calculus - What is the limit of the sequence n!/4^n?

I am trying to find the limit of the sequence by using root test and I don't understand why the limit is not zero?

(the answer is inf).

Answer

By the root test:

$$\begin{array}{rcl}
\displaystyle \limsup_{n\to\infty} \sqrt[n]{a_n} &=& \displaystyle \limsup_{n\to\infty} \sqrt[n]{\dfrac{n!}{4^n}} \\
&=& \displaystyle \dfrac14 \limsup_{n\to\infty} \sqrt[n]{n!} \\
&=& \displaystyle \dfrac14 \limsup_{n\to\infty} \sqrt[n]{\exp\left(n \ln n - n\right)} \\
&=& \displaystyle \dfrac14 \limsup_{n\to\infty} \exp\left(\ln n - 1\right) \\
&=& \displaystyle \dfrac1{4e} \limsup_{n\to\infty} n \\

&=& \infty
\end{array}$$

Hence the sequence diverges to infinity.

integration - Square root of fractional part integral

Does the following integral have a closed form ?
$$\int_{0}^{1}\sqrt{\bigg\{\frac{1}{x}\bigg\}}dx$$

Where $\{x\}$ denotes the fractional part of $x$.

trigonometry - Simple expressions for $sum_{k=0}^ncos(ktheta)$ and $sum_{k=1}^nsin(ktheta)$?

I'm curious if there is a simple expression for

$$
1+\cos\theta+\cos 2\theta+\cdots+\cos n\theta
$$
and
$$
\sin\theta+\sin 2\theta+\cdots+\sin n\theta.
$$
Using Euler's formula, I write $z=e^{i\theta}$, hence $z^k=e^{ik\theta}=\cos(k\theta)+i\sin(k\theta)$.
So it should be that
$$

\begin{align*}
1+\cos\theta+\cos 2\theta+\cdots+\cos n\theta &= \Re(1+z+\cdots+z^n)\\
&= \Re\left(\frac{1-z^{n+1}}{1-z}\right).
\end{align*}
$$
Similarly,
$$
\begin{align*}
\sin\theta+\sin 2\theta+\cdots+\sin n\theta &= \Im(z+\cdots+z^n)\\
&= \Im\left(\frac{z-z^{n+1}}{1-z}\right).

\end{align*}
$$
Can you pull out a simple expression from these, and if not, is there a better approach? Thanks!

Answer

Take the expression you have and multiply the numerator and denominator by $1-\bar{z}$, and using $z\bar z=1$:
$$\frac{1-z^{n+1}}{1-z} = \frac{1-z^{n+1}-\bar{z}+z^n}{2-(z+\bar z)}$$

But $z+\bar{z}=2\cos \theta$, so the real part of this expression is the real part of the numerator divided by $2-2\cos \theta$. But the real part of the numerator is $1-\cos {(n+1)\theta} - \cos \theta + \cos{n\theta}$, so the entire expression is:

$$\frac{1-\cos {(n+1)\theta} - \cos \theta + \cos{n\theta}}{2-2\cos\theta}=\frac{1}{2} + \frac{\cos {n\theta} - \cos{(n+1)\theta}}{2-2\cos \theta}$$

for the cosine case. You can do much the same for the case of the sine function.

elementary number theory - $3$ never divides $n^2+1$

Problem: Is it true that $3$ never divides $n^2+1$ for every positive integer $n$? Explain.

Explanation: If $n$ is odd, then $n^2+1$ is even. Hence $3$ never divides $n^2+1$, when $n$ is odd.

If $n$ is even, then $n^2+1$ is odd. So $3$ could divide $n^2+1$.

And that is where I am stuck. I try to plug in numbers for $n$ but I want a more general form of showing that $3$ can't divide $n^2+1$ when $n$ is even.

Answer

Instead of considering whether $n$ is even or odd, consider the remainder when $n$ is divided by $3$; as an example, if the remainder is $1$, we have $$n = 3k + 1 \implies n^2 + 1 = 9k^2 + 6k + 2$$

which is not divisible by $3$. There are two more cases.

real analysis - Help with convergence tests for series

I have a few questions to ask about series and convergence tests. I have been struggling to study everything fully and if someone can give me advices I will be really thankful.This is what I know so far, and please correct me if I am wrong with anything.

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Definition 1.
For the series $\sum_{n=1}^\infty a_n$ we say that it converges if its sequence of partial sums converges. If it doesn't converge, we say it diverges.
Let's say $S_k$ is the partial sum. If $S_k\to \pm \infty$ we say that
the series diverges.
If $S_k\to$ nothing , we say it diverges as well.

---

Defintion 2.
$\sum_{n=1}^\infty a_n$, if the series converges then the sequence $(a_n)\to 0$ ($n\to \infty$). If $(a_n)\to 0$ then it does not mean that the $\sum_{n=1}^\infty a_n$ converges.

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Examples:
$\sum_{n=1}^\infty \frac{1}{n}$ diverges,
$\sum_{n=1}^\infty \frac{1}{n^\alpha}$ (if $ \alpha$ >1 converges, else diverges) etc.

How do we do tests for the convergence of the following series:

a) $\sum_{n=1}^\infty \arctan\frac{\sqrt n -2}{\sqrt n + 2}$

b) $\sum_{n=1}^\infty \ln (1+\frac{1}{n})$

I have a few questions for the absolute convergence.

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Definition 3.
For the series $\sum a_n$ we say that it converges absolutely if the series $\sum |a_n|$ converges. If the series converges but doesn't converge absolutely, we say it conditionally converges.

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Definition 4.
LEIBNIZ: (alternate series) $\sum (-1)^n \cdot b_n$ , if $b_n \to 0$ , the series converges.
ABELL: $\sum a_n b_n $, if 1) $\sum a_n$ converges
2) sequence $(b_n)$ is decreasing/increasing and bounded, then $\sum a_n b_n $ converges.

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How do we check the absolute and normal convergence of the following series:

a) $\sum_{n=1}^\infty \frac{1}{\sqrt[3]{n^2+1}}\cdot \sin \frac{n\pi}{3}$
b) $\sum_{n=1}^\infty (-1)^n \cdot \tan \frac{3}{\sqrt[4]{n}}$

Thanks.

Answer

Hint for a: I believe that $lim_{n \rightarrow \infty}arctan(.)=arctan(lim_{n \rightarrow \infty}( \frac{\sqrt n -2}{\sqrt n + 2}))=\frac{\pi}{4}.$ Therefore, you can deduce a diverges.

Hint for b: The product of positive real numbers $\prod_{n=1}^{\infty}a_n$ converges iff the sum $\sum_{n=1}^{\infty} log(a_n)$ converges.

If you now look at the sequence of partial products for $\prod_{n=1}^{M}\frac{n+1}{n}=(M+1)$, which tends to infinity as M goes to infinity.

combinatorics - Combinatorially prove that $sum_{i=0}^n {n choose i} 2^i = 3^n $

So I'm not sure at all how to prove things using a combinatorial proof. Where to do I start? What do i need to think about etc. For example how would i prove

$$\sum_{i=0}^n {n \choose i} 2^i = 3^n $$

calculus - Evaluate $int_{0}^{infty}sqrt{frac{ a^2 -y^2 + sqrt{(a^2-y^2)^2+4y^2} }{(a^2-y^2)^2+4y^2}}dy=sqrt{2}pi$

Prove or disprove that for any $a>1$,
$$\int_{0}^{\infty}\sqrt{\frac{ a^2 -y^2 + \sqrt{(a^2-y^2)^2+4y^2} }{(a^2-y^2)^2+4y^2}}dy=\sqrt{2}\pi~.$$

I came across with this integral evaluating inverse Laplace transformation of $\frac{1}{\sqrt{1+s^2}}$ without using any complex integral method. If possible I want to know if there's a way to show that
$$\lim_{T\to\infty}\int_{a-iT}^{a+iT}{ds\over\sqrt{1+s^2}}=2\pi i, a>0$$
without using the residue method.

Actually I met a problem here by the reason explained in this question. Any help appreciated.

Answer

How do you get that a real integral over a real interval will result in an imaginary value?

Use $s=\sinh(t)$ to get
$$
\int\frac1{\sqrt{1+s^2}}=\int \frac{\cosh(t)}{\sqrt{1+\sinh^2(t)}}dt=t+C
$$
to evaluate this integral.

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As to the changed boundaries, you still have to evaluate $$\operatorname{Arsinh}(a+iT)-\operatorname{Arsinh}(a-iT)=2i\,Im(\operatorname{Arsinh}(a+iT)).$$

Since
$$
\sinh(x+iy)=\cos(y)\sinh(x)+i\sin(y)\cosh(x)
$$
one has to solve $\cos(y)\sinh(x)=a$, $\sin(y)\cosh(x)=T$ for $y$. Combined, the equation
$$
T=\sin(y)\sqrt{1+\frac{a^2}{\cos^2(y)}}
$$

has to be solved in such a way that the solution $y(T)$ is continuous in $T$ from $0$ to $\infty$. Since $\sin(y)$ is bounded, large values for $T$ have to be reached by $\cos(y)\approx 0$, which means that $y\in[0,\frac\pi2)$ and the limit for $T\to\infty$ is reached by $y\to\frac\pi2$.