Finding explicit formula of this recursion formula

[Ace.]

Homework Statement

Write an explicit formula for the sequence determined by the following recursion formula.

t$_{1}$= 0; t$_{n}$ = t$_{n-1}$ + $\frac{2}{n(n+1)}$

The Attempt at a Solution

t$_{1}$ = 0

t$_{2}$ = t$_{1}$ + $\frac{2}{2(2+1)}$
t$_{2}$ = $\frac{1}{3}$

t$_{3}$ = t$_{2}$ + $\frac{2}{3(3+1)}$
t$_{3}$ = $\frac{1}{3}$ + $\frac{2}{3(3+1)}$
t$_{3}$ = $\frac{4}{12}$ + $\frac{2}{12)}$
t$_{3}$ = $\frac{1}{2}$

t$_{4}$ = t$_{3}$ + $\frac{2}{4(4+1)}$
t$_{4}$ = $\frac{1}{2}$ + $\frac{2}{20}$
t$_{4}$ = $\frac{3}{5}$My sequence is 0, $\frac{1}{3}$, $\frac{1}{2}$, $\frac{3}{5}$ $\cdots$

How do I make an explicit formula if there is no common difference nor a common ratio?

 

[Curious3141]

Homework Statement

Write an explicit formula for the sequence determined by the following recursion formula.

t$_{1}$= 0; t$_{n}$ = t$_{n-1}$ + $\frac{2}{n(n+1)}$

The Attempt at a Solution

t$_{1}$ = 0

t$_{2}$ = t$_{1}$ + $\frac{2}{2(2+1)}$
t$_{2}$ = $\frac{1}{3}$

t$_{3}$ = t$_{2}$ + $\frac{2}{3(3+1)}$
t$_{3}$ = $\frac{1}{3}$ + $\frac{2}{3(3+1)}$
t$_{3}$ = $\frac{4}{12}$ + $\frac{2}{12)}$
t$_{3}$ = $\frac{1}{2}$

t$_{4}$ = t$_{3}$ + $\frac{2}{4(4+1)}$
t$_{4}$ = $\frac{1}{2}$ + $\frac{2}{20}$
t$_{4}$ = $\frac{3}{5}$


My sequence is 0, $\frac{1}{3}$, $\frac{1}{2}$, $\frac{3}{5}$ $\cdots$

How do I make an explicit formula if there is no common difference nor a common ratio?

Do a partial fraction decomposition on $\frac{2}{n(n+1)}$. Let that be $\frac{A}{n} + \frac{B}{n+1}$ (you determine A and B).

Now $t_n = t_{n-1} + \frac{A}{n} + \frac{B}{n+1}$ and $t_{n-1} = t_{n-2} + \frac{A}{n-1} + \frac{B}{n}$.

Substitute the latter expression into the first and see what happens. Now continue successive substitution until you arrive at $t_1$.

 

[haruspex]

Here's another way. The form n(n+1)(n+2)... (n+r-1) in sums of series is strongly analogous to the form x^r in integration. So for Ʃ1/(n(n+1)) consider ∫dx/x². This gives you a guess for the sum of the series, which you can then refine by taking the difference of two consecutive terms and comparing it with the original.

 

[Ace.]

is that calculus? :$

 

[haruspex]

is that calculus? :$

Yes. I take it you've not done any integration yet.