Science:Math Exam Resources/Courses/MATH103/April 2011/Question 01 (a)

MATH103 April 2011

  •  Q1 (a)  •  Q1 (b)  •  Q1 (c)  •  Q2 (a)  •  Q2 (b)  •  Q2 (c)  •  Q2 (d)  •  Q3 (a)  •  Q3 (b)  •  Q4 (a)  •  Q4 (b)  •  Q5 (a)  •  Q5 (b)  •  Q5 (c)  •  Q5 (d)  •  Q6 (a)  •  Q6 (b)  •  Q7 (a)  •  Q7 (b)  •  Q7 (c)  •  Q8  •  

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Question 01 (a)
For each of the following series, indicate whether or not they converge: $\sum _{n=1}^{\infty }{\frac {2}{n}}$ $\sum _{n=1}^{\infty }{\frac {1}{2^{n}}}$ $\sum _{n=1}^{\infty }2^{n}$ $\sum _{n=1}^{\infty }{\frac {1}{n^{2}}}$ $\sum _{n=2}^{\infty }{\frac {n}{1-n^{3}}}$

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Hint
$\sum _{n=1}^{\infty }{\frac {1}{n}}$ and $\sum _{n=1}^{\infty }{\frac {1}{n^{2}}}$ are both standard. A constant factor ($\not =0$) does not change the convergence behavior. Also recall the geometric series $\sum _{n=0}^{\infty }q^{n}$, and that the convergence here depends on the value of $q$. For the last example, compare it to a classical series above.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. $\sum _{n=1}^{\infty }{\frac {2}{n}}$ is divergent, by the integral test. $\sum _{n=1}^{\infty }{\frac {1}{2^{n}}}$ is convergent. This is a geometric series, equal to ${\frac {1}{1-{\frac {1}{2}}}}-1=1.$ $\sum _{n=1}^{\infty }2^{n}$ is divergent, since $\lim _{n\to \infty }2^{n}=\infty .$ $\sum _{n=1}^{\infty }{\frac {1}{n^{2}}}$ is convergent, by the integral test. $\sum _{n=2}^{\infty }{\frac {n}{1-n^{3}}}$ is convergent. To show this, use the fact that the absolute value of the terms inside the sequence is $\sum _{n=2}^{\infty }{\big |}{\frac {n}{1-n^{3}}}{\big |}=\sum _{n=2}^{\infty }{\frac {n}{n^{3}-1}}$. As all the terms are positive, we can use the comparison test with $\sum _{n=1}^{\infty }{\frac {2}{n^{2}}}$ and show that since this series converges by the p-series test and ${\frac {n}{n^{3}-1}}\leq {\frac {2n}{n^{3}}}={\frac {2}{n^{2}}}$ valid since $2\leq n^{3}$ for all $n\geq 2$ and thus $1\leq {\frac {n^{3}}{2}}=n^{3}-{\frac {n^{3}}{2}}$ (then rearrange this last inequality to get the above displayed equation). Therefore, the series $\sum _{n=2}^{\infty }{\big |}{\frac {n}{1-n^{3}}}{\big |}=\sum _{n=2}^{\infty }{\frac {n}{n^{3}-1}}$ converges. Then, we know that $\sum _{n=2}^{\infty }{\frac {n}{1-n^{3}}}$ is absolutely convergent and hence convergent.

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