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8

$$ \begin{align} \lim_{m\to\infty}\left(\sum_{n=1}^m\sum_{k=1}^{n-1}\frac{(-1)^k}{nk}+H_m\log(2)\right) &=\lim_{m\to\infty}\sum_{n=1}^m\sum_{k=n}^\infty\frac{(-1)^{k-1}}{nk}\\ &=\sum_{n=1}^\infty\sum_{k=n}^\infty\frac{(-1)^{k-1}}{nk}\\ &=\sum_{k=1}^\infty\sum_{n=1}^k\frac{(-1)^{k-1}}{nk}\\ &=\sum_{k=1}^\infty(-1)^{k-1}\frac{H_k}k\\ ...


4

How about this one? $$ (a_n)_{n\in N}=n \pmod k $$


3

If I understand correctly you want to prove that every real number has a unique non-terminating expansion. Suppose $a$ has two non-terminating binary expansions, since they are different they must be different in at least one of the digits, consider the leftmost digit that is different. This exists since the numbers left of the decimal point are a finite ...


3

Concretely consider $I_n := [n, \infty)$. By the existence of $\lfloor x+1 \rfloor$ for all real $x$, you have that for $n>\lfloor x+1 \rfloor$, $x\notin I_n$ (wich is basically the archimedian property you mentioned), thus $$\bigcap_{n\in\mathbb N} [n,\infty) = \emptyset$$ So it is a valid example. A more general class of examples is given by this ...


2

If the sequence $\{x_n\}$ is unbounded, the sequence $\{|x_n|\}$ is unbounded above. Let $n_1 = 1$ and for each $k \ge 2$ select $n_k$ with the property that $|x_{n_{k}}| \ge |x_{n_{k-1}}| + 1$. Then $\{|x_{n_k}|\}$ increases to $\infty$. Now look at the subsequence of positive terms of $\{x_{n_k}\}$ and the subsequence of negative terms. Both are monotone, ...


2

Hint. Take $a = 0$ and $L = 1$, for example. Can you find an example where $|f(x)|$ is constant, but $f(x)$ has no limit at $0$?


2

Let $x = 1/h$, so that now you're interested in $$ \lim_{x \to \infty} \frac{e^{-x}}{1/x} = \lim_{x \to \infty} \frac{x}{e^x} = \lim_{x \to \infty} \frac{1}{e^x} = 0, $$ using l'Hôpital's in the second equality.


2

Series $\displaystyle \sum_{n=1}^{\infty}(a_{n})^a$ is convergence, so there exists $n_0 \in \mathbb{N}$, that: $$\forall n \geq n_0 \; (a_n)^{a}<1$$ So: $$\exists n_0\in \mathbb{N}\ \forall n \geq n_0 \; a_n<1$$ You know that if $1>a_n>0$ you have $\frac{|a_n|}{n} \leq |a_n| \leq |(a_n)^a|$, so: $$\left|\sum_{n=1}^{\infty} ...


1

If $\|x_k\| > k$ for all $k$ and $\{n_k\}$ is an increasing sequence of natural numbers, then $\|x_{n_k}\| > n_k \ge k$ for all $k$ too. In particular every subsequence is unbounded. Unbounded sequences cannot converge.


1

This is a conseqeunce of Fatou's Lemma: $$ \int_X \liminf_{n\to\infty}u_n\,d\mu\le\liminf_{n\to\infty}\int_X u_n\,d\mu, $$ and hence $$ \int_X u\,d\mu=\int_X \liminf_{n\to\infty}u_n\,d\mu\le\liminf_{n\to\infty}\int_X u_n\,d\mu= \lim_{n\to\infty}\int_X u_n\,d\mu=4. $$


1

I presume that $\Delta(x_n)$ is defined as $|x_n - A|$ and $\Delta(y_n)$ as $|y_n - B|$. There is an error in this proof, where $N'$ is defined. $\Delta(y_n)$ should be taken to be less than $\dfrac{\varepsilon B^2}{16(|A| + 1)}$, not $\dfrac{\varepsilon B^2}{16}$. In the definition of $N'$, the parts about $1$ and $|B|/4$ were just there so that there ...


1

First note that $$ \lim_{x\to 1^-} \frac{1}{\log x} =-\infty $$ And $$ \lim_{x\to 1^+} \frac{1}{\log x} =\infty $$ Therefore $$ \lim_{x\to 1} \frac{1}{\log x} \Rightarrow \mbox{does not exist} $$


1

The classic Dirchlet Function, $$f(x) =\left\{\begin{matrix} 1 & x \in \Bbb Q \\ -1& x \notin \Bbb Q \end{matrix}\right..$$ Notice $\lim_{x \to 0} f(x)$ does not exist, but $\lim_{x \to 0} | f(x) | = 1.$


1

I believe the extra conditions to your problem are unnecessary, and what you are essentially asking is if the $ L^2$ limit is the same as the $ C^2$ limit. I claim they are the same for a $ C^0$ limit. To do so you only need to show that their $ L^2 $ difference is 0 on any compact set $ K $. Compare this $ L^2$ difference to the $ L^2$ differences between ...


1

Working with the first equation, notice that: \begin{align*} &~|\mathbf{x-a}| = 2|\mathbf{x-b}| \\ &\iff |\mathbf{x-a}|^2 = 4|\mathbf{x-b}|^2 \qquad\text{since norms are nonnegative} \\ &\iff (\mathbf x - \mathbf a) \cdot (\mathbf x - \mathbf a) = 4[(\mathbf x - \mathbf b) \cdot (\mathbf x - \mathbf b)] \\ &\iff \mathbf x \cdot \mathbf x - ...


1

The proof looks correct. The only thing I noticed was your use of the ceiling function $\lceil$ and $\rceil$. Unless your professor taught limits in this manner, traditionally one might invoke the Archimedean Principle to show that there exists $N \in \mathbb{N}$ such that $N > a$ for some real number $a$. In fact, the proof of the ceiling function is ...


1

Yes, your reasoning is a correct proof.


1

No, it isn't. First of all think: does $\lim \limits_{n \to \infty} \sqrt{n^2 + 1} - 1 = 0$? Your mistake is that assuming that when we take $n > N$, that we will have $\sqrt{n^2 + 1} - 1 < \epsilon$, but actually when we take $n > N$ we will have $\sqrt{n^2 + 1} - 1 > \epsilon$. Here would be a correct proof: Multiply by $\frac{\sqrt{n^2 + 1} ...


1

The typical technique here is to get this into a fraction rather than a subtraction. Here, multiply by the 'conjugate' of the top, so multiply it by $\frac {\sqrt {n^2+1} +n} {\sqrt {n^2+1} +n}$ Then you get $\frac 1 {\sqrt {n^2+1} +n}$. Now you have a limit that is of the form $\frac 1 \infty$, so it goes to 0


1

Example: $$ a_n=\sin\left(\frac{2n+1}{k}\pi\right),\quad n\in\mathbb N. $$


1

Hint: To prove $s_n+t_n$is Cauchy, if I give you an $\epsilon$ you have to be able to find $N$ such that if $m,n \gt N, |(s_n+t_n)_(s_m+t_m)| \lt \epsilon$ But you have been told that each sequence is Cauchy. So if you demand an $N$ such that if $n,m\gt N$ you have $|s_n-s_m| \lt \epsilon_s = \epsilon /2$ and similarly for $\epsilon _t=\epsilon /2$.....


1

By definition (of compactness), given any compact space $X$, any open cover of $X$ has a finite subcover. In general, however, one cannot say before one has an open cover how many open sets one might need to make a subcover. For example, for any $\epsilon > 0$, the set of $\epsilon$-balls in $[0, 1]$ is an open cover, and so admits some finite subcover. ...


1

Let $X$ be a compact Hausdorff space. Assume that there are $k$ distinct points, $x_i,i=1,2,\ldots,k$, in $X$. By Hausdorfness we can find open sets $U_i\subset X, i=1,2,\ldots,k$, such that $x_i\in U_j$ if and only if $i=j$. Consider the closed set $K=X\setminus \bigcup_{i=1}^k U_i$. It is closed and intersects trivially with the closed set ...



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