+249 Compute the sum \(\frac{2}{1 \cdot 2 \cdot 3} + \frac{2}{2 \cdot 3 \cdot 4} + \frac{2}{3 \cdot 4 \cdot 5} + \cdots\)
Forms geometric series as follows:
∑2/[n*(n+1)*(n+2)], n=1 to infinity
Sum = F / (1 - r)
Sum = 1/3 / (1 - 1/3)
Sum = 1/3 / (2/3)
Sum= 1/2
+36915 I was looking at this question earlier, but I could not figure out an 'r' common ratio..... you say it is 1/3....can you explain how you got that? I do not see it....... Thanx !
By using the "Partial Sum Formula", you can easily see that it is converging to 1/2. Since this an infinite series with a sum of 1/2 and 1st.term 1/3, therefore the common ratio will be 1/3:
sum_(n=1)^m 2/(n (n+1) (n+2)) = (m^2+3 m)/(2 (m+1) (m+2))
P.S. You may try the 1st. 1,000,000 terms and you will get:0.499999999999000002999993000015....etc.
+36915 But I still do not see a common ratio of 1/3...
a geometric series may simply be written as
a + a r + a r 2 + a r 3 + ⋯
and 1/3 does not fit the bill.......how did you get 1/3??
No, you cannot see it from the series as written. You can only derive it, indirectly, from infinite deries formula: s=a1/[1 - r]. You can possibly see it by repeated subtractions of the denominators, such as:
3, 12, 30, 60, 105,168, 252, 360, 495, 660, 858
9, 18, 30, 45, 63, 84, 108, 135, 165, 198
9, 12, 15, 18, 21, 24, 27, 30, 33
3, 3, 3, 3, 3, 3, 3, 3............=1/3 common ratio!!!
