I try to determine the convergence of the following power series

$$ \sum_{k=1}^\infty \frac{x^k}{\sqrt{k(k+1)}}$$

I tried it using the ratio test, but I am not sure if this is correct My approach:

$$ \lim_{k\rightarrow \infty} \left|\frac{\frac{1}{\sqrt{(k+1)(k+2)}}}{\frac{1}{\sqrt{k(k+1)}}}\right| = \frac{\sqrt{k(k+1)}}{\sqrt{(k+1)(k+2)}} = \frac{\sqrt{k^2+k}}{\sqrt{k^2+3k+2}} = \frac{k^2+k}{k^2+3k+2} = \frac{k^2(1+\frac{1}{k})}{k^2(1+\frac{3k}{k}+\frac{3}{k^2})} = \frac{1}{1}$$

That would mean that $r=1$ and the series converges for $|x|<1$. I know from WA that the series converges, but is my way to the solution correct?

  • $\begingroup$ In general this looks good but you forgot a square root on the way and the limit as well. If you fix this, it is gonna be ok. But be aware that your limit equals $1/r$ and not $r$ (which is irrelevant in this particular example of course). $\endgroup$ – frog Mar 10 '15 at 15:16
  • $\begingroup$ I am aware that the limit is 1/r, but what do you mean by I forgot the square root on my way? I mean, I just canceled the squares/quadrated both sides of the term. $\endgroup$ – Christoph S Mar 10 '15 at 16:38
  • $\begingroup$ You can't quadrate numerator and denominator: $2/3$ is not equal to $4/9$, for example. $\endgroup$ – zar Sep 25 '16 at 21:40

Not really. We have:
$$\left|\frac{\frac{1}{\sqrt{(k+1)(k+2)}}}{\frac{1}{\sqrt{k(k+1)}}}\right| = \frac{\sqrt{k(k+1)}}{\sqrt{(k+1)(k+2)}} = \frac{\sqrt{k}}{\sqrt{k+2}}.$$
Then let $k\to\infty$. (Your answer is right, of course.)

  • $\begingroup$ Okay, so you just canceled the (k+1) and then to finish I should have to move the sqrt to the whole term and then the limit inside ok. Is my approach, quadrating the terms correct or not? $\endgroup$ – Christoph S Mar 10 '15 at 16:45
  • 1
    $\begingroup$ @ChristophS Since $\sqrt 1 = 1$, your answer is right. If you get another limit, say $2$, then the convergent radius is $1/\sqrt 2$. $\endgroup$ – Eclipse Sun Mar 10 '15 at 23:32

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