Exercises: Cumulative

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Exercises relating to various topics we have studied.

Determine whether the series below converges absolutely, conditionally, or diverges. $\sum _{n=1}^\infty \frac {(-1)^n}{n - e^{-n}}$

Absolute Conditional Diverge

Determine whether the series below converges absolutely, conditionally, or diverges. $\sum _{n=1}^\infty \frac {(-1)^n e^{-n}}{n - e^{-n}}$

Absolute Conditional Diverge

Take absolute values and try direct comparison test.

Try direct comparison to $e^{-n}$ .

Determine whether the series below converges absolutely, conditionally, or diverges. $\sum _{n=1}^\infty \frac {(-1)^n n}{n - e^{-n}}$

Absolute Conditional Diverge

What is the dominant term in the denominator as $n \rightarrow \infty$ ?

The dominant term in the denominator is $n$ .

This means that the terms do not go to zero.

Determine whether the series below converges absolutely, conditionally, or diverges. $\sum _{n=1}^\infty \frac {1}{n^2 + (-1)^n n}$

Absolute Conditional Diverge

What is the dominant term in the denominator as $n \rightarrow \infty$ ?

The dominant term in the denominator is $n^2$ .

Determine whether the series below converges absolutely, conditionally, or diverges. $\sum _{n=1}^\infty \frac {(-1)^n}{n - e^{-n}}$

Absolute Conditional Diverge

Determine whether the series below converges absolutely, conditionally, or diverges. $\sum _{n=1}^\infty \frac {(-1)^n \sqrt {n}}{\ln (n+1)}$

Absolute Conditional Diverge

Which term dominates as $n \rightarrow \infty$ : $\sqrt {n}$ or $\ln (n+1)$ ?

Answer: $\sqrt {n}$ dominates $\ln (n+1)$ as $n \rightarrow \infty$ .

Determine whether the series below converges absolutely, conditionally, or diverges. $\sum _{n=1}^\infty \frac {(-1)^n}{\sqrt {n} + \ln (n+1)}$

Absolute Conditional Diverge

To show that convergence is not absolute, try limit comparison.

The comparison series can be taken to be $1/\sqrt {n}$ in this case.

Determine whether the series below converges absolutely, conditionally, or diverges. $\sum _{n=1}^\infty \frac {1}{n} \ln \left (2 + \frac {1}{n} \right )$

Absolute Conditional Diverge

Try limit comparison with the harmonic series.

Determine whether the series below converges absolutely, conditionally, or diverges. $\sum _{n=1}^\infty \frac {n \cos n \pi }{n^2 - 1}$

Absolute Conditional Diverge

With absolute values, compare to a harmonic series.

How do we know that the alternating series test applies?

Sample Exam Questions

Determine whether the following series converge or diverge. $\text {I: } \sum _{n=1}^\infty \frac {n^3}{n^4+4} \ \ \ \text {II: } \sum _{n=1}^\infty \frac {3^n}{n!} \ \ \ \text {III: } \sum _{n=2}^\infty \frac { \ln \ln n}{\ln n} \ \ \ \text {IV: } \sum _{n=1}^\infty \frac {3n^2}{(n!)^2}$

I & II converge; III & IV diverge I & III converge; II & IV diverge I & IV converge; II & III diverge II & III converge; I & IV diverge II & IV converge; I & III diverge III & IV converge; I & II diverge

Determine whether the following series are convergent or divergent. Justify your answers. $\text {I: } \sum _{n=1}^\infty \frac {n^2-3n}{\sqrt [3]{n^{10}-4n^2}} \ \ \ \ \text {II: } \sum _{n=1}^\infty \frac {(-n)^n}{5^{2n+3}}$

I & II divergent I convergent, II divergent I divergent, II convergent I & II convergent

Determine whether the following series are convergent or divergent. Justify your answers. $\text {I: } \sum _{n=1}^\infty \frac {\arctan n}{n^4} \ \ \ \ \text {II: } \sum _{n=1}^\infty \frac {\sin \frac {1}{n}}{n^2}$

I & II divergent I convergent, II divergent I divergent, II convergent I & II convergent

Determine which of the following series are convergent. For full credit, be sure to explain your reasoning and specify which tests were used. $\text {I: } \sum _{n=2}^\infty 2ne^{-n^2} \ \ \ \ \text {II: } \sum _{n=2}^\infty \frac {n + 2 \ln n}{2 n^4} \ \ \ \ \text {III: } \sum _{n=2}^\infty \frac {n^n}{n!}$

only I only I and II only I and III only II only II and III only III