Binomial Theorem Approximations/Examples/Arbitrary Example 1

Example of Binomial Theorem Approximation

$\paren {1 \cdotp 0 6}^{1/3} \approx 1 \cdotp 019613$

to $6$ decimal places.

Proof

Although this article appears correct, it's inelegant. There has to be a better way of doing it.
In particular: Appallingly non-rigorous. Verbatim from the book.
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$\ds \paren {1 \cdotp 0 6}^{1/3}$

$=$

$\ds \paren {1 + 0 \cdotp 06}^{1/3}$

$\ds $

$\approx$

$\ds 1 + \dfrac 1 3 \paren {0 \cdotp 06} + \dfrac {\paren {\frac 1 3} \paren {-\frac 2 3} } {2!} \paren {0 \cdotp 06}^2 + \dfrac {\paren {\frac 1 3} \paren {-\frac 2 3} \paren {-\frac 5 3} } {3!} \paren {0 \cdotp 06}^3 + \dfrac {\paren {\frac 1 3} \paren {-\frac 2 3} \paren {-\frac 5 3} \paren {-\frac 7 3} } {4!} \paren {0 \cdotp 06}^4$

$\ds $

$\approx$

$\ds 1 + 0 \cdotp 02 - 0 \cdotp 0004 + 0 \cdotp 000133 - 0 \cdotp 00000053$

$\ds $

$=$

$\ds 1 \cdotp 0196128$

$\ds $

$\approx$

$\ds 1 \cdotp 019613$

$\blacksquare$

Sources

1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text I$. Algebra: The Binomial Theorem: Approximations: Example $1$.