Help with Sum ∑n!/(3*4*5...*n)

[RealKiller69]

Homework Statement

∑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 don't know how to continue, help please

 

[LCKurtz]

Homework Statement

∑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 don't know how to continue, help please

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?

 

[RealKiller69]

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?

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

 

[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?

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

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?

 

[RealKiller69]

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?

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
.
.
.
n=∞ Sn=1/3+2/12+3!/5*4*3...n!/(3*4*5*6*...*n)

 

[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.

 

[HallsofIvy]

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 $\frac{1}{2}+ \frac{1}{2}+ \frac{1}{2}+ \cdot\cdot\cdot$.

 

[LCKurtz]

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.

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 $\frac{1}{2}+ \frac{1}{2}+ \frac{1}{2}+ \cdot\cdot\cdot$.

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