Find the sum(1/4)+(4/8)+(8/12)+

  • Thread starter hadi amiri 4
  • Start date
In summary, the OP has shown no work towards the problem and it is unclear if the problem will converge to something nice.
  • #1
hadi amiri 4
98
1
find the sum
(1/4!)+(4!/8!)+(8!/12!)+...
 
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  • #2


That is clearly divergent. Try to simplify it and you will see.
 
  • #3


daudaudaudau said:
That is clearly divergent. Try to simplify it and you will see.
Not so!

An upper bound would be 1 + 1/5^4 + 1/9^4 + 1/13^4 + ... which converges.
 
  • #4


mathman said:
Not so!

An upper bound would be 1 + 1/5^4 + 1/9^4 + 1/13^4 + ... which converges.

Regardless, the OP has shown no work towards the problem.
 
  • #5


I can't see the trick to solve it offhand. We know it converges, but do you know for sure it converges to something nice? This seems like a somewhat-hard problem because the number 4 is actually important.

[tex] \sum _{n=0}^\infty \frac{(k n)!}{( k (n+1) )!}[/tex]

See if k is 1 then it will be the harmonic series which diverges.
 
  • #6


(1/4!)+(4!/8!)+(8!/12!)+(12!/16!)+...
 
  • #7


the answer contains Pi and Ln .
 
  • #8


hadi amiri 4 said:
the answer contains Pi and Ln .

Yes, according to Mathematica the answer is
[tex]
\frac{1}{24}(6\log2-\pi)\approx 0.0423871
[/tex]
Did you solve this manually?
 
  • #9


(1/4!)+(1/8*7*6*5)+(1/12*11*10*9)+...
we can guess the general sentence .after that use definite integrals.
thats all.
 
  • #10


[tex] \sum _{k=0}^\infty \frac{1}{(4k+1)(4k+2)(4k+3)(4k+4)}[/tex]
 
  • #11


hadi amiri 4 said:
[tex] \sum _{k=0}^\infty \frac{1}{(4k+1)(4k+2)(4k+3)(4k+4)}[/tex]

Yes, and using
[tex]
a^{-1}=\int_0^\infty{dxe^{-ax}}
[/tex]
this can be written as
[tex]
\sum_{k=0}^\infty{\int_0^\infty{dx_1e^{-x_1(4k+1)}}\int_0^\infty{dx_2e^{-x_2(4k+2)}}\int_0^\infty{dx_3e^{-x_3(4k+3)}}\int_0^\infty{dx_4e^{-x_4(4k+4)}}}=\int_{[0,\infty)^4}{d^4(x_1,x_2,x_3,x_4)\frac{e^{-x_1-2x_2-3x_3-4x_4}}{1-e^{-4(x_1+x_2+x_3+x_4)}}}
[/tex]
Letting [itex]y_n=e^{-nx_n}[/itex] the last expression becomes
[tex]
\frac{1}{24}\int_{[0,1]^4}{d^4(y_1,y_2,y_3,y_4)}\frac{1}{1-y_1^4y_2^2y_3^{3/4}y_4}
[/tex]
These integrals can be solved explicitly but it's not nice. Did you do it this way?
 
Last edited:
  • #12


[tex]\sum_{k=0}^infty \frac{1} {(4k+1)(4k+2)(4k+3)(4k+4) [/tex]
is equal to
[tex]\sum(\frac{1} {6(4k+1)} - \frac{1} {2(2k+1)} + \frac{1} {2(4k+3)} - \frac{1} {6(4k+1)} )[/tex]
and use integrals
 
  • #13


[tex]\sum(\frac{1} {6(4k+1)} - \frac{1} {2(4k+1)} + \frac{1} {2(4k+3)} - \frac{1} {6(4k+1)} )[/tex]
 

1. What is the sum of the given fractions?

The sum of the given fractions is 17/12 or 1.4166 when rounded to four decimal places.

2. How do you find the sum of fractions with different denominators?

To find the sum of fractions with different denominators, you need to first find a common denominator. This can be done by finding the least common multiple (LCM) of the denominators. Then, convert each fraction to an equivalent fraction with the common denominator and add them together. In this case, the LCM of 4, 8, and 12 is 24, so we convert 1/4 to 6/24, 4/8 to 12/24, and 8/12 to 16/24. The sum is then 34/24 which simplifies to 17/12.

3. Can the sum of fractions with different denominators be simplified?

Yes, the sum of fractions with different denominators can be simplified if possible. In this case, we can simplify 17/12 to 1.4166 by dividing both the numerator and denominator by their greatest common factor, which is 1 in this case.

4. Is there a formula for finding the sum of fractions with different denominators?

Yes, there is a formula for finding the sum of fractions with different denominators. It is: Sum = (a/b) + (c/d) + (e/f) + ... = (ad + bc + ef + ...) / (bd * df * ...)

5. Can you find the sum of fractions without converting them to equivalent fractions?

Yes, it is possible to find the sum of fractions without converting them to equivalent fractions. This can be done by using the formula mentioned in the previous answer. However, converting them to equivalent fractions makes the calculation easier and reduces the chance of error.

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