Absolute convergent: 18 flashcards

A series $\sum a_n$ converges absolutely if the series of absolute values $\sum |a_n|$ converges.

Absolute convergence implies ordinary convergence: if $\sum |a_n|$ converges, then $\sum a_n$ converges.

The alternating harmonic series $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}$ converges conditionally: it converges, but $\sum \frac{1}{n}$ diverges.

Yes. Since $|\sin(n)| \le 1$, we have $\left|\frac{\sin(n)}{n^2}\right| \le \frac{1}{n^2}$. The p-series $\sum \frac{1}{n^2}$ converges, so by comparison, $\sum \left|\frac{\sin(n)}{n^2}\right|$ converges.

If a series converges absolutely, then any rearrangement of its terms converges to the same sum. Conditional convergence does not guarantee this (Riemann rearrangement theorem).

For a series $\sum a_n$, if $\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = L$, then: if $L < 1$, the series converges absolutely; if $L > 1$, it diverges; if $L = 1$, the test is inconclusive.

Compute $\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n \to \infty} \frac{(n+1)/2^{n+1}}{n/2^n} = \lim_{n \to \infty} \frac{n+1}{2n} = \frac{1}{2} < 1$. Thus the series converges absolutely.

For a series $\sum a_n$, if $\lim_{n \to \infty} \sqrt[n]{|a_n|} = L$, then: if $L < 1$, the series converges absolutely; if $L > 1$, it diverges; if $L = 1$, the test is inconclusive.

Using the root test: $\sqrt[n]{|a_n|} = \left(\frac{n}{n+1}\right)^n = \left(1 - \frac{1}{n+1}\right)^n \to e^{-1} < 1$. Hence the series converges absolutely.

No. If $\sum |a_n|$ converges, then $\sum a_n$ converges. Absolute convergence is a stronger condition than convergence.

Absolute convergence: $\sum |a_n|$ converges. Conditional convergence: $\sum a_n$ converges but $\sum |a_n|$ diverges.

The geometric series $\sum_{n=0}^{\infty} \frac{1}{2^n}$ converges absolutely because $\sum \left|\frac{1}{2^n}\right| = \sum \frac{1}{2^n}$ converges (ratio $= 1/2$).

Not necessarily convergent. For example, $\sum \frac{1}{n^2}$ converges absolutely, but $\sum \left(\frac{1}{n}\right)^2 = \sum \frac{1}{n^2}$ converges; however, if $a_n = \frac{(-1)^n}{\sqrt{n}}$, then $\sum a_n$ converges conditionally, not absolutely, and $\sum a_n^2$ diverges. So absolute convergence of $\sum a_n$…

Yes. The theorem: If $\sum |a_n|$ converges, then $\sum a_n$ converges.

For absolutely convergent series, any rearrangement converges to the same sum. For conditionally convergent series, rearrangements can change the sum or even diverge.

It converges conditionally: the alternating series test shows convergence, but $\sum \frac{1}{n}$ diverges, so it does not converge absolutely.

The integral test can be used to test convergence of $\sum |a_n|$ if $|a_n|$ is positive, continuous, and decreasing. If the integral of $|a_n|$ converges, then the series converges absolutely.

False. Convergence does not imply absolute convergence. Example: $\sum \frac{(-1)^n}{n}$ converges but its absolute series diverges.