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.

  • $\begingroup$ 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

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

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


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 $R = 1$.

  • $\begingroup$ 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

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)).

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

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