


2

The ratio test does work here: $$\begin{align} \left|\frac{a_{n+1}}{a_n}\right| &= \left|\frac{x^{n+1}(1-x^{n+1})}{x^n(1-x^n)}\right|\cdot\frac{n}{n+1} \\ &= \left|x\cdot\frac{1-x^{n+1}}{1-x^n}\right|\cdot\frac{n}{n+1} \end{align}$$ If $|x|<1$ then as $n\to\infty$, $x^n\to 0$ and that ratio tends to $x$. If $|x|=1$ then the series obviously ...


2

If $0<x<1$ we have $$ 0<\frac{\sin x}{x^{3/2}}< \frac{1}{x^{1/2}}\quad\text{and}\quad\int_0^1\frac{dx}{x^{1/2}}<\infty\text{ since }1/2<1. $$ On the other hand $$ \Bigl|\frac{\sin x}{x^{3/2}}\Bigr|\le\frac{1}{x^{3/2}}\quad\text{and}\quad\int_1^\infty\frac{dx}{x^{3/2}}<\infty\text{ since }3/2>1. $$


2

You want to use a comparison test. Remember with series that you can disregard a finite number of terms. When $n \geq 2$ $$n^7+13n^5+9n+2 \leq n^7 +13n^7 + 9n^7+n^2 = 24n^7,$$ which implies $$ \frac{1}{n^7+13n^5+9n+2} \geq \frac{1}{n^7 +13n^7 + 9n^7+n^2} \geq \frac{1}{24n^7}.$$ For the numerator, you have $$n^6 + 13n^5 + n + 1 \geq n^6.$$ Therefore ...


2

Hint. You have, as $n \to \infty$, $$ \frac{n^6+13n^5+n+1}{n^7+13n^4+9n+2} = \frac1n+\mathcal{O}{\left(\frac1{n^2} \right)} $$ thus, by the comparison test, the initial series is divergent.


2

Suppose $$\lim_{n \to \infty} |x_n|^{1/n} = L < 1.$$ Then by the root test for infinite series the series $$\sum_{n = 1}^\infty x_n$$ converges absolutely. It is a necessary condition that $\lim_{n \to \infty} x_n = 0$ for a series $\sum_n x_n$ to converge. (By the Cauchy criterion for series.) Hence, $$\lim_{n \to \infty} |x_n| = 0.$$


2

I think you want $X_n$ in your sums instead of $X_0$. Anyway: Case 1: Yes, monotone convergence is sufficient to prove it. Case 2: If $Y$ takes on one sign, or more generally is bounded in one direction or another, then monotone convergence will give you this. In general this could fail. The most general but still practical way to check this that I can ...


2

I post this answer because Dirichlet's test has not been mentioned in any of the previous answers. Let $a_n=(-1)^{n(n-1)/2}$ and $b_n=1/n$. The partial sums of $a_n$ are bounded and $b_n$ is decreasing and converging to $0$. Dirichlet's test implies the series is convergent.


2

If you note $\|\cdot \|_\infty$ the sup ess norm, you have : $$\| u - \tilde{u} \|_{\infty} \leq \| u - u_k\|_{\infty}+ \| u_k - \tilde{u} \|_{\infty}$$ Now let $\epsilon > 0$, As $u_k \to \tilde{u}$ uniformly, there exist $k_1$ such that $\forall k > k_1, \ \| \tilde{u} - u_k\|_{\infty} < \frac{\epsilon}{2}$ As $u_k \to u$ in $W^{1,2}_0$, it ...


2

Just take the sequence: $$a_1=J,\ \ a_n=I\ \ \ \text{for }n>1$$


2

If you want a sequence $a_n$ that starts at $J$ and converges toward $I$ as $n\to\infty$, I think the one that is easiest defined is $$ a_n=I-\frac{I-J}{n},\quad n\geq1 $$


2

The difference quotient is equal to $f'(\xi)$ for some $\xi\in(x,x+1/n)$. Now because of the uniform continuity of the derivative you have that for each $\epsilon >0\exists \delta >0$ that $|f'(x)-f'(y)|\leq \epsilon \forall x,y: |x-y|\leq \delta$. Chose $1/n\leq \delta$ and you have it.


2

The partial sums $C_N$ of the sequence $c_{n\text{ mod } r}$ satisfy $$|C_N| = \left|\sum_{n=0}^Nc_{n\text{ mod }{r}}\right| \leq \sum_{n=0}^{r-1}|c_{n}|$$ and is therefore bounded. Since $a_{n}$ is monotonicly decreasing with $\lim\limits_{n\to\infty} a_n = 0$ we have that Dirichlet's test apply and it follows that the series $\sum c_{n\text{ mod }r}a_n$ ...


2

Numerically, the sequence apparently converges, and to a limit of about $4.48761882$. If we change the value of $f(1)$, then for small enough positive $f(1)$, the sequence converges, but for larger positive $f(1)$, the sequence diverges to $+\infty$. The borderline for "small enough" seems to be somewhere between $1.4$ and $1.5$. We observe that$$f(n+1) ...


1

This is true for any $p\in[1,\infty)$. First, since $\Omega$ is bounded, $L^\infty(\Omega)\subset L^p(\Omega)$. Say $\epsilon>0$. Choose $K\subset\Omega$ with $K$ compact, so that if $g=f\chi_K$ then $$||f-g||_p<\epsilon.$$ Now if $\phi_n$ is a smooth approximate identity with compact support then the convolution $g*\phi_n$ is smooth, ...


1

you just need to show that characteristics of intervals can be approximated in $L^p$ norm, and then extend this result arbitrary measurable sets (using the regularity of the Lebesgue measure). Finally extend this result to simple functions. This is also Lusin Theorem. Whose main idea is to use Urysohn's functions. So the answer of your why question is ...


1

For each $n\in N$ there is a $c_{n}\in (x,x+\frac{1}{n})$ such that $$\tag1n(f(x+1/n)-f(x))=f'(c_{n})$$ Let $\epsilon>0$. Since $f'$ is uniformly continuous on $\mathbb R$ we may choose $\delta >0$ such that $\vert x-y\vert <\delta \Rightarrow \vert f'(x)-f'(y)\vert <\epsilon$. Now, choose $N\in \mathbb N$ so large that $n>N\Rightarrow ...


1

I think the particular values of $m$ and $s$ you are using give rise to unexpected computational difficulties. Perhaps it is just too much to wish for two-place accuracy retrieving $s$ with a million simulated values. Also, taking the ordinary SD of lognormal data may not be the optimal way to estimate parameter $s$ (especially when it is small). I'm not ...


1

How about trying Cauchy-Schwarz: $$ \sum_{n=1}^\infty\frac{\sqrt{a_n}}n \le\left(\sum_{n=1}^\infty a_n\right)^{1/2} \left(\sum_{n=1}^\infty\frac1{n^2}\right)^{1/2} $$ You can use Cauchy-Schwarz to make the estimate in your question. $$ \begin{align} \sum_{k=n}^m\frac{\sqrt{a_k}}k &\le\left(\sum_{k=n}^m a_k\right)^{1/2} ...


1

If the question is about a sequence with first term $I$ and that converges to $J$, take ANY converging sequence, from real $a_0$ to the real limit $a_\infty$, and map these to $J$ and $I$: $$b_n:=(I-J)\frac{a_n-a_0}{a_\infty-a_0}+J.$$ If the question is about a function that, when iterated from $J$, yields $I$ in the limit, take $$a_0=J,\ ...


1

The question is not too clear: what do you mean exactly by "function"? What do you mean "oscillating reasonably close at $+\infty$? Anyways, here's a recipe to generate infinitely many analytic functions $f$ of real (non-negative) variable such that $f(0)=J$ and $\lim_{x\to\infty}f(x)=I$ ($I$ and $J$ being integers is irrelevant). Just start with $$ ...



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