$\sum_{n=1}^{\infty}\frac{(-1)^{\lfloor\log_2n\rfloor}}{\lfloor\log_2n\rfloor+1}(x-x_0)^n$ convergence/divergence

I have a problem with determining whether these series are convergent/divergent at the endpoints of their radii of convergence. None of the tests or approaches I know seems to by applicable here...

$$\sum_{n=1}^{\infty}\frac{(-1)^{\lfloor\log_2n\rfloor}}{\lfloor\log_2n\rfloor+1}(x-x_0)^n$$

$$\sum_{n=1}^{\infty}\frac{(-1)^{\lfloor\log_2n\rfloor+n}}{\lfloor\log_2n\rfloor+1}(x-x_0)^n$$

It is immediate that the radii of convergence are 1. The intervals of convergence are supposedly $\langle x_0-1,x_0+1)$, resp. $(x_0-1,x_0+1\rangle$, which doesn't seem correct, as $\sum_{n=1}^{\infty}\frac{(-1)^{\lfloor\log_2n\rfloor+1}}{\lfloor\log_2n\rfloor+1}$ and $\sum_{n=1}^{\infty}\frac{(-1)^{\lfloor\log_2n\rfloor}}{\lfloor\log_2n\rfloor+1}$ either both converge or diverge.

But the question is: how to determine whether these series convergent/divergent?

$\sum_{n=1}^{\infty}\frac{(-1)^{\lfloor\log_2n\rfloor+1}}{\lfloor\log_2n\rfloor+1}$,$\sum_{n=1}^{\infty}\frac{(-1)^{\lfloor\log_2n\rfloor}}{\lfloor\log_2n\rfloor+1}$,$\sum_{n=1}^{\infty}\frac{(-1)^{\lfloor\log_2n\rfloor+n}}{\lfloor\log_2n\rfloor+1}$,$\sum_{n=1}^{\infty}\frac{(-1)^{\lfloor\log_2n\rfloor+n+1}}{\lfloor\log_2n\rfloor+1}$

Thanks!

There are two different problems here, $x-x_0=1$ and $x-x_0=-1$. We look first at $x-x_0=1$.

So we are interested in the series $$\sum_1^\infty \frac{(-1)^{\lfloor \log_2 n\rfloor}}{\lfloor \log_2 n\rfloor +1 }.$$

Look at the sum of the terms from $n=2^m$ to $n=2^{m+1}-1$, where $m$ is large. Over this interval $\lfloor\log_2 n\rfloor$ does not change. Let us imagine (it makes no real difference) that $m$ is even.

Then the sum of the terms from $n=2^m$ to $n=2^{m+1}-1$ is equal to $\frac{2^m}{m+1}$, which is large. So the partial sums do not have a limit, and therefore the series diverges.

The case $x-x_0=-1$, despite the forbidding appearance, turns out to be pleasantly trivial. Again consider the sum from $n=2^m$ to $n=2^{m+1}-1$. For $m\ge 1$, there is an even number of terms of equal absolute value, and alternating in sign! They cancel. Since the terms have limit $0$, we have convergence.

• I have seen this before and realized that. What confused me is that multiplying by $-1$ we get convergent series. Interesting! What was really helpful was pointing out the Leibniz test. I thought it is not applicable and discarded it, without thinking about modifying it, thanks! Jul 9 '14 at 16:00
• Another modification of Leibniz would prove convergence of $\sum_{n=1}^{\infty}\frac{(-1)^{\lfloor\log_2n\rfloor+n}}{\lfloor\log_2n\rfloor+1}$ I guess? Jul 9 '14 at 16:10
• You are welcome. For the case of the denominator we used, everything is trivial, since there is complete cancellation. The "Leibniz" argument I mentioned is not necessary, though it would be for denominators that did not have such a simple form. Jul 9 '14 at 16:13