Convergence of series: 18 MCQs with answers

If $\lim_{n \to \infty} a_n \neq 0$, then the series $\sum a_n$ diverges.

The n-th term test states that if the limit of the terms does not approach zero, the series cannot converge. The converse (limit zero implies convergence) is false.

$p > 1$

The p-series test states that $\sum \frac{1}{n^p}$ converges if and only if $p > 1$. For $p = 1$ it diverges (harmonic series).

It diverges to infinity.

The harmonic series is a classic example of divergence, even though its terms approach zero.

False

The harmonic series $\sum \frac{1}{n}$ has terms approaching zero but diverges. The n-th term test is only one-way.

Ratio Test

The presence of an exponential term $2^n$ and a polynomial $n^2$ makes the Ratio Test effective, as the ratio simplifies nicely.

$b_n$ is decreasing and $\lim b_n = 0$

The Alternating Series Test requires that the absolute values $b_n$ decrease monotonically to zero.

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

Absolute convergence means the series of absolute values converges. Here, $\sum \frac{1}{n^2}$ converges (p-series with p=2), so the series converges absolutely.

The ratio of the terms of two series as $n \to \infty$

The Limit Comparison Test uses $\lim_{n \to \infty} \frac{a_n}{b_n} = c$ where $0 < c < \infty$ to conclude both series behave the same.

Comparison Test with $\frac{1}{n^2}$

Since $\frac{1}{n^2+1} < \frac{1}{n^2}$ and $\sum \frac{1}{n^2}$ converges, the Comparison Test directly applies.

For $f(x)$ positive, continuous, and decreasing, $\sum f(n)$ and $\int_1^\infty f(x) dx$ either both converge or both diverge.

The Integral Test requires $f$ to be positive, continuous, and decreasing on $[1, \infty)$; then the series and improper integral share convergence behavior.

It converges but not absolutely.

Conditional convergence means the series converges, but the series of absolute values diverges. The alternating harmonic series is a classic example.

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

The series $\sum \frac{1}{\sqrt{n}}$ is a p-series with $p = 1/2 \leq 1$, so it diverges.

Root Test

The presence of the $n$-th power suggests the Root Test, which simplifies to $\lim \frac{2n+1}{3n-1} = \frac{2}{3} < 1$, so the series converges.

$\frac{\pi^2}{6}$

This is the famous Basel problem, solved by Euler, showing the sum is $\frac{\pi^2}{6}$.

False

If the limit in the Ratio Test equals 1, the test is inconclusive, as seen with p-series.

$(-\infty, \infty)$

This is the exponential series, which converges for all real $x$ because the factorial grows faster than any power, so the radius of convergence is infinite.

Converges

Applying the Ratio Test gives $\lim \frac{(n+1)!/(n+1)^{n+1}}{n!/n^n} = \lim \left(\frac{n}{n+1}\right)^n = \frac{1}{e} < 1$, so the series converges.

$0$

Using the Ratio Test, $\lim \left|\frac{(n+1)! x^{n+1}}{n! x^n}\right| = \lim (n+1)|x| = \infty$ for any $x \neq 0$, so the series converges only at $x=0$, radius $0$.