Write \(1/(1+x^2)\) as a power series and find its convergence interval

Key concept: This is a classic geometric-series move in disguise. Once you rewrite the function in the right form, the series and its convergence interval fall out naturally.

Step by step

We start with

$\frac{1}{1+x^2}.$

Step 1: Match the geometric series form

Recall the geometric series identity

$\frac{1}{1-r}=\sum_{n=0}^{\infty} r^n, \qquad |r|<1.$

Here we want to rewrite

$\frac{1}{1+x^2}=\frac{1}{1-(-x^2)}.$

So take

$r=-x^2.$

Then

$\frac{1}{1+x^2}=\sum_{n=0}^{\infty} (-x^2)^n.$

Step 2: Simplify the series

after expanding the power,

$\frac{1}{1+x^2}=\sum_{n=0}^{\infty} (-1)^n x^{2n}.$

Step 3: Find the interval of convergence

The geometric series converges when

$|r|<1.$

Since $r=-x^2$,

$|-x^2|<1 \quad \Longrightarrow \quad x^2<1 \quad \Longrightarrow \quad |x|<1.$

Now check the endpoints:

• At $x=1$:

  $\sum_{n=0}^{\infty} (-1)^n,$

  which diverges.

• At $x=-1$:

  $\sum_{n=0}^{\infty} (-1)^n,$

  which also diverges.

So the interval of convergence is

$(-1,1).$

Final answer

$\frac{1}{1+x^2}=\sum_{n=0}^{\infty} (-1)^n x^{2n}, \qquad |x|<1,$

with interval of convergence

$(-1,1).$

Pitfall alert

The biggest mistake is forgetting that the geometric series condition is on the ratio $r$, not directly on the denominator. Here the denominator is $1+x^2$, so the right substitution is $r=-x^2$.

Another easy miss is assuming the endpoints work because the function itself is defined there. A power series representation can still diverge at the endpoints even when the original function is perfectly finite.

Try different conditions

If the function were

$\frac{1}{1-ax^2},$

then the same method gives

$\frac{1}{1-ax^2}=\sum_{n=0}^{\infty} (ax^2)^n,$

which converges when

$|ax^2|<1 \quad \Longrightarrow \quad |x|<\frac{1}{\sqrt{|a|}}$

for $a\neq 0$. The exact endpoint behavior would still need checking separately.

Further reading

geometric series, power series, interval of convergence