Find a formula for the nth partial sum of the series and use it to find the series' sum if the series converges. [5/12]+[5/23]+[5/3*4]+...+[5/n(n+1)]+...

Question

asked Oct 11, 2020 226k views

1 Answer

Pieter Herroelen · Answer 1 · 2020-10-16T21:57:41+0000

Answer:
S_n=5(1-(1)/(n+1)) ; 5

Explanation:

Given series :
$[(5)/(1\cdot2)]+[(5)/(2\cdot3)]+[(5)/(3\cdot4)]+....+[(5)/(n\cdot(n+1))]$

Sum of series =
$S_n=\sum^(\infty)_(1)\ [(5)/(n\cdot(n+1))]=5[\sum^(\infty)_(1)(1)/(n\cdot(n+1))]$

Consider
$(1)/(n\cdot(n+1))=(n+1-n)/(n(n+1))$

=(1)/(n)-(1)/(n+1)

⇒
$S_n=5\sum^(\infty)_(1)(1)/(n\cdot(n+1))=5\sum^(\infty)_(1)[(1)/(n)-(1)/(n+1)]$

Put values of n= 1,2,3,4,5,.....n

⇒
S_n=5((1)/(1)-(1)/(2)+(1)/(2)-(1)/(3)+(1)/(3)-(1)/(4)+......-(1)/(n)+(1)/(n)-(1)/(n+1))

All terms get cancel but First and last terms left behind.

⇒
S_n=5(1-(1)/(n+1))

Formula for the nth partial sum of the series :

S_n=5(1-(1)/(n+1))

Also,
$\lim_(n \to \infty) S_n = 5(1-(1)/(n+1))$

$=5(1-(1)/(\infty))\\\\=5(1-0)=5$