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.