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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 at 2:33
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3 Answers 3

up vote 1 down vote accepted

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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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ \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}

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

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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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