calculus - The limit of general term in a series

I have the following statement -

If $\sum_{1}^{\infty} a_{n}^2$ converge then $\sum_{1}^{\infty} a_{n}^3$ converge.

Well i know this statement is true , but if can someone explain why

$\lim_{n\rightarrow\infty}(a_{n}^2) = 0$ implies that $\lim_{n\rightarrow\infty}(a_{n}) = 0$

(a fact that help to prove this statement) , Thanks!

abstract algebra - Show the norm map is surjective

Let $K/F$ be an extension of finite field. Show that the norm map $N_{K/F}$ is surjective.

Here is what I have so far:

Since $F$ is a finite field and $K/F$ is a finite extension of degree $n$, so $\operatorname{Gal}(K/F)=\langle\sigma\rangle$, where $\sigma(a)= a^{q}$ with $q=p^{m}=|F|$. In addition, by primitive element theorem, $K=F(\alpha)$ for some $\alpha \in K$.

We want to show $N_{K/F}(\alpha)$ generates $F$. By the definition of norm, we have $$N_{K/F}(\alpha)=\alpha^{1+q+\cdots+q^{n-1}}$$ and since $(1+q+\cdots+q^{n-1})(q-1)=q^n-1$, we have the order of $N_{K/F}(\alpha)$ is divided by $q-1$.

But I want to show $o(N_{K/F}(\alpha))=q-1$ in order to conclude surjectivity. So any hint for how to proceed? Any other methods are also perferred.

Answer

I think you have a lot of the right stuff written down. Let's take $\alpha$ to be a generator for the cyclic [see Lemma 1.6 here for a proof] group $K^*$, so $\alpha$ has order $q^n - 1$. Then, as you say, its norm is
\[
N_{K/F}(\alpha) = \alpha^{1 + q + \cdots + q^{n - 1}} \in F^*.
\]

This norm generates $F^*$ because its order is precisely $q - 1$. This is just group theory: if an element $x$ in a group has order $mk$ then $x^m$ has order $k$.

You could also focus on the size of the kernel, as Mike B suggests. As Brandon points out, these are the roots of a certain polynomial, which gives you an upper bound. And there is an obvious lower bound.

A very useful generalization of this is Hilbert's Theorem 90.

calculus - Limit of a set of fractions

I'm having trouble with this particular exercise in limits, and I just can't seem to find a way to crack it.

I saw a similar exercise online where they used integrals, but it's pretty early in the course so we're only supposed to use basic limit arithmetics and the Squeeze theorem (Oh boy, and I thought it had a bad name in MY language). That said, I already tried the Squeeze theorem and it doesn't work.

$$\lim_{n\to\infty} \left( \frac{1}{1\cdot4}+\frac{1}{4\cdot7}+...+\frac{1}{(3n-2)(3n+1)} \right)$$

What am I missing?

Thanks in advance.

calculus - Critical values of sine function, when undefined

This may be a beginner question, but I really want to understand this good and thorough.

I have this sine function, withing the interval $-3 ≤ x ≤ 3$

$$f(x)=\frac{\sin (\pi x)}{\sin(2\pi x)}$$

My question is regarding local maximum and minimum values wich i assume this function does not have, as all $f'(x)=0$ are undefined? In the same manner this function would also not have a horizontal tangent?

(as far as i understand its possible to transform the function using trig identities, but the questions above is regarding the function in it's original form.)

Would be really greatful for any help.

calculus - Are the partial sums for $sum_{n=1}^{infty}sin(n^a)$ bounded for $ageq1$ and unbounded for $0

I know that the partial sums of $$\sum_{n=1}^{\infty}\sin(n)$$

are bounded between $\frac{\cos\left(\frac{1}{2}\right)-1}{2\sin\left(\frac{1}{2} \right)}$ and $\frac{1+\cos\left(\frac{1}{2} \right)}{2\sin\left(\frac{1}{2}\right)}$.

On the other hand, the partial sums of
$$\sum_{n=1}^{\infty}\sin(\sqrt{n})$$
are unbounded.

I think that the partial sums of $\sum_{n=1}^{\infty}\sin(n^a)$ are bounded for $a \geq 1$ and unbounded for $0< a<1$, but how can I prove this? I think this question involves Euler-Maclaurin sum.

Answer

$\bullet$ $a>1$ is an integer

If $a>1$ is an integer, then this post in MO by Terry Tao solves that the partial sums are not bounded.

$\bullet$ $a>1$ is not an integer

If $a>1$ is not an integer, the same argument by Terry Tao still works, since we have equidistribution of $n^a$ mod $2\pi$, and any sum or difference of $(n+i)^a$ with $1\leq i \leq h$ modulo $2\pi$. Let $X_i$ be the random variable $\sin^a (k+i)$ where $k=1,\ldots n$. Assuming the boundedness of partial sums, we end up having a contradiction that the random variables $X_1, \ldots, X_h$ such that $\mathrm{Var}(X_1+\cdots+X_h)$ is bounded as $h\rightarrow \infty$ by assumption, but $\mathrm{Var}(X_1+\cdots+X_h)\sim h/2$ as $h\rightarrow\infty$.

$\bullet$ $0

The first part of this answer of mine, shows that $\sum_{\alpha can be arbitrarily large. Thus, unboundedness of the partial sums follows.

For a better estimate, we apply Lemma 4.8 of The Theory of the Riemann Zeta-function written by Titchmarsh.

Let $f(x)$ be a real differentiable function in the interval $[a,b]$, let $f'(x)$ be monotonic, and let $|f'(x)|\leq \theta <1$. Then
$$
\sum_{a$$

Taking imaginary part from the lemma and $f(n)=n^a/(2\pi)$, we have

$$
\sum_{n\leq N} \sin(n^a) = \int_{1-}^N \sin(x^a) \ dx + O(1).
$$

The change of variable $x^a=t$ gives
$$
\int_{1-}^N \sin(x^a) \ dx=\int_{1-}^{N^a} \frac1a t^{\frac1a-1}\sin t \ dt.
$$

Applying the integration by parts to the last integral, we obtain an estimate of
$$-\frac1a N^{1-a}\cos(N^a) + O(N^{\max\{0,1-2a\}}).$$
This expression is clearly unbounded. Therefore, the partial sums are unbounded when $0, and $a>1$.

proof writing - Prove that 1/2 + 1/4 + 1/8 ....... = 1

I've often heard that instead of adding up to a little less than one, 1/2 + 1/4 + 1/8... = 1. Is there any way to prove this using equations without using Sigma, or is it just an accepted fact? I need it without Sigma so I can explain it to my little sister.

It is not a duplicate because this one does not use Sigma, and the one marked as duplicate does. I want it to use variables and equations.

Answer

For physical intuition, so you can explain it to your little sister, I will use a 1m long ruler.

Take the ruler an divide it into two equal parts:

$$1=\frac{1}{2}+\frac{1}{2}$$

Take one of the parts you now have, and again divide it in half.

$$=\frac{1}{2}+\frac{1}{4}+\frac{1}{4}$$

Take one of the smaller parts you now have, and again divide it in half.

$$=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{8}$$

Repeat. In general for $n$ a positive integer,

$$=\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2^n} \right)+\frac{1}{2^n}=1$$

So,

$$\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2^n}=1-\frac{1}{2^n}$$

As we let $n$ become a really big (positive) integer, note the sum gets closer and closer to $1$, because $\frac{1}{2^n}$ gets really close to zero (the smallest part of the ruler you have left over gets close to 0 meters in length). We say the sum converges to $1$ in the limit that $n \to \infty$.

elementary number theory - Does $3$ divide $2^{2n}-1$?

Prove or find a counter example of : $3$ divide $2^{2n}-1$ for all $n\in\mathbb N$.

I compute $2^{2n}-1$ until $n=5$ and it looks to work, but how can I prove this ? I tried by contradiction, but impossible to conclude. Any idea ?