Question

Find a formula for the nth partial sum of the series and use it to find the series' sum if the series converges.

Answer 1

Answer: $S_n=5(1-\dfrac{1}{n+1})$ ; 5

Step-by-step explanation:

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

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

Consider $\dfrac{1}{n\cdot(n+1)}=\dfrac{n+1-n}{n(n+1)}$

$=\dfrac{1}{n}-\dfrac{1}{n+1}$

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

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

⇒ $S_n=5(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+......-\dfrac{1}{n}+\dfrac{1}{n}-\dfrac{1}{n+1})$

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

⇒ $S_n=5(1-\dfrac{1}{n+1})$

Formula for the nth partial sum of the series :

$S_n=5(1-\dfrac{1}{n+1})$

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

$=5(1-\dfrac{1}{\infty})\\\\=5(1-0)=5$