Guess the closed form on the following sequence?

any help would be appreciated, have no idea where to start

$u_1 = 2/3$ and $u_{k+1}$ such that:

$$u_k + \frac{1}{(k+2)(k+3)}$$ for all, k are natural numbers

guess a general formula (i.e the closed form) of the sequence

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Do you mean $u_{k+1}=u_k+\frac{1}{(k+2)(k+3)}$? – David H Apr 26 '14 at 2:33

If $u_{k+1}=u_k+\dfrac{1}{(k+2)(k+3)}$, then we have the following sequence $$\left\{\dfrac23,\dfrac23+\dfrac{1}{12},\dfrac23+\dfrac{1}{12}+\dfrac{1}{20},\dfrac23+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30},\dots\right\}=\left\{\dfrac23,\dfrac34,\dfrac45,\dfrac56,\dots\right\}$$ Then the general formula for $k\in\mathbb N$ would be $$\left\{\dfrac{k+1}{k+2}\right\}_{k=1}^\infty$$

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$$\color{#00f}{\large% u_{k} = {k + 1 \over k + 2}\,,\qquad k \geq 1}$$

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Hint: So $u_1=\frac{2}{3}$ and $u_2=\frac{2}{3}+\frac{1}{(3)(4)}$ and $u_3=\frac{2}{3}+\frac{1}{(3)(4)}+\frac{1}{(4)(5)}$ and so on.

Note that $\frac{1}{(3)(4)}=\frac{1}{3}-\frac{1}{4}$ and $\frac{1}{(4)(5)}=\frac{1}{4}-\frac{1}{5}$.

Do a couple more terms and notice the beautiful cancellations (telescoping). In general $\frac{1}{(k+2)(k+3)}=\frac{1}{k+2}-\frac{1}{k+3}$.

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