Search found 177 matches

by Riemann
Wed Aug 01, 2018 9:33 am
Forum: Number theory
Topic: Irrational number
Replies: 1
Views: 6069

Re: Irrational number

Let $p \leq N$ be the last prime. If we prove that between $p$ and $N$ does not exist a number that has $p$ as a factor we are done. So, we need to prove that $2p>N$. But this is exactly what Bertrand's postulate says.
by Riemann
Tue Jun 19, 2018 7:20 pm
Forum: Calculus
Topic: A definite Integral
Replies: 2
Views: 3791

Re: A definite Integral

Are you sure about the upper limit? Should not it be $\frac{\pi}{2}$ ?
by Riemann
Sat May 26, 2018 1:31 pm
Forum: Real Analysis
Topic: A zeta limit
Replies: 2
Views: 4410

Re: A zeta limit

Using the above fact we get that $\zeta^{(n)}(0) \sim -n!$.
by Riemann
Sat May 26, 2018 1:28 pm
Forum: Real Analysis
Topic: A zeta limit
Replies: 2
Views: 4410

Re: A zeta limit

We know that the function $\displaystyle f(z) \equiv \zeta(z) + \frac{1}{1-z}$ is a holomorphic function. The Taylor series around $0$ is $$\zeta(z) + \frac{1}{1-z} = \sum_{n=0}^{\infty} \left( \frac{\zeta^{(n)}(0)}{n!} + 1 \right) z^n$$ which converges forall $z \in \mathbb{C}$ thus $\displaystyle ...
by Riemann
Sat May 26, 2018 1:18 pm
Forum: Real Analysis
Topic: A limit with Euler's totient function
Replies: 1
Views: 3567

Re: A limit with Euler's totient function

We are quoting a theorem by Omran Kouba: Theorem: Let $\alpha$ be a positive real number and let $\{a_n\}_{n \in \mathbb{N}}$ be a sequence of positive real numbers such that $$\lim_{n \rightarrow +\infty} \frac{1}{n^\alpha} \sum_{k=1}^{n} a_k = \ell$$ For every continuous function $f$ on the interv...
by Riemann
Sat May 26, 2018 1:11 pm
Forum: Number theory
Topic: Series with least common multiple.
Replies: 1
Views: 3496

Re: Series with least common multiple.

An elementary approach would be as follows: For all $M \in \mathbb{N}^+$ of the form $M=p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k}$ the number of solutions $$\mathrm{lcm} (m, n) =M$$ is given by $(2a_1+1) (2a_2+1) \cdots (2a_k+1)$. It follows that the given series equals $$\sum_{M =1}^{\infty} \frac{1}{M^...
by Riemann
Sat May 26, 2018 1:02 pm
Forum: Real Analysis
Topic: On an evaluation of an arctan limit
Replies: 1
Views: 3623

Re: On an evaluation of an arctan limit

Why not prove the more general result? Let $f:[0, 1] \rightarrow (0, +\infty)$ be a bounded integrable function. Then: \[\lim_{n \rightarrow +\infty} \frac{1}{n} \sum_{k=1}^{n} \frac{f\left ( \frac{k}{n} \right )}{1+2\sqrt{\frac{1}{n} f\left ( \frac{k}{n} \right )+1}} = \frac{1}{3} \int_{0}^{1} f(x)...
by Riemann
Sat May 26, 2018 1:00 pm
Forum: Competitions
Topic: 16 th Cuban Mathematical Competition of Universities [Problem 5]
Replies: 1
Views: 4256

Re: 16 th Cuban Mathematical Competition of Universities [Problem 5]

Greetings, We are focusing on the $\alpha$' s lying in the interval $(0, 1)$. That is because each term of the series is $1$ periodic. Let $\mathbb{Z} \ni k >0$ and let $n$ be the maximal integer for which it holds \[k-1 <n\alpha < k\] Since it holds that $\left \{ n \alpha \right \} \geq 1-\alpha$ ...
by Riemann
Sat May 12, 2018 4:31 pm
Forum: Analysis
Topic: Logrithmic Integral
Replies: 3
Views: 5027

Re: Logrithmic Integral

The Fourier series is the way to go here. Recall that $$\cos(2kx) = \dfrac{e^{i2kx} + e^{-i2kx}}{2}$$ Hence, \begin{align*} \sum_{k=1}^{\infty} \dfrac{\cos(2kx)}k &= \sum_{k=1}^{\infty} \dfrac{e^{i2kx} + e^{-i2kx}}{2k} \\&= \dfrac12 \big(-\log (1-e^{i2x} )-\log (1-e^{-i2x} ) \big) \\&= -...
by Riemann
Sat May 12, 2018 9:22 am
Forum: Analysis
Topic: Logarithmic and Trigonometric Integral
Replies: 2
Views: 4012

Re: Logarithmic and Trigonometric Integral

A hint is along these lines. Apply the sub $x=\arctan t$ and use the well known fact that

$$\sin \left ( \arctan t \right ) = \frac{t}{\sqrt{t^2+1}}$$

The final answer is $\dfrac{7 \pi^3}{216}$.