Partial Sum & Sum of a Series?
Consider the series :
Sum from n=2, to infinity: (3)/(4n^2+8n+3)
(1) Find the the 18th-partial sum of the given series.
S18 = ?
(2) The sum of the given series is: ?
Hints:
-Find a general partial sum first and take limit.
-The Nth-partial sum of a series is the sum of the first N terms of the series.
The answers are S18 = 54/205 and sum= 3/10.
Any ideas on how to get to these answers?!
Comments
(1)
S18
= summation of 3 / (4n^2 + 8n + 3) from n = 2 to n = 19
= summation of (3/2)(1/(2n + 1) - 1/(2n + 3)) from n = 2 to n = 19
= (3/2) * summation of (1 / (2n + 1)) - (1 / (2n + 3)) from n = 2 to n = 19
= (3/2) * [(1/5) - (1/7) + (1/7) - (1/9) + (1/9) - (1/11) + .... + (1/39) - (1/41)]
= (3/2) * [(1/5) - (1/41)]
= (3/2) * (36/205)
= 54/205
(2)
Sum of given series
= summation of 3 / (4n^2 + 8n + 3) from n = 2 to infinity
= summation of (3/2)(1/(2n + 1) - 1/(2n + 3)) from n = 2 to n = infinity
= (3/2) * summation of (1 / (2n + 1)) - (1 / (2n + 3)) from n = 2 to n = infinity
= lim(k>infinity) (3/2) * summation of (1 / (2n + 1) - (1 / (2n + 3) from n = 2 to n = k
= lim(k->infinity) (3/2) * [(1/5) - (1 / (2k + 3))]
= (3/2) * [(1/5) - 0]
= (3/2) * (1/5)
= 3/10