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
 A: $\newcommand{\+}{^{\dagger}}
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\begin{align}
\color{#c00000}{u_{k + 1}}&=u_{k} + {1 \over \pars{k + 2}\pars{k + 3}}
=\color{#c00000}{u_{k} + {1 \over k + 2} - {1 \over k + 3}}
\\[3mm]&=\pars{u_{k - 1} + {1 \over k + 1} - {1 \over k + 2}} + {1 \over k + 2}
- {1 \over k + 3} = \color{#c00000}{u_{k - 1} + {1 \over k + 1} - {1 \over k + 3}}
\\[3mm]&=\pars{u_{k - 2} + {1 \over k} - {1 \over k + 1}} + {1 \over k + 1}
- {1 \over k + 3}= \color{#c00000}{u_{k - 2} + {1 \over k} - {1 \over k + 3}}=\cdots
\\[3mm]&=
\color{#c00000}{\overbrace{u_{1}}^{\ds{{2 \over 3}}} + {1 \over 3} - {1 \over k + 3}}
={k + 2 \over k + 3} 
\end{align}

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

A: 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 $\frac{1}{(k+2)(k+3)}=\frac{1}{k+2}-\frac{1}{k+3}$. 
A: 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 
$$\left\{\dfrac{k+1}{k+2}\right\}_{k=1}^\infty$$
