Power Series Calculator

A power series centred at $c$ is

$\sum_{n=0}^{\infty} a_n (x - c)^n$

and the question is always the same: for which $x$ does it converge? The answer is always an interval centred on $c$, of half-width $R$, the radius of convergence.

Get $R$ from the ratio test. Form

$L = \lim_{n\to\infty} \left| \frac{a_{n+1}(x-c)^{n+1}}{a_n (x-c)^n} \right|$

and demand $L < 1$. Three outcomes:

• $R = 0$ — converges only at $x = c$ (e.g. $\sum n!\,x^n$)
• $0 < R < \infty$ — converges on $(c-R,\ c+R)$, diverges outside
• $R = \infty$ — converges for every real $x$ (e.g. $\sum x^n/n!$)

The assumption people forget: the ratio test says nothing at $|x - c| = R$, because there $L = 1$ exactly. The two endpoints must be tested separately, and they can behave differently from each other.

Substituting an endpoint turns the power series into an ordinary numerical series, and two standard results usually settle it.

p-series:

$\sum_{n=1}^{\infty} \frac{1}{n^p} \quad\text{converges iff}\quad p > 1$

So $\sum 1/n^{3/2}$ converges and $\sum 1/n$ — the harmonic series, $p = 1$ — diverges, marginally but definitely.

Alternating series test: $\sum (-1)^n b_n$ converges if $b_n$ decreases monotonically to zero. This is why $\sum (-1)^n/n$ converges while $\sum 1/n$ does not; the first is conditionally convergent.

Geometric series, worth recognising instantly:

$\sum_{n=0}^{\infty} ar^n = \frac{a}{1-r} \quad \text{for } |r| < 1$

When it applies: the geometric sum formula needs $|r| < 1$ strictly; at $|r| = 1$ the terms do not shrink and the series diverges by the $n$th term test.

• Reporting an open interval without checking the endpoints — half the marks on a typical exam question live there.
• Assuming both endpoints behave the same way — $\sum (x-2)^n/(n3^n)$ converges at one end and diverges at the other.
• Confusing $p$-series with geometric — $\sum 1/n^p$ has $n$ in the base, $\sum r^n$ has $n$ in the exponent. They converge under completely different conditions.
• Thinking $a_n \to 0$ proves convergence — the harmonic series is the standing counterexample. The $n$th term test can only prove divergence.
• Dropping the absolute values in the ratio test — the limit is of a modulus.
• Forgetting the centre — the interval is centred at $c$, not at $0$, so a radius of $3$ about $c = 2$ gives $(-1, 5)$.