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I'm trying to understand how to handle power series who use floor or ceiling functions in their general term.

For example, the power series $\displaystyle{\sum_{k \geq 1} \left\lfloor \frac{2^k}{(k+1)^2}\right\rfloor}x^k$ is supposed to have a radius of convergence of $\frac{1}{2}$ but I fail to see how to handle this.

I see how the series converges for values less than $\frac{1}{2}$, but why does it diverge for larger values?

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2 Answers 2

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You can also write $$a_k := \left \lfloor \frac{2^k}{(k+1)^2} \right \rfloor $$ and then it means that $$ a_k \leq \frac{2^k}{(k+1)^2} \leq a_k + 1 $$ and then multiplying by $|x|^k$ we get

$$ a_k |x|^k \leq \frac{2^k}{(k+1)^2} |x|^k \leq (a_k + 1) |x|^k $$

Now you can calculate the radius of convergence of the series $$\sum_{k = 1}^{\infty} \frac{2^k}{(k+1)^2} |x|^k $$ and it is equal to $1/2$.

And now you can conclude that the radius of convergence of the series $\sum a_k x^k$ is at least $1/2$ from the leftmost inequality. But using the rightmost inequality you can also see that the radius of convergence of the series $\sum a_k x^k$ cannot exceed $1/2$ because that would contradict the fact that the radius of convergence of the series $$\sum_{k = 1}^{\infty} \frac{2^k}{(k+1)^2} |x|^k $$ is $1/2$.

This is because if the series $\sum a_k x^k$ converges for some $x$ with $|x| > 1/2$ then the series $\sum (a_k + 1) x^k$ also converges because it is equal to the sum of the series $\sum a_k x^k + \sum x^k$, and then this would imply the convergence of the series $$\sum_{k = 1}^{\infty} \frac{2^k}{(k+1)^2} |x|^k $$

also for this $x$.

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  • $\begingroup$ I like this idea of simply squeezing the series, it's easy and elegant. Thanks for the help. $\endgroup$ Mar 28, 2011 at 20:26
  • $\begingroup$ It's interesting to note that in the very last part of your argument, we are actually using the fact that we can use an $x$ value smaller than 1 in order to guarantee the convergence of the $\Sigma x^k$ series. $\endgroup$ Mar 28, 2011 at 20:43
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    $\begingroup$ @David Yes you're right. That's because you can restrict to $|x| < 1$ from the start because otherwise the series diverges. $\endgroup$ Mar 28, 2011 at 20:58
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Notice that when $k$ is large enough, $\displaystyle{\left\lfloor \frac{2^k}{(k+1)^2}\right\rfloor \geq\frac{1}{2}\frac{2^k}{(k+1)^2}}$. From this you can apply the root test to see that the radius of convergence is at most $\frac{1}{2}$.

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  • $\begingroup$ Makes sense, thanks for the hint! $\endgroup$ Mar 28, 2011 at 19:44

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