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Prove that if $\sum_{n=0}^{\infty}na_nx^{n-1}$ converges for $|x| \lt r$, then $\sum_{n=0}^{\infty}a_nx^n$ also converges for $|x| \lt r$, i.e. have the same radius of convergence.

I tried to apply comparison test, but only get absolute convergence, i.e. I know how to prove if $\sum_{n=0}^{\infty}|na_nx^{n-1}|$ converge then $\sum_{n=0}^{\infty}|a_nx^n|$ also converges. But I'm stuck on showing the above property. Could someone give a proof please? Thanks.

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  • $\begingroup$ Hint: ratio test. $\endgroup$ – David Heras Feb 17 '16 at 2:27
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A power series $\sum a_nx^n$ converges absolutely if $|x| < R$ and diverges if $|x| > R$, where the radius of convergence can be found as

$$R = \left(\limsup_{n \to \infty}|a_n|^{1/n}\right)^{-1}.$$

Note that

$$\limsup_{n \to \infty}|na_n|^{1/n} \leqslant \limsup_{n \to \infty}|n|^{1/n} \limsup_{n \to \infty}|a_n|^{1/n}= 1 \cdot \limsup_{n \to \infty}|a_n|^{1/n}, $$

and, since $|a_n| \leqslant |na_n|,$

$$\limsup_{n \to \infty}|a_n|^{1/n} \leqslant \limsup_{n \to \infty}|na_n|^{1/n}.$$

Hence,

$$\limsup_{n \to \infty}|na_n|^{1/n} = \limsup_{n \to \infty}|a_n|^{1/n}.$$

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  • $\begingroup$ Thanks, but isn't this a proof of $\sum_{n=0}^{\infty}|na_nx^{n-1}|$ converges? I'm trying to prove $\sum_{n=0}^{\infty}a_nx^n$ converges. $\endgroup$ – user57891 Feb 17 '16 at 2:41
  • $\begingroup$ Reread the last line. $\sum na_nx^{n-1}$ converges for $|x|<R$ <-> the RHS is finite <-> the left hand side is finite <-> $\sum a_nx^n$ converges for $|x|<R$ $\endgroup$ – Stella Biderman Feb 17 '16 at 2:44
  • $\begingroup$ The two series must have the same radius of convergence. If the first series converges for $|x| < r$ then $r \leqslant R$ and the second series converges here as well. $\endgroup$ – RRL Feb 17 '16 at 2:46
  • $\begingroup$ We are using $\limsup ( a_n b_n) \leqslant \limsup a_n \limsup b_n$ which is not too difficult to show using the definition of $\limsup$. Also $\limsup n^{1/n} = \lim n^{1/n} = 1$ which is also straightforward. $\endgroup$ – RRL Feb 17 '16 at 3:13
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You want to show that, if $R = \left(\limsup_{n \to \infty}|na_n|^{1/n}\right)^{-1} $ then $R = \left(\limsup_{n \to \infty}|a_n|^{1/n}\right)^{-1} $.

This follows from $\lim_{n \to \infty} n^{1/n} = 1 $.

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If $\sum n a_n x^n$ converges for $|x|<r$, then for any $0<r_1<r$ there holds $\lim_{n\to \infty}na_nr_1^n=0$. In particular, for all large enough $n$ you get $|na_nr_1^n|<1$. This means that $|a_n|< 1/(nr_1^n)<1/r_1^n$ for $n$ large enough.

Now if you want to show for a given $x,|x|<r$ that $\sum a_n x^n$ converges, then pick $r_1 with $|x|

The argument is a bit more subtle in the other direction (that the derivative has radius of convergence at least as big as the original), since the inequality would be $|a_n|<n/r_1^n$. So you would end up comparing with $\sum nq^n$ with $0<q<1$. While not a geometric series, it also converges, for example by a ratio test.

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