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You may proceed to simplify the induction hypothesis to make it easier to do the induction step. So your induction hypothesis assume k^{k+1}>(k+1)^k is equivalent to \textstyle(1+\frac{1}{k})^k<k. You can see why by dividing by k^k. Now in order to do the induction step, one only needs to...

\text{You can generalise it more to }\forall\alpha,\beta\in\Bbb R,\lim\limits_{n\rightarrow\infty}\left(\prod_{r=0}^n{n\choose r}\right)^\frac{1}{(n+\alpha)(n+\beta)}=\sqrt e I rewrote the proofs for this more general formula and it is now at...

Yes. There are a lot of old threads with matters not yet resolved which are still worth exploring. Some old threads are closed like the old integration marathons, but most are still open.

I checked the difference in answers on wolframalpha So it looks like you have the same for when x>0. But what about if x<0? Then you can also have (-5\sqrt{5},2(\pi n-\tan^{-1}(2+\sqrt{5}))),n\in\Bbb Z

I put them into wolframalpha and got assuming you only want the real solutions. There are complex ones too but I left them out.

Use blah blah instead of $blah blah$ to avoid Invalid Equation message. \displaystyle \lim_{n\rightarrow \infty} n^{-2}\cdot \sum^{n}_{k=0}\ln\bigg(\binom{n}{k}\bigg) Put into wolframalpha and you get 0.5

You know that in this forum for LaTeX you use blah blah instead of $blah blah$ otherwise you get Invalid Equation message Anyway I fixed it for you. Here were your questions: \int\frac{\sqrt{9x^2+4x+6}}{8x^2+4x+6}dx \text{Evaluation of }\int\frac{x^2(x\sec x+\tan x)}{(x\tan x-1)^2}dx Then...

\text{By the way did you know that for any real number }\alpha, \lim_{n\rightarrow\infty}\left(\prod_{r=0}^n{n\choose r}\right)^\frac{1}{n(n+\alpha)}=\sqrt e\ ?

I might look at that later but if we call your way Method 4 then there is yet another method using integration. Method 5 \text{If }\textstyle p_n=\ln\prod_{r=0}^n{n \choose r}=\ln\prod_{r=1}^nr^{2r-n-1}=\sum_{r=1}^n(2r-n)\ln\frac{r}{n}+n\ln n-\ln n! \text{noting that...

Here is another method using the Stolz–Cesàro theorem. It starts the same as the first method but then goes in a completely different direction, nevertheless ending up at the same answer. Method 3. \begin{aligned}\ln\lim_{n\rightarrow\infty}\left(\prod_{r=0}^n{n\choose...

Solutions. 8. Solve the equation from the areas of the 2 trapezia (5+2)(10)/2=(6+4)(k-4)/2\Rightarrow k=11\therefore D. You can also work it out using the integrals of the lines, but using the areas of the trapezia is easier. 10. \pi\int_0^{64}((2\sqrt...

You can put them all into wolframalpha to get the answers 8. D 10. D 13a. i. 2x^6+\frac{7x^4}{4}-9x+c 13 a ii 8\sqrt x+c 13b 12.5 13e \frac{\pi}{30} 14 a \frac{10}{3} 14b i \frac{32x}{(5x^2+1)^2} 14b ii \frac{x^2-3}{10x^2+2}+c

So to do the actual question: a. \text{Base Step: } 2^1=2>1\text{ but in this case I would also check }2^2=4>2\text{ and you will see why in the induction step} \text{Induction Step:} \text{If for }k>1, 2^k>k\text{ then } 2^{k+1}=2^k\cdot2>k\cdot2=k+1+k-1>k+1\text{ since }k>1 Hence by the...

Yes I think it should be less than or equal to, equal in the case of n=1. If you allow n to be a real number you can get it to be less than 1 but the maximum is \sqrt[e]{e} so for positive integers the maximum is \sqrt[3]{3}\approx1.44 and minimum is 1. So a tighter bound for positive integers...

There is another way to do it using Stirling's formula So here is Method 2: \text{If }p(n)=\left(\prod_{r=0}^n{n \choose r}\right)^\frac{1}{n(n+1)}\text{ then }p(n)^{n(n+1)}=p(n-1)^{n(n-1)}\cdot\frac{n^n}{n!}. \text{ If }L=\lim_{n\rightarrow\infty}p(n)\text{ then }...

The gamma function is just a continuous version of factorial and the digamma and trigamma just come from the derivatives. The hardest function that appeared was the hyperfactorial but I got rid of it pretty quickly by using l'Hôpital's rule. That's a nifty rule to use in limits - even for much...

If I can do it then anyone can do it.

If you assume if it exists then it is 2k for an integer k. But then I'd be more inclined just to add 2. So now 2k+2=2(k+1) is a bigger even integer - contradiction \therefore there is no biggest even integer.

might be easier to use mod 4 arithmetic for n^2=4m+3 If 4 divides n^2-3 then there exists an integer m such that n^2-3=4m\therefore n^2=4m+3 4m+3\equiv3(\mod 4) If n is an integer then there are 4 cases for n^2 n^2\equiv(0(\mod 4))^2=0(\mod 4) n^2\equiv(1(\mod 4))^2=1(\mod 4)...

This is how I did it without wolframalpha. First use a logarithm to turn the product into a sum so it is easier to deal with. Thereafter it is a matter of simplifying the limit as much as possible. Method 1 (see more methods in subsequent posts)...