Absolute convergent: 18 MCQs with answers

The series $\sum |a_n|$ converges.

Absolute convergence requires the series of absolute values to converge.

It converges.

Absolute convergence implies convergence of the original series.

Converges absolutely

By definition, if the series of absolute values converges, the series is absolutely convergent.

Ratio test

The ratio test examines the limit of $|a_{n+1}/a_n|$; if < 1, the series converges absolutely.

Convergent

By definition, absolute convergence means the series of absolute values converges.

True

Absolute convergence implies convergence of the original series.

False

Conditional convergence means the series converges but the absolute value series diverges.

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

The series of absolute values is $\sum 1/n^2$, a p-series with p=2 > 1, hence converges.

Absolutely convergent

Since $|\sin(n)| \le 1$, the absolute value series is bounded by $\sum 1/n^2$, which converges.

Every absolutely convergent series is convergent.

Absolute convergence is a stronger condition than convergence; it implies convergence.

Conditionally convergent

The absolute value series is the harmonic series, which diverges, but the alternating series converges by the alternating series test.

Absolutely convergent

The absolute value series is a geometric series with ratio 1/2, which converges.

$\sum_{n=1}^{\infty} \frac{1}{n}$

The harmonic series $\sum 1/n$ diverges, so it is not absolutely convergent.

It converges but not absolutely.

Conditional convergence means the series converges, but the series of absolute values diverges.

If $\sum |a_n|$ diverges, then $\sum a_n$ must diverge.

A series can converge conditionally even if the absolute value series diverges (e.g., alternating harmonic series).

Conditionally convergent

The absolute value series is $\sum 1/n^{0.5}$, a p-series with p=0.5 < 1, which diverges. The alternating series converges by the alternating series test.

Absolutely convergent

The absolute value series is $\sum 1/n^3$, a p-series with p=3 > 1, hence converges.

$\sum_{n=1}^{\infty} \frac{1}{n^2+1}$

The series $\sum 1/(n^2+1)$ converges by comparison to $\sum 1/n^2$, so it is absolutely convergent.