Finding a closed form for $\sum_{n = 1}^{\infty} \frac{1}{(1 + x^2)^{n}}$

Question

So I got the sum

$$\sum_{n = 1}^{\infty} \frac{1}{(1 + x^2)^{n}}$$ and I want a closed form.

Can I just do

$$\frac{1}{1 - r} = \frac{1}{1 - \frac{1}{1 + x^2}} = \frac{1}{x^2} + 1 $$

for a closed form?

Answer 1

The formula that you are using for the sum of a geometric series, namely that $$ \sum_{n=0}^\infty ar^n = \frac{a}{1-r}, $$ is valid only when $|r|<1$. In your case, the common ratio $r$ is $$ \frac{1}{1+x^2}, $$ which is less than $1$ for all $x \neq 0$. So, for $x \neq 0$, you can express your series as $$ \frac{\frac{1}{1+x^2}}{1-\frac{1}{1+x^2}} = \frac{1}{x^2}. $$ Note that you missed the initial term $a = \frac{1}{1+x^2}$ in your computation.

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When $x = 0$, the series diverges because it is the infinite sum $1 + 1 + 1 + \cdots$.

Answer 2

Another way of solving it is by bringing it to the standard form, then you can use the formula for infinite geometric series considering $x \ne 0$ take $n=m+1$, so you have:

$$ \sum_{m = 0}^{\infty} \frac{1}{(1 + x^2)^{m+1}}= \frac{1}{(1 + x^2)} \sum_{m = 0}^{\infty} \frac{1}{(1 + x^2)^{m}}=\frac{1}{(1 + x^2)}\cdot \frac{(1 + x^2)}{ x^2}=\frac{1}{x^2}$$