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Jan
28
awarded  Refiner
Jan
28
revised Is it possible to compute the $\limsup$ of $x_n$ where $x_n$ is the $n^{th}$ digit of $\pi$?
edited title
Jan
28
revised Is it possible to compute the $\limsup$ of $x_n$ where $x_n$ is the $n^{th}$ digit of $\pi$?
added 215 characters in body
Jan
28
answered Is it possible to compute the $\limsup$ of $x_n$ where $x_n$ is the $n^{th}$ digit of $\pi$?
Jan
27
revised How to evaluate $\lim _{x\to \infty }\left(x\sqrt{x^2-x}-x^2\cos\left(\frac{1}{\sqrt{x}}\right)\right)$?
added 1 character in body; edited title
Jan
27
answered How to evaluate $\lim _{x\to \infty }\left(x\sqrt{x^2-x}-x^2\cos\left(\frac{1}{\sqrt{x}}\right)\right)$?
Jan
25
revised Probability that a number has $m$ indistinct factors
added 11 characters in body
Jan
24
revised Probability that a number has $m$ indistinct factors
edited tags
Jan
24
revised Probability that a number has $m$ indistinct factors
added 40 characters in body
Jan
24
comment Probability that a number has $m$ indistinct factors
@Lovsovs: It's called "little-o" notation. For $f,g$ two functions where $g(x)>0$ we write $f(x)=o(g(x))$ to mean that $$\lim_{x\rightarrow \infty}\frac{f(x)}{g(x)}=0.$$ (Sometimes this is defined when $x\rightarrow 0$, depending on the context of the question) You can think of this as $g(x)$ dominating $f(x)$. For example, for any fixed $k$, we have that $x^k=o(e^x)$ or for any $k,\epsilon>0$ we have that $(\log x)^k=o(x^{\epsilon}).$
Jan
24
revised Probability that a number has $m$ indistinct factors
added 104 characters in body
Jan
24
answered Probability that a number has $m$ indistinct factors
Jan
23
awarded  Inquisitive
Jan
22
asked Dealing with a difficult sum of binomial coefficients, $\sum_{l=0}^{n}\binom{n}{l}^{2}\sum_{j=0}^{2l-n}\binom{l}{j} $
Jan
21
comment How, if at all, does pure mathematics benefit from $2^{74207281}-1$ being prime?
See: (1) primes.utm.edu/notes/faq/why.html (2) en.wikipedia.org/wiki/Largest_known_prime_number (3) en.wikipedia.org/wiki/Great_Internet_Mersenne_Prime_Search
Jan
21
revised Prove convergence of: $ \sum_{n=1}^\infty\frac{(-1)^n\cdot\sqrt{n}}{(n+1)\cdot2^n}\cdot(x-3)^n $
added 17 characters in body
Jan
21
comment How I can find limit of a given problem?
Here's a another way to do it: This limit equals $$\lim_{n\rightarrow \infty} \left(\binom{2n}{n} \frac{n!}{n^n}\right)^{1/n}$$and $\frac{1}{2n+1}4^{n}\leq \binom{2n}{n}\leq 4^{n},$ so we see that the limit equals $$4 \lim_{n\rightarrow \infty} \left(\frac{n!}{n^n}\right)^{1/n}=\frac{4}{e}$$ by Sterling's approximation.
Jan
21
answered Prove convergence of: $ \sum_{n=1}^\infty\frac{(-1)^n\cdot\sqrt{n}}{(n+1)\cdot2^n}\cdot(x-3)^n $
Jan
21
revised Limits: How to evaluate $\lim\limits_{x\rightarrow \infty}\sqrt[n]{x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0}}-x$
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Jan
21
awarded  Yearling