Sum the series below:

[MATH]\sum_{n=4}^{999}\dfrac{1}{\sqrt[3]{n^2-2n+1}+\sqrt[3]{n^2+2n+1}+\sqrt[3]{n^2-1}}.[/MATH]

Question 1: Do you think the expression on the denominator can be factored?

Yes.

No.

This is a really rich way to asking problem so we make the students to think deep before simply giving out the answer yes or no so easily.

We might not be so experienced enough to tell if the expression on the denominator $(\sqrt[3]{n^2-2n+1}+\sqrt[3]{n^2+2n+1}+\sqrt[3]{n^2-1})$ can or cannot be factored.

But we can be smart enough to work with some figures, i.e., we could replace $n$ by $2$ into the expression and see if it will generate some insight/perspective from there. It is way more simpler to work with figures than with variables.

When $n=2$, we have:

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

$=\sqrt[3]{2^2-2(2)+1}+\sqrt[3]{2^2+2(2)+1}+\sqrt[3]{2^2-1}$

$=\sqrt[3]{1}+\sqrt[3]{9}+\sqrt[3]{3}$

$=\sqrt[3]{1}+\sqrt[3]{3}+\sqrt[3]{9}$

Okay, we're asked in fact if we could factor the expression $\sqrt[3]{1}+\sqrt[3]{3}+\sqrt[3]{9}$.

Nope, there is no common term here but we do smell a rat, we need to substitute $n$ by $3$ this one time to check:

When $n=3$, we have:

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

$=\sqrt[3]{3^2-2(3)+1}+\sqrt[3]{3^2+2(3)+1}+\sqrt[3]{3^2-1}$

$=\sqrt[3]{4}+\sqrt[3]{16}+\sqrt[3]{8}$

$=\sqrt[3]{4}+\sqrt[3]{8}+\sqrt[3]{16}$

$=2^{\frac{2}{3}}+2^{\frac{3}{3}}+2^{\frac{4}{3}}$

Hey, this time we see the common factor, and that is $2^{\frac{2}{3}}$.

Therefore, we can rewrite

$\begin{align*}\sqrt[3]{3^2-2(3)+1}+\sqrt[3]{3^2+2(3)+1}+\sqrt[3]{3^2-1}&=2^{\frac{2}{3}}+2^{\frac{3}{3}}+2^{\frac{4}{3}}\\&=2^{\frac{2}{3}}(1+2^{\frac{1}{3}}+2^{\frac{2}{3}})\end{align*}$

Hence the answer for this problem is a yes.

Something that I must mention it here is that I don't see much help after we gotten the factored form of the general term for the series. This is very important, if our first step doesn't help much, then we need to think of other productive solution and ditch the current one for good.

Question 2: If we're asked to rewrite the general term of the series, which of the following concept would we use next?

A. Clearing the Fraction

B. Rationalizing the Denominator

C. Partial Fraction

Answer:

Previously, we have been successfully to factor the denominator's expression when we took $n=3$ so that we have:

$\begin{align*}\dfrac{1}{\sqrt[3]{3^2-2(3)+1}+\sqrt[3]{3^2+2(3)+1}+\sqrt[3]{3^2-1}}&=\dfrac{1}{2^{\frac{2}{3}}+2^{\frac{3}{3}}+2^{\frac{4}{3}}}\\&=\dfrac{1}{2^{\frac{2}{3}}(1+2^{\frac{1}{3}}+2^{\frac{2}{3}})}\end{align*}$

Can you tell by now that we are actually dealing with a geometric series here?

$\Large 2^{\frac{2}{3}}+2^{\frac{3}{3}}+2^{\frac{4}{3}}=2^{\frac{2}{3}}\overbrace{+}^{\times 2^{\frac{1}{3}}}2^{\frac{3}{3}}\overbrace{+}^{\times 2^{\frac{1}{3}}}2^{\frac{4}{3}}$

Looking back to work with the given expression in $n$, we see that we need to rearrange the term such that they are arranged in the correct order so it could be deemed to be a geometric series:

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

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

To check if we have it written correctly, we check if they have the same common ratio:

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

$\dfrac{n^2-1}{n^2+2n+1}\overbrace{=}^{?}\dfrac{n^2-2n+1}{n^2-1}$

$\dfrac{(n+1)(n-1)}{(n+1)^2}\overbrace{=}^{?}\dfrac{(n-1)^2}{(n+1)(n-1)}$

Simplifying and cross multiplying to get:

$\dfrac{(n-1)}{(n+1)}\overbrace{=}^{?}\dfrac{(n-1)}{(n+1)}$

$(n-1)(n+1)=(n-1)(n+1)$

So we have written the terms in the correct order.

Note that the first term $(a)$ and the common ratio $(r)$ are $\sqrt[3]{n^2+2n+1}$ and $\left(\dfrac{n-1}{n+1}\right)^{\frac{1}{3}}$ respectively and also, $\sqrt[3]{n^2+2n+1}+\sqrt[3]{n^2-1}+\sqrt[3]{n^2-2n+1}$ is the sum of the first three terms which can be summarized using the sum formula for the geometric series where $S_n=\dfrac{a(1-r^n)}{1-r}$, since $r=\left(\dfrac{n-1}{n+1}\right)^{\frac{1}{3}}=\left(1-\dfrac{2}{n+1}\right)^{\frac{1}{3}}\lt1$.

Hence,

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

$=\dfrac{a(1-r^3)}{1-r}$

$=\dfrac{\sqrt[3]{n^2+2n+1}(1-\left(\left(\dfrac{n-1}{n+1}\right)^{\frac{1}{3}}\right)^3}{1-\left(\dfrac{n-1}{n+1}\right)^{\frac{1}{3}}}$

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

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

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

$=\dfrac{2(n+1)}{(n+1)((n+1)^{\frac{1}{3}}-(n-1)^{\frac{1}{3}})}$

$=\dfrac{2}{(n+1)^{\frac{1}{3}}-(n-1)^{\frac{1}{3}}}$

Therefore the original general term $\dfrac{1}{\sqrt[3]{n^2-2n+1}+\sqrt[3]{n^2+2n+1}+\sqrt[3]{n^2-1}}$ becomes:

$\dfrac{1}{\dfrac{2}{(n+1)^{\frac{1}{3}}-(n-1)^{\frac{1}{3}}}}=\dfrac{(n+1)^{\frac{1}{3}}-(n-1)^{\frac{1}{3}}}{2}$

And we are asked to find

[MATH]\begin{align*}\sum_{n=4}^{999}\dfrac{1}{\sqrt[3]{n^2-2n+1}+\sqrt[3]{n^2+2n+1}+\sqrt[3]{n^2-1}}&=\sum_{n=4}^{999} \dfrac{(n+1)^{\frac{1}{3}}-(n-1)^{\frac{1}{3}}}{2}\\&=\sum_{n=4}^{999}\dfrac{1}{2}((n+1)^{\frac{1}{3}}-(n-1)^{\frac{1}{3}})\end{align*}[/MATH]

Phew...after working with all this, we can be sure that what we are doing is to rationalizing the denominator.

The answer, is therefore B.

Question 3: Do you think the trigonometric substitution for the variable $x$ would be useful in finding the sum for this problem?

Yes.

No.

Answer:

There is no way that trigonometric could be of use in this particular problem. In fact, there is rarely any connection between trigonometric substitution and sum of series sort of problem.

Therefore, the answer is a No.

Question 4: Will you suspect that there is the possibility that this series telescopes?

No, there is no way this series telescopes.

This is definitely a telescoping series.

Answer:

More often than not, questions of this sort that come out in the International Mathematical Olympiad contests are to test our ability to recognize and hence prove the series telescopes. It makes sense because in any competition, some of the problems are designed in such a way that there is a smart solution for each of them so that would save students lots of precious time to handle the more intriguing problems.

Therefore, the answer is B.

Question 5: If this is a telescoping sum, would you consider to let the sum as S, and then cube both sides of the equation in your next step?

Yes.

No.

Answer:

This problem aims at checking if we are familiar with the properties of the formula that involves finite sum.

Given

[MATH]S=\sum_{n=4}^{999}\dfrac{1}{2}((n+1)^{\frac{1}{3}}-(n-1)^{\frac{1}{3}})[/MATH] and

[MATH]S^3\ne \sum_{n=4}^{999} \left(\dfrac{1}{2}((n+1)^{\frac{1}{3}}-(n-1)^{\frac{1}{3}})\right)^3[/MATH]

Therefore we know cubing both sides of the equation is meaningless and the answer is hence the definite No.

Question 6: If this is a telescoping sum that the general term of the series can take the form such as f(n)-f(n-1), what would you do to find that second form to represent the general term?

A. I don't think this is a doable question.

B. I would try working with the first few terms by replacing n with $2$, $3$ or any other numbers to see where that leads me.

Answer:

Like what I have already done in question 3, you can make an educated guess that if the expression written in variable doesn't give us enough hint what to do in our best next step, then we could try to work with numbers rather than variables, as it often gives us another way of looking at the problem.

The answer is therefore B.

Question 7: If this is a telescoping sum, does there exist a closed form expression for this sum?

Yes.

No.

Answer:

Any telescoping sum will have closed form expression therefore the answer is a "Yes".

Question 8: What is the answer to this problem?

A. $-\dfrac{\sqrt[3]{5}}{2}-\dfrac{\sqrt[3]{6}}{2}+5$

B. $-\dfrac{\sqrt[3]{3}}{2}-\dfrac{\sqrt[3]{4}}{2}+5+\dfrac{\sqrt[3]{999}}{2}$

C. $-\dfrac{\sqrt[3]{3}}{2}+5$

Answer:

Picking up where we left off (from question 3), we see that:

[MATH]\sum_{n=4}^{999}\dfrac{1}{\sqrt[3]{n^2-2n+1}+\sqrt[3]{n^2+2n+1}+\sqrt[3]{n^2-1}}[/MATH]

[MATH]=\sum_{n=4}^{999} \dfrac{(n+1)^{\frac{1}{3}}-(n-1)^{\frac{1}{3}}}{2}[/MATH]

[MATH]=\sum_{n=4}^{999}\dfrac{1}{2}((n+1)^{\frac{1}{3}}-(n-1)^{\frac{1}{3}})[/MATH]

$=\dfrac{1}{2}[(5^{\frac{1}{3}}-3^{\frac{1}{3}})+(6^{\frac{1}{3}}-4^{\frac{1}{3}})+(7^{\frac{1}{3}}-5^{\frac{1}{3}})+(8^{\frac{1}{3}}-6^{\frac{1}{3}})+\cdots$

$\,\,\,\,\,\,\,\,\,\,\,\,+(998^{\frac{1}{3}}-996^{\frac{1}{3}})+(999^{\frac{1}{3}}-997^{\frac{1}{3}})+(1000^{\frac{1}{3}}-998^{\frac{1}{3}})]$

$=\dfrac{1}{2}[(\cancel{5^{\frac{1}{3}}}-3^{\frac{1}{3}})+(\cancel{6^{\frac{1}{3}}}-4^{\frac{1}{3}})+(\cancel{7^{\frac{1}{3}}}-\cancel{5^{\frac{1}{3}}})+(\cancel{8^{\frac{1}{3}}}-\cancel{6^{\frac{1}{3}}})+\cdots$

$\,\,\,\,\,\,\,\,\,\,\,\,+(\cancel{998^{\frac{1}{3}}}-\cancel{996^{\frac{1}{3}}})+(999^{\frac{1}{3}}-\cancel{997^{\frac{1}{3}}})+(1000^{\frac{1}{3}}-\cancel{998^{\frac{1}{3}}})]$

$=\dfrac{1}{2}[-3^{\frac{1}{3}}-4^{\frac{1}{3}}+999^{\frac{1}{3}}+1000^{\frac{1}{3}}]$

$=\dfrac{1}{2}[-3^{\frac{1}{3}}-4^{\frac{1}{3}}+999^{\frac{1}{3}}+10]$

Therefore, B is the correct answer.

Question 9. Please rate the difficulty on this problem on a scale from 1 to 5 with 5 being the highest level of difficulty and 1 being the lowest level of difficulty.

This question is more like a survey problem to check the overall students' capability to deal with less straightforward telescoping series and to see if students could simplify the general term so they know the series actually telescopes.

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