# Search results

1. ### Binomial Ratio

Evaluation of the sum \displaystyle \frac{\sum^{r}_{k=0}\binom{n}{2k}\cdot \binom{n-2k}{r-k}}{\sum^{n}_{k=r}\binom{n}{k}\cdot \binom{2k}{2r}(0.75)^{n-k}\cdot (0.5)^{2k-2r}}\ \ ,\ \ (n\geq 2r)
2. ### Determinant of matrix of order 0*0

A have a doubt : If \displaystyle A=[a_{ij}]_{0\times 0} is a matrix . Then \displaystyle |A| =
3. ### Exponential and factorial inequality

Thanks Tywebb. I have tried like this way I am assuming \displaystyle n=2k, k\geq 3, k\in\mathbb{Z} So we have \displaystyle \frac{n!}{\bigg(\frac{n}{2}\bigg)^n}=2\prod^{\frac{n}{2}-1}_{r=1}\frac{\bigg(\frac{n}{2}-r\bigg)\bigg(\frac{n}{2}+r\bigg)}{\frac{n}{2}\cdot...
4. ### Exponential and factorial inequality

Thanks cossine. Using your hint , We use Stirling approximation \displaystyle n!\approx \bigg(\frac{n}{e}\bigg)^{n}\sqrt{2\pi n}\approx \bigg(\frac{n}{e}\bigg)^n. So we have \displaystyle \frac{n}{\bigg(n!\bigg)^{\frac{1}{n}}}\approx e>2\Longrightarrow n!<\bigg(\frac{n}{2}\bigg)^n
5. ### Exponential and factorial inequality

Proving the result \displaystyle \frac{n^n}{n!}>2^n, n>6
6. ### limit with consecutive factorials

Thanks Tywebb.
7. ### limit with consecutive factorials

Thanks Tywebb. Can we solve it without using Stolz and Stirling approximation like definite integration If yes then please explain me.
8. ### limit with consecutive factorials

Evaluation of \displaystyle \lim_{n\rightarrow \infty}\bigg(((n+1)!)^{\frac{1}{n+1}}-(n!)^{\frac{1}{n}}\bigg)
9. ### binomial limit

Thanks Tywebb for different Methods.
10. ### binomial limit

Evaluation of \displaystyle \lim_{n\rightarrow \infty}\bigg[\binom{n}{0}\cdot \binom{n}{1}\cdot \binom{n}{2}\cdots \cdots \binom{n}{n}\bigg]^{\frac{1}{n(n+1)}}
11. ### Irrational Integration

(a)\; $Evaluation of$\int\frac{\sqrt{9x^2+4x+6}}{8x^2+4x+6}dx$(b)\;$Evaluation of $\int\frac{x^2(x\sec x+\tan x)}{(x\tan x-1)^2}dx.$

19. ### Sequence and series

Thanks camelrider.
20. ### Sequence and series

$Let$\displaystyle a_{n} = n+\frac{1}{n}$for$n=1,2,3,4,....20$and$p=\frac{1}{20}\sum^{20}_{n=1}a_{n}And $q = \frac{1}{20}\sum^{20}_{n=1}\frac{1}{a_{n}}$. Then proving $q\in \left(0,\frac{21-p}{21}\right)$. Trial p = \frac{1}{20}\left((1+2+3+\cdot +20)+1+\frac{1}{2}+\frac{1}{3}+\cdots...