# Radius of convergence of power series $\sum_{n=0}^{\infty} \frac{(2x-5)^n}{n^2}$

Consider the below example: $$\sum_{n=0}^{\infty}\frac{(2x-5)^n}{n^2},\qquad c_n=\frac{2^n}{n^2},\qquad R=2^{-1}=\frac{1}{2}.$$

\begin{align*} \lim_{n\to\infty}\frac{c_{n+1}}{c_n} &= \lim_{n\to\infty}\frac{2^{n+1}/(n+1)^2}{2^n/n^2} \\ &=\lim_{n\to\infty}\frac{2n^2}{(n+1)^2}\\ &= \lim_{n\to\infty}\frac{2}{(1+\frac{1}{n})^2}\\ &= 2. \end{align*}

How come the expression $$\sum_{n=0}^{\infty} \frac{(2x-5)^n}{n^2}$$ does not appear to be used? Is $c_n$ derieved from that? How?

UPDATE

In another example I have:

$$\sum_{n=1}^{\infty} \frac{n^2(x-3)^{n+1}}{5^n}, \qquad c_{n+1}=\frac{n^2}{5^n}$$

If I try to express the series in

$$\sum_{n=1}^{\infty} c_n (x-a)^n$$

I get

$$\frac{n^2 (x-3)}{5^n} \cdot (x-3)^n, \qquad c_n = \frac{n^2 (x-3)}{5^n}$$

Or can the $(x-3)$ be removed somehow?

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You should start your sum at $n=1$, not $n=0$ (to avoid division by 0). – David Mitra Mar 11 '12 at 2:04

If I understand you correctly, the answer is "yes". The $c_n$ are the coefficients of the power series.

Given the power series $\sum\limits_{n=1}^\infty c_n(x-a)^n$, the radius of convergence can be calculated from the limit $\lim\limits_{n\rightarrow\infty}{ |c_{n+1}|\over| c_n|}$, provided this limit exists (the radius of convergence is the reciprocal of this limit). Your series is not quite in the right form to use this, though. You need to write it as $$\sum\limits_{n=1}^\infty { (2x- 5)^n\over n^2}= \sum\limits_{n=1}^\infty {2^n(x-2.5)^n\over n^2} =\sum\limits_{n=1}^\infty {{2^n\over n^2}(x-2.5)^n };$$ so, $c_n={2^n\over n^2}$.

For your second example, $$\tag{1} \sum\limits_{n=1}^\infty {n^2 (x-3)^{n+1}\over 5^n},$$ think of the formula $\lim\limits_{n\rightarrow\infty}{ |c_{n+1}|\over| c_n|}$ as taking the limit of "the absolute value of a general coefficient of the series divided by the absolute value of the preceding coefficient". Keep in mind the coefficients are the numbers in front of the $(x-2)^k$ terms.

So for the series given in $(1)$, the coefficients are ${n^2\over 5^n}$ and you'd evaluate $\lim\limits_{n\rightarrow\infty}{ n^2/5^{n }\over (n-1)^2/5^{n-1}}$. You would not include any expressions containing $x-3$ when using the formula.

Note that in $(1)$, you have $c_{n+1}= {n^2\over 5^n}$, as in your lecture notes (so $c_n={(n-1)^2\over 5^{n-1}}$).

What you did in your attempt is not the correct approach. If you wanted to write the series so that $x-2$ was raised to the index power, you'd have to reindex the series (though, there is no reason to do this): $$\sum\limits_{n=1}^\infty {n^2 (x-3)^{n+1}\over 5^n} =\sum\limits_{m=2}^\infty {(m-1)^2 (x-3)^{m}\over 5^{m-1}}.$$ Here $c_m={(m-1)^2\over5^{m-1}}$.

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Thanks for clarifying my doubts, but I have nother question where my attempt is different from whats given in the lecture notes, please see update if you can :) – Jiew Meng Mar 11 '12 at 8:18

Hint:

Let $$a_{0}+a_{1}z+a_{2}z^{2}+a_{3}z^{3}+a_{4}z^{4}+\cdots$$ be a power-series, and consider the series $$a_{1}+a_{2}z^{1}+a_{3}z^{2}+a_{4}z^{3}+\cdots$$ which is obtained by differentiating the power series term by term. The derived series has the same circle of convergence as the original series. Similarly the series $$\sum _{n =0}^{\infty }\dfrac {a_{n}z^{n+1}} {n+1},$$ obtained by integrating the original power-series term by term, has the same circle of convergence as $\sum _{n=0}^{n=\infty }a_{n}z^{n}$.

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Should the second series be $a_1+2 a_2 z^1 + 3 a_3 z^2 + 4 a_4 z^3 + \cdots$? – Antonio Vargas Mar 11 '12 at 15:42
@AntonioVargas Yes it is meant, thank you for correcting me. – Hardy Mar 11 '12 at 19:48