Lecture 30 Convergence Tests Part 5

Text References: Course notes pp. 135-136 & Rogawski 10.5

30.1 Recap

Last time, we learned about the Alternating Series Test, ASET, and absolute vs conditional convergence.

Exercise 30.1 Show that \(\displaystyle \sum_{k=0}^\infty \frac{(-1)^k}{\sqrt{k!}}\) converges and approximate its sum correct to \(2\) decimal places.

Solution. We will prove convergence using the Alternating Series Test:

  • \(\lim_{k\to\infty} \dfrac{1}{\sqrt{k!}}=0\)
  • \(a_{k+1}<a_{k}\) since \(\dfrac{1}{\sqrt{(k+1)!}}<\dfrac{1}{\sqrt{k!}}\)

In order to estimate the sum, we can use ASET. The sum being correct to \(2\) decimal places means that we want to find a value of \(k\) such that \(|s-s_k|\leq a_{k+1}=\dfrac{1}{\sqrt{(k+1)!}}<10^{-2}\).

We’re going to have to do a bit of trial and error here. We have

\[\begin{align*} a_5 &= \frac{1}{\sqrt{5!}} \approx 0.09129\\ a_6 &= \frac{1}{\sqrt{6!}} \approx 0.03727 \\ a_7 &= \frac{1}{\sqrt{7!}} \approx 0.01409 \\ a_8 &= \frac{1}{\sqrt{8!}} \approx 0.00498 \\ \end{align*}\]

Note that \(a_8\) does not affect the first two decimal places, so approximations with this error value will be correct to \(2\) decimal places. We have \(s_7\approx 0.43488\), and this is correct to two decimal places.

30.2 Learning Objectives

  • Apply the Ratio Test appropriately to determine the convergence of a series.
  • Apply the Root Test appropriately to determine the convergence of a series.

30.3 The Ratio Test

The Ratio Test is very effective to determine whether a series is absolutely convergent.

Theorem 30.1 Suppose that \(\displaystyle \lim_{k\to\infty} \left | \dfrac{a_{k+1}}{a_k}\right |=L\).

  • If \(L<1\), then the series \(\displaystyle \sum a_k\) is absolutely convergent.
  • If \(L>1\), then the series \(\displaystyle \sum a_k\) is divergent.
  • If \(L=1\), then the test fails; the series could be absolutely convergent, conditionally convergent, or divergent.

Exercise 30.2 Determine whether \(\displaystyle \sum_{k=0}^\infty (-1)^k\frac{k!}{1000^k}\) converges.

Solution. Applying the root test with \(a_k= \dfrac{k!}{1000^k}\) and \(a_{k+1}=\dfrac{(k+1)!}{1000^{k+1}}\), we get

\[\begin{align*} \lim_{k\to\infty}\left | \frac{a_{k+1}}{a_k}\right| &= \lim_{k\to\infty}\left | \frac{\frac{(k+1)!}{1000^{k+1}}}{\frac{k!}{1000^k}}\right | \\ &= \lim_{k\to\infty} \frac{(k+1)!1000^k}{k!1000^{k+1}} \\ &= \lim_{k\to\infty} \frac{k+1}{1000} = \infty >1 \end{align*}\]

Therefore, by the Ratio Test, the series is divergent.

30.4 The Root Test

Similar to the Ratio Test, the Root Test is used to determine whether a series is absolutely convergent.

Theorem 30.2 Suppose that \(\displaystyle \lim_{k\to\infty} \sqrt[k]{|a_k|}=L\).

  • If \(L<1\), then the series \(\displaystyle \sum a_k\) is absolutely convergent.
  • If \(L>1\), then the series \(\displaystyle \sum a_k\) is divergent.
  • If \(L=1\), then the test fails; the series could be absolutely convergent, conditionally convergent, or divergent.

Exercise 30.3 Determine whether the series \(\displaystyle \sum_{k=1}^\infty \left(\frac{k}{2k+3} \right)^k\) converges.

Solution. Applying the Root Test, we have

\[\begin{align*} \lim_{k\to\infty} \sqrt[k]{|a_k|} &= \lim_{k\to\infty}\sqrt[k]{\left|\frac{k}{2n+3}\right|^k} \\ &= \lim_{k\to\infty} \frac{k}{2k+3}\\ &= \frac{1}{2}<1 \end{align*}\]

Therefore, by the Root Test, the series is absolutely convergent.