Question

Determine whether the series is absolutely convergent,
conditionally convergent, or divergent.
1. Sum of (-1)^(n-1)(n)/(n^2+4) with n=1 -> infinite
2. Sum of (n!)/(100^n) with n=1 -> infinite
3. Sum of (n^10)/((-10)^(n+1)) with n=1 -> infinite
4. Sum of (3-cos(n))/(n^(2/3)-2) with n=1 -> infinite

Answer 1

Answer on Question #54902 – Math – Calculus

Determine whether the series is absolutely convergent, conditionally convergent, or divergent.

1. $\sum_{n=1}^{\infty} \frac{(-1)^{(n-1)(n)}}{n^2 + 4}$

2. $\sum_{n=1}^{\infty} \frac{n!}{100^n}$

3. $\sum_{n=1}^{\infty} \frac{n^{10}}{(-10)^{n+1}}$

4. $\sum_{n=1}^{\infty} \frac{3 - \cos(n)}{\frac{2}{n^2 - 2}}$

Solution

1. $\sum_{n=1}^{\infty} \frac{(-1)^{(n-1)(n)}}{n^2 + 4} \leq \sum_{n=1}^{\infty} \left| \frac{(-1)^{(n-1)(n)}}{n^2 + 4} \right| = \sum_{n=1}^{\infty} \frac{1}{n^2 + 4} \leq \sum_{n=1}^{\infty} \frac{1}{n^2}.$

This is an example of series $\sum_{n=1}^{\infty} \frac{1}{n^p}$ , which is convergent if $p > 1$ . In our case $p = 2$ . Thus the series $\sum_{n=1}^{\infty} \frac{1}{n^2}$ is convergent and $\sum_{n=1}^{\infty} \frac{(-1)^{(n-1)(n)}}{n^2 + 4}$ is absolutely convergent.

2. Using the Ratio Test,

$$\lim _ {n \to \infty} \left| \frac {a _ {n + 1}}{a _ {n}} \right| = \lim _ {n \to \infty} \left| \frac {\frac {(n + 1) !}{1 0 0 ^ {n + 1}}}{\frac {n !}{1 0 0 ^ {n}}} \right| = \lim _ {n \to \infty} \left| \frac {n + 1}{1 0 0} \right| = \infty .$$

Thus, the series is divergent.

3. $\sum_{n=1}^{\infty} \frac{n^{10}}{(-10)^{n+1}} \leq \sum_{n=1}^{\infty} \left| \frac{n^{10}}{(-10)^{n+1}} \right| = \frac{1}{10} \sum_{n=1}^{\infty} \frac{n^{10}}{10^n}$

Using the Ratio Test,

$$\lim _ {n \to \infty} \left| \frac {a _ {n + 1}}{a _ {n}} \right| = \lim _ {n \to \infty} \left| \frac {\frac {(n + 1) ^ {1 0}}{1 0 ^ {n + 1}}}{\frac {n ^ {1 0}}{1 0 ^ {n}}} \right| = \lim _ {n \to \infty} \left| \frac {\left(1 + \frac {1}{n}\right) ^ {1 0}}{1 0} \right| = \frac {1}{1 0}.$$

Therefore, since $\frac{1}{10} < 1$ , the Ratio Test says that the series $\sum_{n=1}^{\infty} \frac{n^{10}}{10^n}$ converges.

That's why, by the Comparison Test, the series $\sum_{n=1}^{\infty} \frac{n^{10}}{(-10)^{n+1}}$ is absolutely convergent.

4.

$$\sum_{n=1}^{\infty} \frac{3 - \cos(n)}{n^{\frac{2}{3}} - 2} > \sum_{n=1}^{\infty} \frac{3 - 1}{n^{\frac{2}{3}}} = 2 \sum_{n=1}^{\infty} \frac{1}{n^{\frac{2}{3}}}.$$

This is an example of series $\sum_{n=1}^{\infty} \frac{1}{n^p}$, which is convergent if $p > 1$ and divergent if $p \leq 1$.

In our case $p = \frac{2}{3}$. Thus, the series $2\sum_{n=1}^{\infty}\frac{1}{\frac{2}{n^3}}$ is divergent.

So, by the direct comparison test, the series $\sum_{n=1}^{\infty} \frac{3 - \cos(n)}{n^{\frac{2}{3}} - 2}$ is divergent too.

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