Question

Let

A = 1/(1*3) + 1/(3*5) + 1/(5*7) + ...

and

B = 1/(2*4) + 1/(4*6) + 1/(6*8) + ...



Find A/B.



Hint: Consider the sum A + B.  Can you relate this sum to A or B?

Answer 1

AAAaannndd the bot gets it wrong yet again!
clearly, the term 1/1 does not appear in A nor in B.
Since A + B = ∞, A/B = 1/2 ??? Does not follow.

However, since A = 1/2 and B = 1/4, using well-known results,
A/B = 2

Answer 2

What utter total nonsense by the bot

rewriting the series:
A = 1*3 + 3*5 + 5*7 + .... + (2n-1)(2n+1)
= 1*3 + 3*5 + 5*7 + .... + 4n^2 - 1
using our well-knows formulas for the summation of infinite series
∑ 4n^2 - 1 from n = 1 to k = 4k(k+1)(2k+1)/6 - k
= (2/3)(2k^3 + 3k^2 + k) - k

In the same way,
B = 2*4 + 4*6 + 6*8 + ... + 2n(2n+2) or 4n^2 + 4n for the last term
∑ 4n^2 + 4n from n = 1 to k
= (2/3)(2k^3 + 3k^2 + k) + 4k(k+1)/2

if we now take LIMIT A/B as k ---> ∞
the leading highest term for both A and B is (4/3)k^3
so A/B = 1

furthermore as k ---> ∞ , A = B

confirmed with this simplistic computer program

10 a = 0 : b = 0
20 for n = 1 to 50000
30 a = 4*n*n-1+a
40 b = 4*n*n+4*n+b
70 next n
80 print a/b,a,b
>run
0.99997 1.666717E+14 1.666767E+14

Answer 3

Please ignore my post, it is as silly as the bot's
I misread the question, it was an interesting exercise nevertheless

Answer 4

I revised my computer program with a minor change necessary
and the new results confirm ooblecks answers

10 a = 0 : b = 0
20 for n = 1 to 50000
30 a = 1/(4*n*n-1)+a
40 b = 1/(4*n*n+4*n)+b
70 next n
80 print a/b,a,b
>run
2.00002 0.499995 0.249995