Recent content by juantheron

Let p(x)=2x^6+4x^5+3x^4+5x^3+3x^2+4x+2. Let \displaystyle I_{k}=\int^{\infty}_{0}\frac{x^k}{p(x)}dx where 0<k<5. Then value of k for which \displaystyle I_{k} is smallest.

Evaluation of $\displaystyle \lim_{n\rightarrow \infty}e^{-n}\sum^{n}_{k=0}\frac{n^k}{k!}$

Thanks friends for yours fantastic solutions i have solved it using Integration. after seeing above solutions by opalg and klass van have seems that my solution is partial (Not fully satisfactory)

Solution Put \begin{equation*} I_{n}=\sqrt{n}\int_{0}^{\pi/4}\cos^{2n-2}(x)\,\mathrm{d}x. \end{equation*} Via the substitutions $ y=\sin x $ and $ y=\frac{z}{\sqrt{n-1}} $ we get \begin{gather*} I_{n}=\sqrt{n}\int_{0}^{\pi/4}(1-\sin^2(x))^{n-1}\,\mathrm{d}x =...

Evaluation of \displaystyle \lim_{n\rightarrow \infty}\sum^{n}_{k=1}\bigg(\frac{k}{n^2}\bigg)^{\frac{k}{n^2}+1}

Finding $$\lim_{n\rightarrow \infty}\sqrt{n}\int^{\frac{\pi}{4}}_{0}\cos^{2n-2}(z)dz$$

Solution Let $\displaystyle I = \int\sqrt{\frac{(1-\sin x)(2-\sin x)}{(1+\sin x)(2+\sin x)}}dx$ We can write $\displaystyle \sqrt{\frac{1-\sin x}{1+\sin x}} = \sqrt{\frac{1-\sin x}{1+\sin x}\times \frac{1+\sin x}{1+\sin x}} = \frac{\cos x}{1+\sin x}$So we get $\displaystyle I = \int\frac{\cos...

Evaluation of $\displaystyle \int \sqrt{\frac{\sin^2 x-3\sin x+2}{\sin^2 x+3\sin x+2}}dx$

Thanks https://mathhelpboards.com/members/olinguito/ My solution is almost same as you.

Finding all natural number ordered pair $(x,y)$ for which $\displaystyle \binom{x}{y} = 2020.$

Thanks Ifdahl for nice solution Here is mine Using $n!=n(n-1)! = [(n-1)+1](n-1)! = (n-1)(n-1)!+(n-1)(n-2)!$ So $(n-1)(n-1)!+(n-1)(n-2)!<(n-1)(n-1)!+(n-1)(n-2)!+(n-1)(n-3)!+\cdots (n-1)1!\;\forall n\geq 4$ So $n!<(n-1)\bigg[(n-1)!+(n-2)!+(n-3)!+\cdots +2!+1!\bigg]\cdots \cdots (1)$ And...

Finding value of $\displaystyle \bigg\lfloor \frac{2020!}{1!+2!+3!+\cdots +2019!}\bigg\rfloor$

$\displaystyle x_{1} = \frac{1}{3}.$ $\displaystyle x_2=\dfrac{4}{9}$. $\displaystyle x_3=\dfrac{52}{81}\in\left(\dfrac{5}{8},\dfrac{2}{3}\right)$.$\displaystyle x_4>\left(\dfrac{5}{8}\right)^2+\dfrac{5}{8}=\dfrac{65}{64}>1$; $\displaystyle...

Consider the sequence $a_{n}$ given by $\displaystyle a_{1} = \frac{1}{3}$ and $\displaystyle a_{k+1}=a^2_{k}+a_{k}$ for $k\geq 2$and Let $\displaystyle S = \frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots \cdots +\frac{1}{a_{2008}}$. Then $\lfloor S \rfloor $ is