## frx 2 years ago 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.

1. suvesh253

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

2. mukushla

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

3. suvesh253

sorry cant find a0

4. suvesh253

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

5. suvesh253

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

6. suvesh253

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

7. suvesh253

did you understand frx?

8. frx

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

9. frx

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

10. mukushla

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

11. suvesh253

yeah what mukushla is saying is better i think

12. frx

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

13. frx

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

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