Reputation
11,590
Top tag
Next privilege 15,000 Rep.
Protect questions
Badges
4 13 29
Newest
 Pundit
Impact
~210k people reached

21h
comment when $\lim_{n\to\infty }\sum_{k=0}^n\binom{n}{k}x^k$ exist?
your first approach is legitimate
2d
comment what is the max possible combinations of 1 2 3 4 5 6 without repeating
The number of ways to seat 6 people at the round table. By rotating each person you get the same seating. Hence $\frac{6!}{6}$.
2d
comment How to calculate $ \sum_{n=1}^{15}n(n!) = ? $
I'm practically devastated
2d
comment How to calculate $ \sum_{n=1}^{15}n(n!) = ? $
Use the property of factorial and cancel out the $S_n$ term
2d
comment How to calculate $ \sum_{n=1}^{15}n(n!) = ? $
$1+ 2! + \ldots n! + (n+1)! = \sum_{k=1}^{n} (k+1)! + 1$
2d
revised How to calculate $ \sum_{n=1}^{15}n(n!) = ? $
edited body
2d
comment How to calculate $ \sum_{n=1}^{15}n(n!) = ? $
1) Define $S_n$, 2) Add the next term on both sides, 3) Manipulate RHS with sm algebra, 4)Get the result
2d
comment How to calculate $ \sum_{n=1}^{15}n(n!) = ? $
explain what exactly?
2d
answered Big O notation: ratio of two $O(\cdot)$'s is $O(\cdot)$ of the ratio?
Feb
11
answered Given two real functions, $f$ and $g$, if $|f(x)|<1$ then $|g(f(x))|<g(1)$? Why?
Feb
10
revised Finding closed form expression for a multiple sum.
deleted 2 characters in body
Feb
10
comment Is this a special probability distribution?
This is called exponential distribution with parameter $\frac{1}{\theta}$
Feb
10
answered What is the probability that nobody receives the same ranking twice?
Feb
10
answered How to calculate $ \sum_{n=1}^{15}n(n!) = ? $
Feb
9
comment Proof $x$, $1+nx≤ (1+x)^n$
Did you mean $\binom{n}{k}$?
Feb
8
comment Solution to $\frac{d}{d\frac{1}{x}} x$
math.stackexchange.com/questions/21199/…
Feb
7
comment Proof based on definition of big-$O$
You can in fact show a stronger result: $n! = o(n^n)$
Feb
7
comment A closed form for the following Series
Sophomore's dream might help...
Feb
5
comment How to simplify this equation to solve for m?
$pm^2+cm -x = 0$ and use the solution to quadratic equation
Feb
4
revised Combinatorial argument for $\sum\limits_{k=i}^{n}\binom{n}{k}\binom{k}{i} = \binom{n}{i}2^{n-i}$
edited tags; edited title