All Questions
21
questions
10
votes
0
answers
201
views
Finding $\frac{3}{1!+2!+3!}+\frac{4}{2!+3!+4!} +\frac{5}{3!+4!+5!}+\cdots+\frac{2008}{2006!+2007!+2008!}$ [duplicate]
How to find :
$$\frac{3}{1!+2!+3!}+\frac{4}{2!+3!+4!} +\frac{5}{3!+4!+5!}+\cdots+\frac{2008}{2006!+2007!+2008!}$$
- 101
6
votes
4
answers
828
views
Summing reciprocal logs of different bases
I recently took a math test that had the following problem:
$$
\frac{1}{\log_{2}50!} + \frac{1}{\log_{3}50!} + \frac{1}{\log_{4}50!} + \dots + \frac{1}{\log_{50}50!}
$$
The sum is equal to 1. I ...
- 171
6
votes
2
answers
144
views
How to show $\sum\limits_{r=0}^n \frac{1}{r!} \lt\left (1 + \frac{1}{n}\right)^{n+1}$ for all $n \ge 1$?
Using the binomial expansion, it is quite is easy to show that $$\left(1+\frac{1}{n}\right)^n \le \sum_{r=0}^{n} \frac{1}{r!} $$ for all $n\in\mathbb{Z^+}$, with equality holds when $n=1.$ (Can it be ...
5
votes
3
answers
2k
views
Digit in units place of $1!+2!+\cdots+99!$
There isn't much I can add to the question description to expand upon the title. I came across this in a multiple choice test. The options were $3$, $0$, $1$ and $7$. I am absolutely stumped. Any ...
- 8,867
5
votes
2
answers
91
views
If , $\frac{5}{7}=\frac{a_2}{2!}+\frac{a_3}{3!}+.........+\frac{a_7}{7!}$ . For , $a_i\in\mathbb Z^+$ & $a_i<i$ . Find $a_2+a_3+......+a_7$?
$Q.$ If , $$\frac{5}{7}=\frac{a_2}{2!}+\frac{a_3}{3!}+.........+\frac{a_7}{7!}$$ For , $a_i\in\mathbb Z^+$ & $a_i<i$ . Find $a_2+a_3+......+a_7$ ?
MY APPROACH :
We have , $$\frac{5}{7}=\frac{...
user904199
5
votes
1
answer
195
views
Summation of factorials.
How do I go about summing this :
$$\sum_{r=1}^{n}r\cdot (r+1)!$$
I know how to sum up $r\cdot r!$ But I am not able to do a similar thing with this.
- 1,567
3
votes
4
answers
288
views
Prove if $n \in \mathbb N$, then $\frac{1}{2!}+\cdots+\frac{n}{(n+1)!}=1-\frac{1}{(n+1)!}$ [duplicate]
Prove if $n \in \mathbb N$, then $\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots+\frac{n}{\left(n+1\right)!} = 1-\frac{1}{\left(n+1\right)!}$
So I proved the base case where $n=1$ and got $\frac{1}{2}$...
- 3,763
3
votes
1
answer
145
views
Find $\sum_{r=1}^{20} (-1)^r\frac{r^2+r+1}{r!}$.
Calculate $$\sum_{r=1}^{20} (-1)^r\frac{r^2+r+1}{r!}\,.$$
I broke the sum into partial fractions and after writing 3-4 terms of the sequence I could see that it cancels but I wasn't able to arrive at ...
- 1,107
3
votes
2
answers
690
views
How many values of $n$ are there for which $n!$ ends in $1998$ zeros?
How many values of $n$ are there for which $n!$ ends in $1998$ zeros?
My Attempt:
Number of zeros at end of $n!$ is
$$\left\lfloor \frac{n}{5}\right\rfloor+\left\lfloor\frac{n}{5^2}\right\rfloor+\...
- 9,599
3
votes
1
answer
223
views
Factorial Identity - True or False?
Let $x$ and $y$ be positive integers.
Then, is
\begin{align}
\frac{x^{xy}}{(xy)!} = \sum_{k_1+...+k_x = xy} \frac{1}{(k_1)!...(k_x)!}
\end{align}
true, where $k_1$, ..., $k_x$ are all positive ...
- 680
2
votes
1
answer
231
views
How to simplify $\sum_{k=1}^nk\cdot k!$ [duplicate]
How do I go about simplifying this:
$$\sum_{k=1}^nk \cdot k!$$
Wolfram alpha tells me it's the same as $(n+1)!-1$ but I don't see how.
- 1,303
2
votes
2
answers
380
views
I am stuck on proving $\frac1{2!}+\frac2{3!}+\dots+\frac{n}{(n+1)!}=1-\frac1{(n+1)!}$ by induction, could anyone check my work?
I will skip the Base Case step.
This is the questions.
Use mathematical induction to prove that$$\frac{1}{2!}+\frac{2}{3!}+\cdots+\frac{n}{(n+1)!}=1-\frac{1}{(n+1)!}$$for all integers $n\ge 1$.
...
user317568
2
votes
2
answers
95
views
Double sum factorial manipulation
$$\sum_{B = 0}^{n-1} \sum_{A = 0}^{n-B-1} \frac{(n-1)!}{B!(n-B-1)!} \frac{(n-B-1)!}{A!(n-B-A-1)!} \frac{A}{A+B+1}$$
This is driving me nuts! Is there anyway to reduce
$$\sum_{B = 0}^{n-1} \sum_{A = ...
- 23
1
vote
1
answer
664
views
Find the value of n if:
$$\sum_{k=0}^n (k^{2}+k+1) k! = (2007).2007!$$
How to approach this problem? In need of ideas. Thank you.
- 105
1
vote
2
answers
136
views
Question about Binomial Sums [duplicate]
Prove that for any $a \in \mathbb{R}$
$$\sum_{k=0}^n (-1)^{k}\binom{n}{k}(a-k)^{n}=n!$$
I rewrote the sum as
$$\sum_{k=0}^n \left((-1)^{k}\binom{n}{k} \sum_{i=0}^n (-1)^{i}a^{n-i} k^{i} \right)$$
...
- 5,719