Help with Sum ∑n!/(3*4*5...*n)

In summary, the given expression is a sum of terms that involve factorials and a denominator that starts at 3 and increases up to n. However, this does not make sense for n=1 or n=2. If the denominator had an additional factor of 2, the resulting expression would simply be a sum of 1/2, which can be simplified to n/2.
  • #1
RealKiller69
11
1

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
 
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  • #2
RealKiller69 said:

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?
 
  • #3
LCKurtz said:
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 "
 
  • #4
LCKurtz said:
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 said:
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?
 
  • #5
LCKurtz said:
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)
 
  • #6
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.
 
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  • #7
You appear to be copying things that you really do not understand. You say that the summand is [itex]\frac{n!}{3*4*\cdot\cdot\cdot n}[/itex] but that \certainly implies that n is at least 5: [itex]\frac{5!}{3*4*5}+ \frac{6!}{3*4*5*6}+ \cdot\cdot\cdot[/itex].

Of course, n! means n(n-1)(n- 2)...(3)(2)(1) so that, for any n [itex]\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}[/itex] so, if I am interpreting this correctly this is just the sum [itex]\frac{1}{2}+ \frac{1}{2}+ \frac{1}{2}+ \cdot\cdot\cdot[/itex].
 
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  • #8
LCKurtz said:
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 said:
You appear to be copying things that you really do not understand. You say that the summand is [itex]\frac{n!}{3*4*\cdot\cdot\cdot n}[/itex] but that \certainly implies that n is at least 5: [itex]\frac{5!}{3*4*5}+ \frac{6!}{3*4*5*6}+ \cdot\cdot\cdot[/itex].

Of course, n! means n(n-1)(n- 2)...(3)(2)(1) so that, for any n [itex]\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}[/itex] so, if I am interpreting this correctly this is just the sum [itex]\frac{1}{2}+ \frac{1}{2}+ \frac{1}{2}+ \cdot\cdot\cdot[/itex].

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

FAQ: Help with Sum ∑n!/(3*4*5...*n)

1. What is the formula for calculating Sum ∑n!/(3*4*5...*n)?

The formula for calculating Sum ∑n!/(3*4*5...*n) is n!/3, where n is the number of terms in the series.

2. How do I simplify Sum ∑n!/(3*4*5...*n)?

To simplify Sum ∑n!/(3*4*5...*n), you can first cancel out any common factors in the numerator and denominator. Then, you can use the formula for calculating the factorial of a number to simplify the remaining terms.

3. Can I use a calculator to find the value of Sum ∑n!/(3*4*5...*n)?

Yes, you can use a calculator to find the value of Sum ∑n!/(3*4*5...*n). Simply input the value of n and use the factorial function on your calculator to calculate n!. Then, divide n! by 3 to get the final result.

4. How does the value of n affect the value of Sum ∑n!/(3*4*5...*n)?

The value of n directly affects the value of Sum ∑n!/(3*4*5...*n). As n increases, the value of n! also increases, resulting in a higher value for the sum. On the other hand, as n decreases, the value of n! decreases, resulting in a lower value for the sum.

5. Can you provide an example of how to solve Sum ∑n!/(3*4*5...*n)?

As an example, let's say we want to find the value of Sum ∑n!/(3*4*5...*n) when n=5. First, we calculate 5! = 120. Then, we divide 120 by 3 to get the final result of 40. Therefore, Sum ∑n!/(3*4*5...*n) when n=5 is equal to 40.

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