MHB Summing a Series of Cubic Roots

[anemone]

Sum the series below:

$$\sum_{n=1}^{999}\dfrac{1}{a_n}$$ where $a_n=\sqrt[3]{n^2-2n+1}+\sqrt[3]{n^2+2n+1}+\sqrt[3]{n^2-1}$.

 

[Euge]

Here is my solution.

Spoiler

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

$$\sum_{n = 1}^{999} \frac{1}{a_n} = \frac{1}{2}\sum_{n = 1}^{999} (\sqrt[3]{n+1} - \sqrt[3]{n-1}) = \frac{1}{2}\sum_{n = 1}^{999} [(\sqrt[3]{n+1} - \sqrt[3]{n}) + (\sqrt[3]{n} - \sqrt[3]{n-1})]$$
$$ = \frac{1}{2}[(\sqrt[3]{1000} - \sqrt[3]{1}) + (\sqrt[3]{999} - \sqrt[3]{0})] = \frac{9 + 3\sqrt[3]{37}}{2}.$$

 

[anemone]

Here is my solution.

Spoiler

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

$$\sum_{n = 1}^{999} \frac{1}{a_n} = \frac{1}{2}\sum_{n = 1}^{999} (\sqrt[3]{n+1} - \sqrt[3]{n-1}) = \frac{1}{2}\sum_{n = 1}^{999} [(\sqrt[3]{n+1} - \sqrt[3]{n}) + (\sqrt[3]{n} - \sqrt[3]{n-1})]$$
$$ = \frac{1}{2}[(\sqrt[3]{1000} - \sqrt[3]{1}) + (\sqrt[3]{999} - \sqrt[3]{0})] = \frac{9 + 3\sqrt[3]{37}}{2}.$$

Thanks Euge for participating in this IMO Problem from China!

Spoiler

For me, it took me a fraction of time to realize $a_n$ is a geometric series that consists of the first three terms. :)

 

[Euge]

Hi anemone,

Spoiler

What do you mean when you say that $a_n$ is a geometric series? There isn't a common ratio between consecutive terms.

 

[anemone]

Hi anemone,

Spoiler

What do you mean when you say that $a_n$ is a geometric series? There isn't a common ratio between consecutive terms.

Hi Euge,

Spoiler

If I rewrite $a_n$ in such a way that it now becomes $a_n=\sqrt[3]{n^2+2n+1}+\sqrt[3]{n^2-1}+\sqrt[3]{n^2-2n+1}$, then I noticed that

$\dfrac{\sqrt[3]{n^2-1}}{\sqrt[3]{n^2+2n+1}}=\dfrac{\sqrt[3]{n^2-2n+1}}{\sqrt[3]{n^2-1}}$

Therefore, $\sqrt[3]{n^2+2n+1}+\sqrt[3]{n^2-1}+\sqrt[3]{n^2-2n+1}$ is a geometric series with:

• the first term $\sqrt[3]{n^2+2n+1}=(n+1)^{\frac{2}{3}}$ and
• the common ratio of $\left(\dfrac{n-1}{n+1}\right)^{\frac{1}{3}}$.

Therefore, $a_n=\dfrac{((n+1)^{\frac{2}{3}})\left(1-\left(\left(\dfrac{n-1}{n+1}\right)^{\frac{1}{3}}\right)^3\right)}{1-\left(\dfrac{n-1}{n+1}\right)^{\frac{1}{3}}}=\dfrac{2}{(n+1)^{\frac{1}{3}}-(n-1)^{\frac{1}{3}}}$

 

[Euge]

Hi anemone,

Spoiler

Thanks for clarifying. Now I understand what you mean. (Nod)