Tag Info

Hot answers tagged

18

Suppose $\{s_i : i\in\mathcal I\}$ is a family of positive numbers.$^\dagger$ We can define $$ \sum_{i\in\mathcal I} s_i = \sup\left\{ \sum_{i\in\mathcal I_0} s_i : \mathcal I_0 \subseteq \mathcal I\ \&\ \mathcal I_0 \text{ is finite.} \right\} $$ (If both positive and negative numbers are involved, then we have to talk about a limit rather than about a ...


6

I just want to seek ways that have nothing to do with $\ln (\sin x)$. Hint. You may consider $$ I(a):=\int_0^\infty\frac{\arctan (ax)}{x(x^2+1)}\:\mathrm dx,\quad 0<a<1, \tag1 $$ and obtain $$ I'(a)=\int_0^\infty\frac1{(x^2+1)(a^2x^2+1)}\:\mathrm dx. $$ By using partial fraction decomposition, we have $$ ...


5

We have the next proposition Proposition 1. Let $X$ be an at most countable set, and let $f\colon X\to\mathbf R$ be a function. Then the series $\sum_{x\in X} f(x)$ is absolutely convergent if and only if $$\sup\left\{\sum_{x\in A}|f(x)|:A\subseteq X, A\text{ finite}\right\}<\infty.$$ Inspired by this proposition, we may now define the concept ...


4

We can write $\displaystyle \sum^{n}_{r=1}\frac{1}{2r-1} = 1+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{2n-1}+\left[\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+....+\frac{1}{2n}\right]-\left[\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+....+\frac{1}{2n}\right]$ So $\displaystyle \sum^{n}_{r=1}\frac{1}{2r-1} = \sum^{2n}_{r=1}\frac{1}{r}-\frac{1}{2}\sum^{n}_{r=1}\frac{1}{r}$ Now ...


3

Hint: write $$\lim\limits_{x\to 0}\frac{2^x-1}{x}=\lim\limits_{x\to 0}\frac{2^x-2^0}{x-0}.$$ This should hopefully look familiar to you (if not: think of the definition of the derivative).


3

Before providing my solution, I'd must admit that Oliver Oloa provides the way to calculate this integral. I merely provide a different approach, using Fourier transforms. First a comment. I tried to use a symmetry argument saying that $$ \int_0^{+\infty}f(x+1/x)\arctan x\frac{dx}{x} =\frac{\pi}{4}\int_0^{+\infty} f(x+1/x)\frac{dx}{x}, $$ but I was not ...


3

We can appeal to the sum of a geometric series as follows: $$\begin{align} \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{2^{2n-1}}&=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{2^{2n+1}}\\\\ &=\frac12\sum_{n=0}^{\infty}\left(\frac{-1}{4}\right)^n\\\\ &=\frac12\frac{1}{1+\frac14}\\\\ &=\frac25 \end{align}$$


3

HINT: $$z\cdot S=z\sum_{k=0}^\infty \frac{2^{2k}z^{2k-1}}{(2k)!}=z\sum_{k=0}^\infty\dfrac{(2z)^{2k}}{(2k)!}$$ Now $e^y=\sum_{r=0}^\infty\dfrac{y^r}{r!}$ $2\cosh(y)=e^y+e^{-y}=?$


3

The group of real numbers $(\Bbb R,+,0,-)$ with its usual topology is a topological group: The addition $+:\Bbb R\times\Bbb R \to \Bbb R$ (where $\Bbb R\times \Bbb R$ has the product topology) and the negation $-:\Bbb R \to \Bbb R$ are continuous maps. These properties of $\Bbb R$ are exactly what leads to formulas like $$\lim(x_n+y_n) = x+y\quad \text{ and ...


2

We may also use creative telescoping. Since: $$\left(\frac{1}{n}-\frac{1}{n+1}\right)\leq\frac{1}{n^2}\leq \left(\frac{1}{n-1}-\frac{1}{n}\right)\tag{1}$$ it follows that: $$ \frac{1}{n+1}\leq\sum_{k>n}\frac{1}{k^2}\leq \frac{1}{n}, \tag{2}$$ hence the limit is $\color{red}{1}$ by squeezing.


2

Maybe it's more useful to use $$\sum_{k\leq n}\omega\left(k\right)=\sum_{p\leq n}\left\lfloor \frac{n}{p}\right\rfloor $$ then $$\frac{1}{n}\sum_{k\leq n}\omega\left(k\right)-\sum_{p\leq n}\frac{1}{p}=\frac{1}{n}\sum_{p\leq n}\left\lfloor \frac{n}{p}\right\rfloor -\sum_{p\leq n}\frac{1}{p} $$ $$=\frac{1}{n}\sum_{p\leq n}\frac{n}{p}-\sum_{p\leq ...


2

Let $a_n = (1 + 1/n)^n.$ We want to show $a_{n+1} - a_{n} \geq \dfrac{1}{n(n+1)}$ for large $n$. $\dfrac{a_{n+1}}{a_n} = \left(1 + \dfrac{1}{n}\right) \left(1 - \dfrac{1}{(n+1)^2}\right)^{n+1}.$ The RHS can be expanded as $\left(1 + \dfrac{1}{n}\right) \left(1 - \dfrac{1}{(n+1)^2}\right)^{n+1} = \dfrac{n+1}{n} \times \left( \underbrace{1 - ...


2

You don't need any formula. Start with: $$\begin{matrix} 2&&8&&27&&85&&260\\ &6&&19&&58&&175\\ &&13&&39&&117 \end{matrix}$$ and observe each term in the third line is thrice the previous term. Hence we may conjecture the continuation of this third line will be: $$\dots\quad ...


2

Let $H$ be a positive unlimited integer of nonstandard analysis. Then, for example, the sum $$ \sum_{n=1}^H n = \frac{H(H+1)}{2}$$ is a sum of uncountably many positive numbers... but it's a hyperfinite nonstandard sum, so it exists by the usual methods of nonstandard analysis. The sum is unlimited, though. Other sums can be finite: e.g. $$ \sum_{n=1}^H ...


2

For $x$ in some interval $[a,b]$, $x^n e^\zeta \leq b^n e^b$. Now $e^b$ is just some constant, and can be ignored. You are now looking at $$ \lim_{n \to \infty} \frac{b^n}{n!},$$ which can be handled in very many ways. For a reference on these, see the answers to the question Prove that $\lim \limits_{n \to \infty} \frac{x^n}{n!} = 0$, $x \in \Bbb R$.


1

$e^\xi$ is just a constant. Use Stirling's approximation to write $$ \lim_{n\to \infty} \frac{x^n}{n!}= \lim_{n\to \infty} \frac{x^n}{\sqrt{2 \pi n}\left(\frac{n}{e}\right)^n}= \frac{1}{\sqrt{2\pi}}\lim_{n\to \infty} \left(\frac{ex}{n}\right)^n\frac{1}{\sqrt{n}} $$ For $n>ex$ the term in the brackets is $<1$ and then its clear that the limit is 0.


1

Let $a_1=200 000$ and $q=75000$. Use the formulas of arithmetic progression. First of all find $a_{12}$. Also find minimal $k$ such that $a_k\geqslant 600000$


1

Consider the series for $\cosh (2z)$ and divide it by $z$....


1

Let the GP be- $a,ar,ar^2,ar^3$. Sum will be- $\frac{a(1-r^4)}{1-r}=30$ Squares of terms- $a^2,a^2r^2,a^2r^4,a^2r^6.$ Sum will be-$\frac{a^2(1-r^8)}{1-r^2}=340$. $$a(1+r^2)(1+r)=30$$ $$a^2(1+r^2)(1+r^4)=340$$ $$\frac{45}{(1+r^2)(1+r)^2}=\frac{17}{(1+r^4)}$$ $$45+45r^4=17(1+r^2)(1+r^2+2r)=17(1+r^2+2r+r^2+r^4+2r^3)$$ $$14+14r^4=17r^2+17r+17r^3$$ The two ...


1

HINT: $$5\{a+(7-1)d\}=6\{a+(2-1)d\}$$ and $$\dfrac52\{2a+(5-1)d\}=\dfrac{65}2$$ where $a$ is the first term, $d$ is the common difference So, we have two simultaneous equations with two unknowns


1

Since $e^{-x}$ decreases so quickly (and with $\dfrac 1{x^3}$ decreasing quickly for small $x>0$) we may deduce that following integral is approximate (and bounded!) by : \begin{align} 2\int_{c\theta}^c x^{-3}e^{-x}dx&\approx \int_{c\theta}^\infty 2\,x^{-3}e^{-x}dx,\quad(*)\\ &\approx \left(\frac 1{(c\theta)^2}-\frac ...


1

Although Dirichlet's test per se does not apply, it seems like a Good Thing to note that the proof of Dirichlet does apply. Sum by parts and you're done. In more detail: Let $(\epsilon_j)$ be the sequence of plus and minus ones, so we're considering convergence of $$\sum_j\frac{\epsilon_j}{j}.$$Let $\sigma_n=\sum_{j=1}^n\epsilon_j$. Dirchlet does not apply ...



Only top voted, non community-wiki answers of a minimum length are eligible