Answer: [tex]S_n=5(1-\dfrac{1}{n+1})[/tex] ; 5
Step-by-step explanation:
Given series : [tex][\dfrac{5}{1\cdot2}]+[\dfrac{5}{2\cdot3}]+[\dfrac{5}{3\cdot4}]+....+[\dfrac{5}{n\cdot(n+1)}][/tex]
Sum of series = [tex]S_n=\sum^{\infty}_{1}\ [\dfrac{5}{n\cdot(n+1)}]=5[\sum^{\infty}_{1}\dfrac{1}{n\cdot(n+1)}][/tex]
Consider [tex]\dfrac{1}{n\cdot(n+1)}=\dfrac{n+1-n}{n(n+1)}[/tex]
[tex]=\dfrac{1}{n}-\dfrac{1}{n+1}[/tex]
⇒ [tex]S_n=5\sum^{\infty}_{1}\dfrac{1}{n\cdot(n+1)}=5\sum^{\infty}_{1}[\dfrac{1}{n}-\dfrac{1}{n+1}][/tex]
Put values of n= 1,2,3,4,5,.....n
⇒ [tex]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})[/tex]
All terms get cancel but First and last terms left behind.
⇒ [tex]S_n=5(1-\dfrac{1}{n+1})[/tex]
Formula for the nth partial sum of the series :
[tex]S_n=5(1-\dfrac{1}{n+1})[/tex]
Also, [tex]\lim_{n \to \infty} S_n = 5(1-\dfrac{1}{n+1})[/tex]
[tex]=5(1-\dfrac{1}{\infty})\\\\=5(1-0)=5[/tex]
