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

[hadi amiri 4]

find the sum
(1/4!)+(4!/8!)+(8!/12!)+...

 

[daudaudaudau]

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

 

[mathman]

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.

 

[Tac-Tics]

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.

 

[maze]

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.

$$\sum _{n=0}^\infty \frac{(k n)!}{( k (n+1) )!}$$

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

 

[hadi amiri 4]

(1/4!)+(4!/8!)+(8!/12!)+(12!/16!)+...

 

[hadi amiri 4]

the answer contains Pi and Ln .

 

[Pere Callahan]

the answer contains Pi and Ln .

Yes, according to Mathematica the answer is
$$\frac{1}{24}(6\log2-\pi)\approx 0.0423871$$
Did you solve this manually?

 

[hadi amiri 4]

(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.

 

[hadi amiri 4]

$$\sum _{k=0}^\infty \frac{1}{(4k+1)(4k+2)(4k+3)(4k+4)}$$

 

[Pere Callahan]

$$\sum _{k=0}^\infty \frac{1}{(4k+1)(4k+2)(4k+3)(4k+4)}$$

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

 

[hadi amiri 4]

$$\sum_{k=0}^infty \frac{1} {(4k+1)(4k+2)(4k+3)(4k+4)$$
is equal to
$$\sum(\frac{1} {6(4k+1)} - \frac{1} {2(2k+1)} + \frac{1} {2(4k+3)} - \frac{1} {6(4k+1)} )$$
and use integrals

 

[hadi amiri 4]

$$\sum(\frac{1} {6(4k+1)} - \frac{1} {2(4k+1)} + \frac{1} {2(4k+3)} - \frac{1} {6(4k+1)} )$$