# Does a_n = (n+1)^(1/3)-n^(1/3) converge?

Apr 10, 2017

As $n \to \infty$ the individual terms tend to $0$, but the sum tends to $\infty$.

#### Explanation:

Given:

${a}_{n} = {\left(n + 1\right)}^{\frac{1}{3}} - {n}^{\frac{1}{3}}$

Use the difference of cubes identity:

${a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$

with $a = {\left(n + 1\right)}^{\frac{1}{3}}$ and $b = {n}^{\frac{1}{3}}$ as follows:

${\lim}_{n \to \infty} {a}_{n} = {\lim}_{n \to \infty} \left({\left(n + 1\right)}^{\frac{1}{3}} - {n}^{\frac{1}{3}}\right)$

$\textcolor{w h i t e}{{\lim}_{n \to \infty} {a}_{n}} = {\lim}_{n \to \infty} \frac{\left({\left(n + 1\right)}^{\frac{1}{3}} - {n}^{\frac{1}{3}}\right) \left({\left(n + 1\right)}^{\frac{2}{3}} + {\left(n + 1\right)}^{\frac{1}{3}} {n}^{\frac{1}{3}} + {n}^{\frac{2}{3}}\right)}{{\left(n + 1\right)}^{\frac{2}{3}} + {\left(n + 1\right)}^{\frac{1}{3}} {n}^{\frac{1}{3}} + {n}^{\frac{2}{3}}}$

$\textcolor{w h i t e}{{\lim}_{n \to \infty} {a}_{n}} = {\lim}_{n \to \infty} \frac{\left(n + 1\right) - n}{{\left(n + 1\right)}^{\frac{2}{3}} + {\left(n + 1\right)}^{\frac{1}{3}} {n}^{\frac{1}{3}} + {n}^{\frac{2}{3}}}$

$\textcolor{w h i t e}{{\lim}_{n \to \infty} {a}_{n}} = {\lim}_{n \to \infty} \frac{1}{{\left(n + 1\right)}^{\frac{2}{3}} + {\left(n + 1\right)}^{\frac{1}{3}} {n}^{\frac{1}{3}} + {n}^{\frac{2}{3}}}$

$\textcolor{w h i t e}{{\lim}_{n \to \infty} {a}_{n}} = 0$

We also find:

${\sum}_{n = 1}^{N} {a}_{n} = {\sum}_{n = 1}^{N} \left({\left(n + 1\right)}^{\frac{1}{3}} - {n}^{\frac{1}{3}}\right)$

$\textcolor{w h i t e}{{\sum}_{n = 1}^{N} {a}_{n}} = {\sum}_{n = 1}^{N} {\left(n + 1\right)}^{\frac{1}{3}} - {\sum}_{n = 1}^{N} {n}^{\frac{1}{3}}$

$\textcolor{w h i t e}{{\sum}_{n = 1}^{N} {a}_{n}} = {\sum}_{n = 2}^{N + 1} {n}^{\frac{1}{3}} - {\sum}_{n = 1}^{N} {n}^{\frac{1}{3}}$

$\textcolor{w h i t e}{{\sum}_{n = 1}^{N} {a}_{n}} = {\left(N + 1\right)}^{\frac{1}{3}} + \textcolor{red}{\cancel{\textcolor{b l a c k}{{\sum}_{n = 2}^{N} {n}^{\frac{1}{3}}}}} - \textcolor{red}{\cancel{\textcolor{b l a c k}{{\sum}_{n = 2}^{N} {n}^{\frac{1}{3}}}}} - {1}^{\frac{1}{3}}$

$\textcolor{w h i t e}{{\sum}_{n = 1}^{N} {a}_{n}} = {\left(N + 1\right)}^{\frac{1}{3}} - 1$

Hence:

${\sum}_{n = 1}^{\infty} {a}_{n} = {\lim}_{N \to \infty} \left({\left(N + 1\right)}^{\frac{1}{3}} - 1\right) = \infty$