# Help with this sum

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1. Jan 13, 2016

### RealKiller69

1. The problem statement, all variables and given/known data

∑n!/(3*4*5...*n)

s1=1/3
sn=1/3+2/(4*3)+3!/(5*4*3)+...+n!/(3*4*5*...n)

so i multiplied the sum with 1/2sn=1/6+1/(4*3)+1/(5*4)+1/(6*5).....+1/((n+2)(n-1))

got blocked here,i dont know how to continue, help please

2. Jan 13, 2016

### LCKurtz

What are the indices on the sum? The denominator makes no sense when n = 1. If the denominator had an additional factor of 2, what would you have?

3. Jan 13, 2016

### RealKiller69

they are multiply symbols," * "=" x "

4. Jan 13, 2016

### LCKurtz

I knew that. But you didn't answer either of my questions. What is the range of the summation? What about n = 1? What about my last question?

5. Jan 13, 2016

### RealKiller69

the range is infinite
n=1 S1=1/3
n=2 S2=1/3+2/12
n=3 S3=1/3+2/12+3!/5*4*3
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.
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n=∞ Sn=1/3+2/12+3!/5*4*3...n!/(3*4*5*6*...*n)

6. Jan 13, 2016

### LCKurtz

You still aren't getting it. Your nth term is $\frac{n!}{3\cdot4\cdot...\cdot n}$. That denominator is 3 times 4 times... up to n. It starts at 3 and works up to n. It makes no sense if n=1 or n=2 because you can't start at 3 and work up to 1 or 2.

Also, you keep ignoring my last question. If there was a 2 in that denominator, how would it be different from the numerator? You can do some simplification.

7. Jan 13, 2016

### HallsofIvy

Staff Emeritus
You appear to be copying things that you really do not understand. You say that the summand is $\frac{n!}{3*4*\cdot\cdot\cdot n}$ but that \certainly implies that n is at least 5: $\frac{5!}{3*4*5}+ \frac{6!}{3*4*5*6}+ \cdot\cdot\cdot$.

Of course, n! means n(n-1)(n- 2)...(3)(2)(1) so that, for any n $\frac{n!}{3*4*\cdot\cdot\cdot*n}= \frac{1*2*3*4\cdot\cdot\cdot n}{3*4*5\cdot\cdot\cdot n}= \frac{1}{2}$ so, if I am interpreting this correctly this is just the sum $\frac{1}{2}+ \frac{1}{2}+ \frac{1}{2}+ \cdot\cdot\cdot$.

8. Jan 13, 2016

### LCKurtz

Or, you can just wait and HallsofIvy will simplify it for you.