Exercises: Series Comparison Tests

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Exercises relating to the direct and limit comparison tests for series.

Suppose you wished to use the Direct Comparison Test to establish convergence of the series $\sum _{n=1}^\infty \frac {1}{n^4 + n}.$ Which of the following options below would be valid series for comparison? Select all that apply.

$\displaystyle \sum _{n=2}^\infty \frac {1}{n^5}$ $\displaystyle \sum _{n=2}^\infty \frac {1}{n^4}$ $\displaystyle \sum _{n=2}^\infty \frac {1}{n^3}$ $\displaystyle \sum _{n=2}^\infty \frac {1}{n^2}$ $\displaystyle \sum _{n=2}^\infty \frac {1}{n}$

Expand for feedback on $\displaystyle \sum _{n=2}^\infty \frac {1}{n^5}$ :

This comparison wouldn’t be valid because $\frac {1}{n^5} \leq \frac {1}{n^4+n} \ \ \text { for all } n \geq 2,$ which is the wrong direction for the inequality when showing convergence.

Expand for feedback on $\displaystyle \sum _{n=2}^\infty \frac {1}{n^4}$ :

This comparison is valid because $\frac {1}{n^4+n} \leq \frac {1}{n^4} \ \ \text { for all } n \geq 2,$ which is the proper direction for the comparison inequality when showing convergence. Also key is that the $p$ -series for $p=4$ is convergent.

Expand for feedback on $\displaystyle \sum _{n=2}^\infty \frac {1}{n^3}$ :

This comparison is valid because $\frac {1}{n^4+n} \leq \frac {1}{n^3} \ \ \text { for all } n \geq 2,$ which is the proper direction for the comparison inequality when showing convergence. Also key is that the $p$ -series for $p=3$ is convergent.

Expand for feedback on $\displaystyle \sum _{n=2}^\infty \frac {1}{n^2}$ :

This comparison is valid because $\frac {1}{n^4+n} \leq \frac {1}{n^2} \ \ \text { for all } n \geq 2,$ which is the proper direction for the comparison inequality when showing convergence. Also key is that the $p$ -series for $p=2$ is convergent.

Expand for feedback on $\displaystyle \sum _{n=2}^\infty \frac {1}{n}$ :

This comparison is not valid because the series (known as the harmonic series) is not convergent, so no comparisons to it can establish convergence.

Suppose you wished to use the Direct Comparison Test to establish convergence of the series $\sum _{n=2}^\infty \frac {1}{n^4 - n}.$ Which of the following options below would be valid series for comparison? Select all that apply.

$\displaystyle \sum _{n=2}^\infty \frac {1}{n^4}$ $\displaystyle \sum _{n=2}^\infty \frac {1}{n^3}$ $\displaystyle \sum _{n=2}^\infty \frac {1}{n^2}$

Expand for feedback on $\displaystyle \sum _{n=2}^\infty \frac {1}{n^4}$ :

This comparison is not valid because $\frac {1}{n^4} \leq \frac {1}{n^4-n^2} \ \ \text { for all } n \geq 2,$ which is the wrong direction for the comparison inequality when showing convergence.

Expand for feedback on $\displaystyle \sum _{n=2}^\infty \frac {1}{n^3}$ :

This comparison is valid because $\frac {1}{n^4-n} \leq \frac {1}{n^3} \ \ \text { for all } n \geq 2$ (because $n^4 - n = n(n^3-1)$ and $n^3 - 1$ is always greater than $n^2$ whenever $n \geq 2$ .) which is the proper direction for the comparison inequality when showing convergence. Also key is that the $p$ -series for $p=3$ is convergent.

Expand for feedback on $\displaystyle \sum _{n=2}^\infty \frac {1}{n^2}$ :

This comparison is valid because $\frac {1}{n^4-n} \leq \frac {1}{n^2} \ \ \text { for all } n \geq 2,$ (because $n^4 - n = n(n^3-1)$ and $n^3 - 1$ is always greater than $n$ whenever $n \geq 2$ .) which is the proper direction for the comparison inequality when showing convergence. Also key is that the $p$ -series for $p=2$ is convergent.

Using only the comparison functions listed here: $\sum _{n=1}^\infty \frac {1}{1+\sqrt {n}}, \qquad \sum _{n=1}^\infty \frac {1}{n+1}, \qquad \sum _{n=1}^\infty \frac {1}{n^2}, \qquad \sum _{n=1}^\infty e^{-n},$ which of the following series can be proved to either diverge or converge using the Direct Comparison Test? (Select all that apply.)

$\displaystyle \sum _{n=1}^\infty \frac {1}{n^{3/2}+n^{1/2}}$ $\displaystyle \sum _{n=1}^\infty \frac {1}{n^2 + n^{1/2}}$ $\displaystyle \sum _{n=1}^\infty \frac {1}{\ln (n+1)}$ $\displaystyle \sum _{n=2}^\infty \frac {1}{n^2-1}$ $\displaystyle \sum _{n=1}^\infty \frac {1}{e^n + 1}$ $\displaystyle \sum _{n=3}{\infty } \frac {1}{2n+3}$

For the series $\displaystyle \sum _{n=1}^\infty \frac {1}{n^{3/2}+n^{1/2}}$ :

This series is convergent, but none of the allowed comparisons will work: Only $1/(n+1)$ and $1/(1+\sqrt {n})$ are larger than $1/(n^{3/2}+n^{1/2})$ , but neither of those series converges.

For the series $\displaystyle \sum _{n=1}^\infty \frac {1}{n^2 + n^{1/2}}$ :

Comparison to $1/n^2$ works fine in this case.

For the series $\displaystyle \sum _{n=1}^\infty \frac {1}{\ln (n+1)}$ :

$\ln (n+1) \leq n+1$ so $1/\ln (n+1) \geq 1/(n+1)$ , so actually diverges by direct comparison.

For the series $\displaystyle \sum _{n=1}^\infty \frac {1}{n^2-1}$ :

This series textitis convergent, but none of the convergent comparison options $1/n^2$ and $e^{-n}$ is greater than $1/n^2$ .

For the series $\displaystyle \sum _{n=1}^\infty \frac {1}{e^n + 1}$ :

The direct comparison $1/(e^n + 1) \leq e^{-n}$ works fine to show convergence of the series.

For the series $\displaystyle \sum _{n=3}{\infty } \frac {1}{2n+3}$ :

This is a divergent series, but direct comparison with the allowed options doesn’t work because $1/(2n+3)$ is smaller than the two divergent options $1/(n+1)$ and $1/(1+\sqrt {n})$ .

Suppose you wished to use the Limit Comparison Test to establish convergence of the series $\sum _{n=1}^\infty \frac {1}{n^4 + n}.$ Which of the following options below would be valid series for comparison? Select all that apply.

Expand for feedback on $\displaystyle \sum _{n=2}^\infty \frac {1}{n^5}$ :

This comparison wouldn’t be valid because $1/n^5$ is too small: the limit of $1/(n^4+n)$ divided by $1/n^5$ is infinite, which is an inconclusive case.

Expand for feedback on $\displaystyle \sum _{n=2}^\infty \frac {1}{n^4}$ :

This comparison is valid because the limit $1/(n^4-n)$ divided by $1/n^4$ is $1$ , which means that they both converge (because a $p$ -series for $p=4$ is convergent).

Expand for feedback on $\displaystyle \sum _{n=2}^\infty \frac {1}{n^3}$ :

This comparison is valid because $1/n^3$ is a convergent $p$ -series which is much larger than $1/(n^4+n)$ , i.e., $\lim _{n \rightarrow \infty } \frac {\frac {1}{n^4+n}}{\frac {1}{n^3}} = \lim _{n \rightarrow \infty } \frac {n^3}{n^4+n} = 0.$

Expand for feedback on $\displaystyle \sum _{n=2}^\infty \frac {1}{n^2}$ :

Also works for the same reason as $1/n^3$ : it’s a convergent $p$ -series and the original series divided by the comparison series goes to zero.

Expand for feedback on $\displaystyle \sum _{n=2}^\infty \frac {1}{n}$ :

This comparison is not valid because the series (known as the harmonic series) is not convergent, so no comparisons to it can establish convergence.

Suppose you wished to use the Limit Comparison Test to establish convergence of the series $\sum _{n=2}^\infty \frac {1}{n^4 - n}.$ Which of the following options below would be valid series for comparison? Select all that apply.

$\displaystyle \sum _{n=2}^\infty \frac {1}{n^4}$ $\displaystyle \sum _{n=2}^\infty \frac {1}{n^3}$ $\displaystyle \sum _{n=2}^\infty \frac {1}{n^2}$

For the Limit Comparison Test, non-dominant terms don’t matter, so the results will be the same here as they were for the series $1/(n^4+n)$ .

Using only the comparison functions listed here: $\sum _{n=1}^\infty \frac {1}{1+\sqrt {n}}, \qquad \sum _{n=1}^\infty \frac {1}{n+1}, \qquad \sum _{n=1}^\infty \frac {1}{n^2}, \qquad \sum _{n=1}^\infty e^{-n}$ , which of the following series can be proved to either diverge or converge using the Limit Comparison Test? (Select all that apply.)

For the series $\displaystyle \sum _{n=1}^\infty \frac {1}{n^{3/2}+n^{1/2}}$ :

This series is convergent, but none of the allowed comparisons will work: Only $1/(n+1)$ and $1/(1+\sqrt {n})$ are larger than $1/(n^{3/2}+n^{1/2})$ , but neither of those series converges.

For the series $\displaystyle \sum _{n=1}^\infty \frac {1}{n^2 + n^{1/2}}$ :

Comparison to $1/n^2$ works fine in this case.

For the series $\displaystyle \sum _{n=1}^\infty \frac {1}{\ln (n+1)}$ :

$\ln (n+1) \leq n+1$ so $1/\ln (n+1) \geq 1/(n+1)$ , so the limit $\frac {n+1}{\ln (n+1)} \rightarrow \infty \text { as } n \rightarrow \infty$ so we get divergence by the divergence of the comparison series.

For the series $\displaystyle \sum _{n=1}^\infty \frac {1}{n^2-1}$ :

Here the non-dominant term $-1$ in the denominator is no longer relevant, so limit comparison to $1/n^2$ works fine.

For the series $\displaystyle \sum _{n=1}^\infty \frac {1}{e^n + 1}$ :

The direct comparison $1/(e^n + 1) / e^{-n} \rightarrow 1$ works fine to show convergence of the series.

For the series $\displaystyle \sum _{n=3}{\infty } \frac {1}{2n+3}$ :

Limit comparison with $1/(n+1)$ works great: $1/(2n+3)$ divided by $1/(n+1)$ equals $\frac {n+1}{2n+3} \rightarrow \frac {1}{2} \text { as } n \rightarrow \infty .$