Series of functions $\log(1+x^{2n})$

Question

Does the series of functions $\sum_{n=0}^{+\infty} \log(1+x^{2n})$ converges for $|x|<1$? Obviously I've observed that for $x=0$ the series is convergent and for $x\ge1$ is divergent. How should I proceed? Thank you.

Answer 1

Given the series of functions $$ \sum_{n=0}^{+\infty} \log(1+x^{2n}) $$ you're interested in its convergence for $|x| < 1$. You've already observed the behavior at $x = 0$ and $x \geq 1$, which is a good start.

To study its convergence for $|x| < 1$, let's use the properties of the logarithm and the fact that the domain in question does not contain any non-positive integers.

Since $\log(1 + t) < t$ for $t > 0$ (you can see this from the series expansion of the logarithm or by considering the graph of $y = \log(1+t)$ and $y = t$), we can say: $$ \log(1 + x^{2n}) < x^{2n} $$

Now, consider the sum of the right-hand side: $$ \sum_{n=0}^{+\infty} x^{2n} $$ This is a geometric series with ratio $x^2$. Since $|x| < 1$, $|x^2| < 1$ as well, so this series converges (to $\frac{1}{1-x^2}$).

By the comparison test, if $\sum_{n=0}^{+\infty} x^{2n}$ converges and $0 \leq \log(1+x^{2n}) < x^{2n}$ for all $n$, then $\sum_{n=0}^{+\infty} \log(1+x^{2n})$ also converges for $|x| < 1$.

Hence, the given series converges for all $|x| < 1$.

I hope this helps!