Euge said:Here is my solution.
Note
$$a_n = (n-1)^{2/3} + (n + 1)^{2/3} + (n-1)^{1/3}(n+1)^{1/3} = \frac{(n+1) - (n-1)}{\sqrt[3]{n+1} - \sqrt[3]{n-1}} = \frac{2}{\sqrt[3]{n+1} - \sqrt[3]{n-1}}.$$
Therefore
$$\sum_{n = 1}^{999} \frac{1}{a_n} = \frac{1}{2}\sum_{n = 1}^{999} (\sqrt[3]{n+1} - \sqrt[3]{n-1}) = \frac{1}{2}\sum_{n = 1}^{999} [(\sqrt[3]{n+1} - \sqrt[3]{n}) + (\sqrt[3]{n} - \sqrt[3]{n-1})]$$
$$ = \frac{1}{2}[(\sqrt[3]{1000} - \sqrt[3]{1}) + (\sqrt[3]{999} - \sqrt[3]{0})] = \frac{9 + 3\sqrt[3]{37}}{2}.$$
Euge said:Hi anemone,
What do you mean when you say that $a_n$ is a geometric series? There isn't a common ratio between consecutive terms.
The formula for summing a series of cubic roots is ∑(n=1 to k)∛(an), where n represents the index and k represents the number of terms in the series.
The purpose of summing a series of cubic roots is to find the total value of a sequence of numbers raised to the third power, or cubic roots. This can be useful in various mathematical and scientific calculations.
The main difference between summing a series of cubic roots and summing a series of square roots is that cubic roots are raised to the third power, while square roots are raised to the second power. This means that cubic roots are generally larger numbers and will result in a larger sum.
Summing a series of cubic roots can be applied in various fields such as engineering, physics, and finance. For example, it can be used to calculate the total force exerted on an object in a mechanical system or to find the total amount of energy required for a chemical reaction.
One common mistake to avoid when summing a series of cubic roots is to forget to include all the terms in the series. Another mistake is to mix up the order of operations, such as adding before taking the cubic root. It is also important to double-check calculations and use a calculator if necessary to avoid any errors.