Series

## Series
            ---
            ## Infinite series
            We want to generalise the concept of $\displaystyle \sum\_{k=1}^n a_k$ to give meaning to $\displaystyle \sum\_{k=1}^\infty a_k$.

Define by $\displaystyle \sum\_{k=1}^\infty a_k = \lim_{n\to\infty} \sum\_{k=1}^n a_k$.

So the *partial sums* $s_n = \displaystyle \sum\_{k=1}^n a_k$ are important, because $\displaystyle \sum\_{k=1}^\infty a_k = \lim_{n\to\infty} s_n$.

#### Definition
            An infinite series is *convergent*/*divergent* if $(s_n)\_{n=1}^\infty$ is *convergent*/*divergent*.
            #### Example
            $\displaystyle \sum\_{k=1}^\infty \frac1k$ is divergent, by earlier example.
            
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            ## Example of convergent infinite series
            In
            $\displaystyle \sum\_{k=1}^\infty \frac1{k(k+1)}$, the terms in the sum are $\displaystyle \frac1{k(k+1)} = \frac1k-\frac1{k+1}$. The partial sums are

$s_n = \left[ \frac11-\frac12 \right] + \left[ \frac12-\frac13 \right] + \ldots + \left[ \frac1n-\frac1{n+1} \right] = 1 - \frac1{n+1} \to 1$ as $n\to\infty$.

So $\displaystyle \sum\_{k=1}^\infty \frac1{k(k+1)}=1$.
            
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            ## Telescoping sum calculation
            $s_n = \left[ \frac11-\frac12 \right] + \left[ \frac12-\frac13 \right] + \ldots + \left[ \frac1n-\frac1{n+1} \right] = 1 - \frac1{n+1} \to 1$ as $n\to\infty$.

Here we used the cancellation
            \\[\begin{alignedat}{8}
                &\tfrac11 & &- \tfrac12 \\\\
                & & &+ \tfrac12 & & - \tfrac13 \\\\
                & & & & &+ \tfrac13 & & - \tfrac14 \\\\
                & & & & & & & & \hspace{1em}\vdots\hspace{1em} \\\\
                & & & & & & & & & &+ \tfrac1{n-1} & & - \tfrac1n \\\\
                & & & & & & & & & & & &+ \tfrac1{n} & & - \tfrac1{n+1} = \tfrac11 - \tfrac1{n+1}.
            \end{alignedat}\\]
            
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            ## Telescoping sum calculation
            Consider $\sum\_{k=1}^\infty \frac1{k(k+2)}=\frac12\sum\_{k=1}^\infty \frac2{k(k+2)}=\frac12\sum\_{k=1}^\infty \left[\frac1k-\frac1{k+2}\right]$.

Therefore,
            $2s_n = \left[ \frac11-\frac13 \right] + \left[ \frac12-\frac14 \right] + \ldots + \left[ \frac1n-\frac1{n+2} \right]$.
            
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            ## Telescoping sum calculation
            $2s_n = \left[ \frac11-\frac13 \right] + \left[ \frac12-\frac14 \right] + \ldots + \left[ \frac1n-\frac1{n+2} \right] = 1+\tfrac12 -\frac1{n+1} -\frac1{n+2} \to \frac32$ as $n\to\infty$.

Here we used the cancellation
            \\[\begin{alignedat}{14}
                &\tfrac11 & & & &- \tfrac13 \\\\
                & & &+ \tfrac12 & & & & - \tfrac14 \\\\
                & & & & &+ \tfrac13 & & & & - \tfrac15 \\\\
                & & & & & & &+ \tfrac14 & & & & - \tfrac16 \\\\
                & & & & & & & & & & & & \hspace{1em}\vdots\hspace{1em} \\\\
                & & & & & & & & & & & & & &+ \tfrac1{n-1} & & & & - \tfrac1{n+1} \\\\
                & & & & & & & & & & & & & & & &+ \tfrac1{n} & & & & - \tfrac1{n+2} = \tfrac11+\tfrac12 - \tfrac1{n+1} - \tfrac1{n+2}.
            \end{alignedat}\\]
            So $\sum\_{k=1}^\infty \frac1{k(k+2)}=\frac12\sum\_{k=1}^\infty \frac2{k(k+2)}=\frac12\sum\_{k=1}^\infty \left[\frac1k-\frac1{k+2}\right]=\frac12\times\frac32=\frac34$.