0
$\begingroup$

I need to find the radius of convergence of the following power series: $$\sum_{n=2}^{\infty} \dfrac{x^{n}}{n^2-n}$$

I know the radius of convergence must be 1 from an online calculator. However to show this I must show the following: $$\lim_{n \to \infty}sup |\dfrac{1}{n^2-n}|^{1/n} = 1$$

I'm not sure how I'm supposed to go about doing this! Following this the radius of convergence would be that to the power of -1, and hence 1.

$\endgroup$
  • $\begingroup$ Sorry but "$\sum\limits_{n=2}^{\infty} \frac{x^{n}}{n^2-n} = 1$" is not a series (but "$\sum\limits_{n=2}^{\infty} \frac{x^{n}}{n^2-n}$" is). $\endgroup$ – Did Mar 29 '16 at 22:09
  • $\begingroup$ Whoops! Corrected. $\endgroup$ – Uq'''12wn1F12u2x3uW31H1JBk9m Mar 29 '16 at 22:17
1
$\begingroup$

It is easier to use the ratio test; rewrite the sequence as: $a_n = \frac1{n(n-1)}$. Then,

$$\left| \frac{a_{n+1}}{a_n} \right| = \frac{n-1}{n+1}$$

(Edit):

So$$\lim \left| \frac{a_{n+1} x^{n+1}}{a_n x^n} \right| = |x|$$

If $|x| < 1$, then.. and if $|x| > 1$, then.. Conclusion $R = 1$.

$\endgroup$
  • $\begingroup$ I'm not sure how this helps me find the radius of convergence? I thought the ratio test is just for straight up series and not power series! $\endgroup$ – Uq'''12wn1F12u2x3uW31H1JBk9m Mar 29 '16 at 22:11
  • $\begingroup$ @Amir A power series is a perfectly good series. The ratio test allows you to conclude convergence if $|x| < 1$ and divergence if $|x| > 1$, so the radius of convergence must be $1$. $\endgroup$ – Bungo Mar 29 '16 at 22:37
1
$\begingroup$

As the answer by Ahmed Hussein shows, the ratio test is easier to apply in this problem. However, with a bit of effort, we can also use the root test.

For $n \geq 3$, define $x_n = (n^2 - n)^{1/n} - 1$ and note that $x_n > 0$. Then $$n^2 - n = (x_n + 1)^n = \sum_{k=0}^{n}{n \choose k}x_n^k$$ by the binomial theorem. Since the summands are nonnegative, the sum is at least as large as the $k=3$ term, so $$n^2 - n \geq {n \choose 3}x_n^3 = \frac{n(n-1)(n-2)}{6}x_n^3 \geq 0$$ which means that $$0 \leq x_n^3 \leq \frac{6(n^2 - n)}{n(n-1)(n-2)} = \frac{6}{n-2}$$ and consequently $$0 \leq x_n \leq \left(\frac{6}{n-2}\right)^{1/3}$$ Since the right-hand side converges to zero as $n \to \infty$, we have $x_n \to 0$ by the sandwich theorem. Recalling the definition of $x_n$, this means that $$\lim_{n \to \infty}(n^2 - n)^{1/n} = 1$$ and so also $$\limsup_{n \to \infty} \left|\frac{1}{n^2 - n}\right|^{1/n} = \left|\frac{1}{\lim_{n \to \infty}(n^2 - n)^{1/n}}\right| = 1$$

Note: I borrowed the basic idea of this argument from Baby Rudin's proof that $\lim_{n \to \infty}n^{1/n} = 1$ (Theorem 3.20(c)).

$\endgroup$
  • $\begingroup$ A more involved but awesome second way of looking at this - Thank you! $\endgroup$ – Uq'''12wn1F12u2x3uW31H1JBk9m Mar 29 '16 at 23:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.