Question

The number sequence
\[a _{0},a _{1},a2...\]
is defined by the recursion formula:
\[a _{1}=1\]
\[a _{n+1}=a _{n}−\frac{ 1 }{ n(n+1) },n≥1\]
Find the closed formula for the sequence and prove it by induction.

Answer 1

we will get a0 as 0 if we put n =1

Answer 2

ok this is what we do when we have just \(a_{n+1}\) and \(a_n\) in recursion formula
\[a _{n+1}-a _{n}=\frac{ 1 }{ n(n+1) }\]\[a _{n}-a _{n-1}=\frac{ 1 }{ n(n-1) }\]\[...\]\[a_2-a_1=\frac{1}{2}\]and add them all

Answer 3

sorry cant find a0

Answer 4

what mukushala has done is correct then you will get a(n+1) = 1 - sum of terms of rhs

Answer 5

for finding the sum of terms on rhs write each term like this\[\frac{ 1 }{(n)(n-1) } = \frac{ n - (n-1) }{(n)(n-1) } = \frac{ 1 }{ n-1 } - \frac{ 1 }{ n }\]

Answer 6

now when you add consecutive terms will get cancelled and you will be left with \[1 - \frac{ 1 }{ n+1 }\]

Answer 7

did you understand frx?

Answer 8

Well, not really yet but thank you, will have to sit down and think about what you guys just wrote:)

Answer 9

I can't see what the closed formula is? Is it \[1-\frac{ 1 }{ n+1}\]?

Answer 10

another way to seeing this is writing it in the form\[a _{n+1}-a _{n}=-\frac{1}{n(n+1)}=\frac{ 1 }{ n+1 }-\frac{1}{n}\]and guessing closed form like \[a_n=\frac{1}{n}\]then prove it by induction because it says find the closed formula(even by guessing) then prove by induction

Answer 11

yeah what mukushla is saying is better i think

Answer 12

I really don't get it :/ I get the first part when subtractin a_n from both sides but is 1/(n+1)-1/n equal to the first ecpression, I can't see it

Answer 13

\[\frac{ 1 }{ n+1 }-\frac{ 1 }{ n }\]

Looking at this expression, why should i guess the closed form is ?\[a _{n}=\frac{ 1 }{ n }\]