# How do you determine if a_n=(1-1/8)+(1/8-1/27)+(1/27-1/64)+...+(1/n^3-1/(n+1)^3)+... converge and find the sums when they exist?

Jan 29, 2018

The sum of the first $n$ terms is given by:

${S}_{n} = 1 - \frac{1}{n + 1} ^ 3$

The sum to infinity is given by:

$S = 1$

#### Explanation:

Let us denote the ${n}^{t h}$ term of the sequence by:

${u}_{n} = \frac{1}{n} ^ 3 - \frac{1}{n + 1} ^ 3$

Then we can denote the partial finite sum by ${S}_{n}$ so that:

${S}_{n} = {\sum}_{r = 1}^{n} \setminus {u}_{r}$
$\setminus \setminus \setminus \setminus = \left(1 - \frac{1}{8}\right) + \left(\frac{1}{8} - \frac{1}{27}\right) + \ldots + \left(\frac{1}{n} ^ 3 - \frac{1}{n + 1} ^ 3\right)$

And then the infinite sum, by $S$ where:

$S = {\sum}_{r = 1}^{\infty} \setminus {u}_{r}$

First we can derive an expression for ${S}_{n}$, as we can write the sum of the first $n$ terms as follows:

${S}_{n} = {u}_{1} + {u}_{2} + \ldots {u}_{n}$

$\setminus \setminus \setminus \setminus \setminus = \left\{1 - \frac{1}{8}\right\} +$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left\{\frac{1}{8} - \frac{1}{27}\right\} +$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left\{\frac{1}{27} - \frac{1}{64}\right\} +$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \vdots$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left\{\frac{1}{n} ^ 3 - \frac{1}{n + 1} ^ 3\right\}$

This is a difference sum, and we can see that almost all terms will cancel with other terms:

${S}_{n} = \left\{1 - \cancel{\textcolor{red}{\frac{1}{8}}}\right\} +$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left\{\cancel{\textcolor{red}{\frac{1}{8}}} - \cancel{\textcolor{b l u e}{\frac{1}{27}}}\right\} +$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left\{\cancel{\textcolor{b l u e}{\frac{1}{27}}} - \cancel{\frac{1}{64}}\right\} +$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \vdots$
 \ \ \ \ \ \ \ \ \ \ {(cancel(1/n^3)-1/(n+1)^3 }

After which we are left with:

${S}_{n} = 1 - \frac{1}{n + 1} ^ 3$

And for the sum, $S$, we have:

$S = {\lim}_{n \rightarrow \infty} {S}_{n}$
$\setminus \setminus = {\lim}_{n \rightarrow \infty} 1 - \frac{1}{n + 1} ^ 3$
$\setminus \setminus = 1 - 0$
$\setminus \setminus = 1$