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$$\sum_{n=2}^\infty(\cos((n^{3}+\sqrt{n}+7)^{\frac{1}{3}})-\cos((n^{3}-2\sqrt{n}+3)^{\frac{1}{3}}))$$

Check convergence.

Please verify my answer below.

$$\sum_{n=2}^\infty(\cos((n^{3}+\sqrt{n}+7)^{\frac{1}{3}})-\cos((n^{3}-2\sqrt{n}+3)^{\frac{1}{3}}))=$$

$$=\sum_{n=2}^\infty-2\sin\left(\frac{\sqrt[3]{n^{3}+\sqrt{n}+7}-\sqrt[3]{n^{3}-2\sqrt{n}+3}}{2}\right)\cdot\sin\left(\frac{\sqrt[3]{n^{3}+\sqrt{n}+7}+\sqrt[3]{n^{3}-2\sqrt{n}+3}}{2}\right)$$

Checking arbirary convergence

$\sin\left(\frac{\sqrt[3]{n^{3}+\sqrt{n}+7}+\sqrt[3]{n^{3}-2\sqrt{n}+3}}{2}\right)\in\langle-1,1\rangle$

$=\sum_{n=2}^\infty|-2\sin\left(\frac{\sqrt[3]{n^{3}+\sqrt{n}+7}-\sqrt[3]{n^{3}-2\sqrt{n}+3}}{2}\right)\cdot\sin\left(\frac{\sqrt[3]{n^{3}+\sqrt{n}+7}+\sqrt[3]{n^{3}-2\sqrt{n}+3}}{2}\right)|\leq2\sum_{n=1}^\infty\left|\sin\left(\frac{\sqrt[3]{n^{3}+\sqrt{n}+7}-\sqrt[3]{n^{3}-2\sqrt{n}+3}}{2}\right)\right|$ [1]

Checking convergence of

[2] $\frac{1}{2}\sum_{n=2}^\infty\sqrt[3]{n^{3}+\sqrt{n}+7}-\sqrt[3]{n^{3}-2\sqrt{n}+3}=\frac{1}{2}\sum_{n=2}^\infty\frac{4+3\sqrt{n}}{\sqrt[3]{(n^{3}+\sqrt{n}+7)^{2}}+\sqrt[3]{n^{3}+\sqrt{n}+7} \sqrt[3]{n^{3}-2\sqrt{n}+3}+\sqrt[3]{(n^{3}-2\sqrt{n}+3)^{2}}}$

$$\frac{1}{2}\sum_{n=2}^\infty\frac{4+3\sqrt{n}}{\sqrt[3]{(n^{3}+\sqrt{n}+7)^{2}}+\sqrt[3]{n^{3}+\sqrt{n}+7}\cdot\sqrt[3]{n^{3}-2\sqrt{n}+3}+\sqrt[3]{(n^{3}-2\sqrt{n}+3)^{2}}}<\frac{4\sqrt{n}}{n^{2}}=\frac{4}{n^{1.5}}$$

From the comparison test, it converge

Checking product test

$$\frac{\sin\left(\frac{\sqrt[3]{n^{3}+\sqrt{n}+7}-\sqrt[3]{n^{3}-2\sqrt{n}+3}}{2}\right)}{\frac{\sqrt[3]{n^{3}+\sqrt{n}+7}-\sqrt[3]{n^{3}-2\sqrt{n}+3}}{2}}\rightarrow1$$

Limit exist and is in $\mathbb{R}$ so from [2], [1] converge. If our series arbitrary converge, it also converge.

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1  
Tip: for writing you sums, you can save a lot of work for yourself by using \sum_{n=2}^\infty = $\sum_{n=2}^\infty$. –  amWhy Jan 9 '13 at 14:33
    
My code come from LyX editor, where that's automatically generated. Anyway thanks for comment! –  Joggi Jan 9 '13 at 14:35
    
@Joggi: agree with amWhy, the Lyx formats terribly here, and as I edited another of your posts for clarity, it was a lot of work for 2 pts! Please at least run a script that does the conversion if you like automation. –  Ron Gordon Jan 9 '13 at 14:42
    
I changed several instances of \overset{\infty}{\underset{n=2}{\sum}} to \sum_{n=2}^\infty. The LyX editor is an imbecile. –  Michael Hardy Jan 9 '13 at 16:16
    
@Michael: I think the small difference is whether the indices appear above and below the summation symbol or next to it. However, a better solution would be \sum\limits_{n=2}^\infty instead of the monstrosity of oversets and undersets. –  Willie Wong Jan 9 '13 at 16:41

1 Answer 1

Let me suggest some simplifications. The $n$th term is the difference of two cosines and the function cosine is $1$-Lipschitz hence each term is at most the difference of the two cube roots. Furthermore, the derivative of the cube root function is decreasing hence for every $a\gt b\gt0$, $$ a^{1/3}-b^{1/3}=\int_b^a\frac{\mathrm dx}{3x^{2/3}}\leqslant\frac{a-b}{3b^{2/3}}. $$ Applying this to $a_n=n^3+\sqrt{n}+7$ and $b_n=n^3-2\sqrt{n}+3$ yields $$ |\cos(a_n^{1/3})-\cos(b_n^{1/3})|\leqslant a_n^{1/3}-b_n^{1/3}\leqslant\frac{a_n-b_n}{3b_n^{2/3}}. $$ Now, $a_n-b_n\sim3\sqrt{n}$ and $b_n\sim n^3$ hence the RHS is equivalent to $$ \frac{3\sqrt{n}}{3(n^3)^{2/3}}=\frac1{n\sqrt{n}}. $$ The series $\sum\limits_n\frac1{n\sqrt{n}}$ converges hence the series $\sum\limits_n\left[\cos(a_n^{1/3})-\cos(b_n^{1/3})\right]$ converges absolutely.

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