Series Convergence Test Calculator

A series $\sum a_n$ converges when its sequence of partial sums has a finite limit:

$S_N = \sum_{n=1}^{N} a_n, \qquad \sum_{n=1}^{\infty} a_n = \lim_{N \to \infty} S_N$

If that limit fails to exist, the series diverges. A test tells you which of these happens; it does not usually tell you the sum.

Two benchmark families anchor everything else:

$\sum_{n=1}^{\infty} ar^{n-1} \text{ converges} \iff |r| < 1, \qquad \sum_{n=1}^{\infty} \frac{1}{n^p} \text{ converges} \iff p > 1$

The harmonic series $\sum \frac{1}{n}$ is the boundary case $p = 1$, and it diverges — slowly, but definitely.

Absolute convergence: if $\sum |a_n|$ converges then $\sum a_n$ does too. The converse fails, and a series like $\sum \frac{(-1)^n}{n}$ is conditionally convergent.

Every test has conditions. Applying one outside its hypotheses produces a confident but meaningless answer.

Choosing quickly

1. Check $\lim a_n$ first — if it is not $0$, you are done.
2. Factorials or $n$-th powers $\Rightarrow$ ratio or root test.
3. A quotient of polynomials $\Rightarrow$ limit comparison with $1/n^{p}$, using the difference of the leading degrees as $p$.
4. An easily integrated $f(n)$ $\Rightarrow$ integral test.
5. Alternating signs $\Rightarrow$ test $\sum |a_n|$ first, then the alternating series test.

• Concluding convergence from $a_n \to 0$: the harmonic series is the standing counterexample. The nth-term test can only prove divergence.
• Using the ratio test on a rational function: it always returns $L = 1$ there, which is inconclusive. Use limit comparison instead.
• Skipping the decreasing hypothesis: the integral test and the alternating series test both require it, and it can fail for small $n$ even when it holds eventually.
• Comparing in the wrong direction: a series larger than a divergent one diverges; a series larger than a convergent one tells you nothing.
• Confusing sequence and series: $a_n \to 0$ says the sequence converges, not the series.