# Tag Info

3

Consider the function $f(n)$ such that $$f(n) = \left \{ \begin{array}[ll] (-1 & \textrm{ if } n \textrm{ mod } 8 \equiv 0,1,2,3 \\ 1 & \textrm{ if } n \textrm{ mod } 8 \equiv 4,5,6,7 \end{array} \right.$$

3

I will assume that you really are starting at $2$. If the start is different, modification is easy. Let $g(n)=\left\lfloor\frac{n+2}{4}\right\rfloor$. Here $\lfloor x\rfloor$ is the "floor" function, the greatest integer $\le x$. Then your sequence is given by $(-1)^{g(n)}$.

3

You have proved that. In mathematics, the limit of a sequence is the value that the terms of a sequence "tend to". If such a limit exists, the sequence is called convergent. To show that $a_n$ converges to zero, it is enough to show that for any $\epsilon>0$ there exists $N$ such that if $n>N$, $a_n<\epsilon$. So, just pick ...

2

Modified to start at $n=2$:$$f(n)=(-1)^{\frac{(n+2)!}{(n-2)!4!}}$$ This will give you your desired sequence. Here is a plot of this function.

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Note that for $x$ small, the sum is close to $\pi / 2$ and yet as $x \to 0$, for fixed $n$ you have $\sin^2 (nx) / n^2 x \simeq (nx)^2/n^2x = x$. Thus as $x \to 0$ you need more and more terms in your series to get up to a particular fixed constant $c < \pi / 2$, so the convergence is not uniform.

2

There's at least one such sequence for each subset $A\subseteq \mathbb N$, namely $$f(n) = \begin{cases} n/2 & \text{if n is even} \\ 5 & \text{if n is odd and }\frac{n-1}2\in A \\ 7 & \text{otherwise} \end{cases}$$ On the other hand, each function $f:\mathbb N\to\mathbb N$ (surjective or not) can be encoded as a subset of $\mathbb N$, for ...

2

Well if you mean sequences in $\mathbb{Z}^+$ then the answer is pretty straightforward: as many as the real numbers. This is easy to see: take any real number in its decimal representation and intertwine $\mathbb{Z}^+$ saying that if $a= a_1 \dots a_k \dots$ then $x_{2n} = a_{n}$ and $x_{2n+1}=n$ is a sequence that contains $\mathbb{Z}^+$. This is ...

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It is because the sum of total search time to search each number in the set once is $$T_{total}=\sum_{k=1}^{N}k=\frac{N(N+1)}{2}$$ And then the average search time is $$T_{average}=\frac{T_{total}}{N}=\frac{N+1}{2}\approx \frac{N}{2}$$ So the point is that $\frac{N}{2}$ is based on the assumption that he/she would not try to search smartly even if numbers ...

1

In a general manner, consider $\big( f_n \big)_{n \in \mathbb{N}}$ a series of continuous functions such that $f_n \to f$ uniformly in a compact interval $[a,b]$ . In your example, $$f_n = \sum\limits_{i = 1}^n F_n(x)$$ represent the partial sums with $f_n(x) \to \sum\limits_{i = 1}^{\infty} F_n (x) = F(x)$. We want to prove ...

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Domination of the absolute value of the general term by the general term of an absolutely convergent series is a valid criterion. You may want to state explicitly that x is real because you are using it in your proof. The theorem is true for complex x but then you need to modify the proof.

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On $[0,1)$, the sequence converges pointwise to the continuous function $x \mapsto 0$. But not uniformly because: $$\lim_{n \to \infty} \lim_{x \to 1} \frac{x^n}{1 + x^n} = \frac12 \neq \lim_{x \to 1} \lim_{n \to \infty} \frac{x^n}{1 + x^n} = 0$$

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Let $e_k = (0,0,0,\ldots,0,0,1,0,0,\ldots)$ where the $1$ is in the $k$th place and all other entries are $0$. This belongs to your proposed space. The set $\{e_k : k=1,2,3,\ldots\}$ is linearly independent.

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Put $\displaystyle R_n(x)=\sum_{k\geq n}\frac{\sin(kx))^2}{k^2x}$ for $x>0$. If your series is uniformly convergent on $I=]0,\delta]$ for some $\delta >0$, then for every $\varepsilon>0$ there exists an $N$ such that for $n\geq N$ and $x\in I$, we have $R_n(x)<\varepsilon$. Now we have $$R_n(x)\geq ... 1 Do you want to find the N such that \frac{1}{N^2}<\varepsilon? Take \sqrt{} both sides to get \frac{1}{N}<\sqrt \varepsilon, then N>\frac{1}{\sqrt\varepsilon} is needed to reach \frac{1}{N^2}<\varepsilon with \varepsilon>0 but small as anyone wishes. 1 I can tell what you intended, though the argument is not quite clear. You may argue this way: Given any \varepsilon > 0, we have \frac{1}{n^{2}} < \varepsilon if n > \sqrt{\frac{1}{\varepsilon}}; taking N := \lceil \sqrt{\frac{1}{\varepsilon}} \rceil + 1 suffices. 1 For any positive integer n, we have T_n-S_n = g(T_n,S_n) where g:\mathbb R^2\to \mathbb R is the map (x,y)\mapsto x-y. Given t\in\mathbb R, it is clear that$$g^{-1}((-\infty,t]) = \{(x,y):x-y\leqslant t\}  is a Lebesgue-measurable set in $\mathbb R^2$, and so $g$ is a measurable function. Since $\sigma(g(T_n,S_n))\subset \sigma(T_n,S_n)$, ...

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