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To find radius of convergence of geometric series $$\sum_{n=1}^\infty a_n$$ I need to use ratio/root test to find $|L|<1$

To find radius of convergence of power series $$\sum_{n=1}^\infty c_n (x-a)^n$$ I am supposed to find the limit $L$ of just the constant term $c_n$?

$$\sum_{n=1}^\infty \frac{(2x-5)^n}{n^2}, \qquad c_n = \frac{2^n}{n^2}, \qquad R = 2^{-1}=1/2$$

How do I know what is a constant? Free of $n$? Or I should focus on getting it into the form $\sum_{n=1}^\infty c_n (x-a)^n$. Thats what I did below

$$\sum_{n=1}^\infty \frac{(x-1)^{n-3} + (x-1)^{n-1}}{4^n + 2^{2n-1}}$$

I first factorized what I can

$$ = \sum_{n=1}^\infty (x-1)^n \cdot \frac{(x-1)^{-3} + (x-1)^{-1}}{4^n(1 + 2^{-1})}$$

So I guess $$c_n=\frac{(x-1)^{-3} + (x-1)^{-1}}{4^n(1 + 2^{-1})}$$

Then $$L=|\frac{c_{n+1}}{c_n}|$$ But what do I do with the $x$?

Correct method is supposed to be knowing its a geometric series of common ratio $\frac{x-1}{4}$. Then radius is |x-1|<4. But here, don't I need to get the radius of $x$ alone, without the $-1$?

Anyways, I think the main thing I am confused about is when to use which method. For geometric series, I am doing the ratio/root test for the whole $a_n$ while in power series, I am "separating" the $a_n$ into $c_n (x-a)^n$ form? And doing test on the constant terms ($c_n$)?

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up vote 4 down vote accepted

The ratio test is a more general approach than the "recognize it as a geometric sum test" - obviously because the latter won't work if the sum is not, in fact, a geometric series - but they are both valid.

One thing to note is that if you compute $|c_{n+1}/c_n|$ using your definitions of $c_n$ in the example you give, the numerator $(x-1)^{-3}+(x-1)^{-1}$ does not depend on $n$ and so it will disappear when we compute the ratio of the two coefficients. Same with $(1+2^{-1})$ in the denominator.

Define $f_n:= b_n(x-a)^n$ and consider the power series

$$f(x)=\sum_{n=0}^\infty f_n=\sum_{n=0}^\infty b_n(x-a)^n.$$

If $\lim\limits_{n\to\infty} |b_{n+1}/b_n|=L$ then our "second" test says the radius of convergence is $R=1/L$. Why? Well,

$$\lim_{n\to\infty}\left|\frac{f_{n+1}}{f_n}\right| =\lim_{n\to\infty}\left|\frac{b_{n+1}(x-a)^{n+1}}{b_n(x-a)^n}\right|=L|x-a|,$$

so the series $\sum\limits_{n=0}^\infty f_n$ converges if $\big|L|x-a|\big|<1$ by our "original" ratio test, equivalently $|x-a|<R$, which defines a circle of radius $R$ (note: $a$ is the center of the circle and tells us nothing about the radius $R$). This exhibits how the "secondary" test, of just comparing the coefficients of $(x-a)^n$, is actually a shortcut to applying the original ratio test, where we test the sequence $f_n$, so to speak.

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Without using ratio test u can also find radius of convergence in this example,

$\sum_{n=1}^\infty c_n$ $\frac{(2x-5)^n} {n^2}$

Comparing the given series with power series,


We have, $c_n=\frac{1}{2^n n^2}$ and $x_0=(\frac{5}{2})^n$

$R=\lim_{n\to \infty}|c_n|^\frac{1}{n}$

$\ \ \ =\lim_{n\to \infty}|\frac{1}{2[(n)^\frac{1}{n}]^2}| $

$\ \ \ R=\frac{1}{2}$

$\ \ \ R=\frac{1}{r}$

$\ \ \ r=2 $

$\therefore$ radius of convergence is 2

Moreover,interval of convergence $(x_0-r,x_0+r)$

In this example root test is very useful because here constant having power n

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