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Jun
10
answered Prove $\lim\limits_{n\to\infty}\frac{1}{\sqrt[n]{n!}}=0$
Jun
4
answered Evaluate $\lim\limits_{n\to\infty}(\frac{3}{x^{4n}}+1)^\frac{1}{n},|x|<1,n\in\mathbb{N}$
Jun
2
comment Is my string of inequalities correct?
Let $a = -s$ to see why
Jun
1
answered convergence sequence $\left(\frac{\ln(n+p)}{\ln(n)}\right)^{n\ln(n)}$
May
31
answered Show that $f(z)$ is constant
May
28
revised prove that $\lim_{x \rightarrow 0^+}\frac{1}{x} \int_0^x\sin(\frac{\pi}{t})dt =0$
edited body
May
28
answered prove that $\lim_{x \rightarrow 0^+}\frac{1}{x} \int_0^x\sin(\frac{\pi}{t})dt =0$
May
28
awarded  Enlightened
May
28
awarded  Nice Answer
May
25
answered Limit of $\frac{f\left(x+g\left(x\right)\right)-f\left(g\left(x\right)\right)}{x}$ as $x\to 0$
May
23
comment How do I calculate $\lim_{n \to \infty} n^\frac{1}{n} (n+1)^\frac{1}{n+1} … (2n)^\frac{1}{2n}$
Thanks, corrected.
May
23
revised How do I calculate $\lim_{n \to \infty} n^\frac{1}{n} (n+1)^\frac{1}{n+1} … (2n)^\frac{1}{2n}$
added 29 characters in body
May
23
answered How do I calculate $\lim_{n \to \infty} n^\frac{1}{n} (n+1)^\frac{1}{n+1} … (2n)^\frac{1}{2n}$
May
11
comment Use mathematical induction to prove $ (2n)!\geq 2^n(n!)^2$ for $n \in \mathbb{N}$
I added an inductive argument
May
11
revised Use mathematical induction to prove $ (2n)!\geq 2^n(n!)^2$ for $n \in \mathbb{N}$
added 324 characters in body
May
11
comment Use mathematical induction to prove $ (2n)!\geq 2^n(n!)^2$ for $n \in \mathbb{N}$
I wasn't sure if you were being asked to use induction or you were just asked the question and you were attempting an inductive proof. At any rate, I still think this is worth pointing out.
May
11
comment Decay of Fourier Transform
The decreasing condition on the Fourier transform's absolute value is incompatible with the conclusion also.
May
11
answered Use mathematical induction to prove $ (2n)!\geq 2^n(n!)^2$ for $n \in \mathbb{N}$
May
11
comment Decay of Fourier Transform
That really doesn't change the issue.
May
11
comment Decay of Fourier Transform
Since $|\hat{f}(y)|^2$ is decreasing, you can't have $|\hat{f}(0)| = 1$ and $|\hat{f}(y)| < 1$ for $y < 0$.