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I'm a recent high school graduate living in Los Angeles, California.


2d
comment The number of prime divisors of any number
Prove that the smallest natural number divisible by $n$ primes is $2^{n}$. It then follows that there can be no natural numbers less then $2^{n}$ with either $n$ or more prime factors because otherwise that would contradict our first statement.
2d
revised The number of prime divisors of any number
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2d
reviewed Approve suggested edit on Subsequence of a sequence converging to its lim sup and lim inf
Aug
28
revised Evaluating a series with the Möbius function and greatest common divisor.
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Aug
27
comment Summing various rearrangements of $1-\frac12+\frac13-\frac14+\cdots$
It would probably be helpful if instead of truncating the series' by hand you wrote them out in a more standardized notation so we can easily see the pattern being displayed.
Aug
26
comment A sum containing harmonic numbers $\sum_{n=1}^\infty\frac{H_n}{n^3\,2^n}$
@ClaudeLeibovici Yes, but not his sum: $\sum_{n=1}^\infty\frac{1}{n^s2^n}$
Aug
26
answered Examples of Functions f and g such that lim f(x)g(x) exist but lim f(x) and lim g(x) doesnt
Aug
26
revised A sum containing harmonic numbers $\sum_{n=1}^\infty\frac{H_n}{n^3\,2^n}$
[Edit removed during grace period]
Aug
26
comment A sum containing harmonic numbers $\sum_{n=1}^\infty\frac{H_n}{n^3\,2^n}$
If you're going to do it in terms of a definite integral involving polylogarithms, you could take note that: $$H_{n}=\int_{0}^1\frac{1}{1-x}(1-x^n) dx$$ Then divide through by $n^32^n$ and sum over $\mathbb{N}$ so that: $$\sum_{n=1}^\infty \frac{H_{n}}{n^32^n}=\int_{0}^1\frac{(Polylog[3,1/2]-Polylog[3,x/2])}{x-1}dx$$
Aug
26
comment A sum containing harmonic numbers $\sum_{n=1}^\infty\frac{H_n}{n^3\,2^n}$
@Conifold For cases $1$,$2$ and $3$ yes. But for integers larger then $3$ no explicit formula are known according to: en.wikipedia.org/wiki/Polylogarithm. See right above "Relationships to other functions"
Aug
26
comment A sum containing harmonic numbers $\sum_{n=1}^\infty\frac{H_n}{n^3\,2^n}$
Some lower order ones if you're interested: $$\sum_{n=1}^\infty\frac{H_{n}}{n2^n}=\frac{\pi^2}{12}$$ $$\sum_{n=1}^\infty\frac{H_{n}}{n^22^n}=\zeta(3)-\frac{\pi^2\ln(2)}{12}$$ Also $$\sum_{n=1}^\infty\frac{1}{n^32^n} =\frac{\ln(2)^3}{6}-\frac{\pi^2\ln(2)}{12}+\frac{7}{8}\zeta(3)$$ $$\sum_{n=1}^\infty \frac{H_{n}}{n^3}=\frac{\pi^4}{72}$$
Aug
22
comment Show that there are infinitely many integers $n$ with a given number of divisors
Then it would not be dependent on $n$, so it would be no. Not yes.
Aug
22
comment Show that there are infinitely many integers $n$ with a given number of divisors
What is $k$? Is it dependent on $n$?
Aug
21
comment Set of points of $[0,1)$ that have a unique binary expansion
1.) Is false for example take the equivalence: $$0.1=0.01111...$$ Because note that: $$\frac{1}{2}=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\cdots$$
Aug
21
revised Sum of $1+\frac{1}{2}+\frac{1\cdot2}{2\cdot5}+\frac{1\cdot 2\cdot 3}{2\cdot 5\cdot 8}+\cdots$
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Aug
21
comment Sum of $1+\frac{1}{2}+\frac{1\cdot2}{2\cdot5}+\frac{1\cdot 2\cdot 3}{2\cdot 5\cdot 8}+\cdots$
@JackD'Aurizio That can be simplified.
Aug
21
comment Sum of $1+\frac{1}{2}+\frac{1\cdot2}{2\cdot5}+\frac{1\cdot 2\cdot 3}{2\cdot 5\cdot 8}+\cdots$
Use hypergeometric functions. The whole thing can be written in terms of trigonometric functions, and logarithms.
Aug
16
comment “Real world” applications of rational functions
matheducators.stackexchange.com ?
Aug
8
answered How to compute $\lim_{n \to \infty} (2^n + 3^n)^{\frac{1}{n}}$?
Aug
8
comment If $~a^3 + b^3 = c^3~$ has nonzero integer solutions and $~c-b~$ is a cubic number and $~c-b \neq 1$
Not really that related but the case for exponent $4$ can be proved in a pretty elementary manner using Fermat's method of descent. Anyway though, from what I know about Will Jagy if he says there probably isn't an elementary proof then I wouldn't hold out too much hope on finding one.