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Consider the initial value problem, $$u''+(x+1)u'-u=0 \ \ \ \ \ \ u(-1)=2, u'(-1)=0.$$ I wish to find the radius of convergence of the power series solution (about $x=-1$).

My attempt:

I have found that the recurrence relation is $$A_{k+2}=-\frac{(k-1)A_k}{(k+2)(k+1)} \ \ k\geq 0$$ To find the radius of convergence, I use the ratio test (where $z=x+1$), $$\lim_{k\to\infty}\left|\frac{A_{k+2}z^{k+2}}{A_kz^k}\right|=\lim_{k\to\infty}\left|\frac{1-k}{(k+2)(k+1)}\right||z^2|=0.$$ Now by the ratio test, this converges absolutely as $0<1$. The answer I have been given states the radius of convergence is $\infty$. Why is this?

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    $\begingroup$ Take another look at the radius of convergence using the ratio test. $\endgroup$
    – MSDG
    Sep 8 '18 at 9:42
  • $\begingroup$ The $z^k$ near the end of the last line should be $|z|^2$. $\endgroup$ Sep 8 '18 at 9:44
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    $\begingroup$ @sobi It says that "Note that r = 1/0 is interpreted as an infinite radius", Does this effectively answer my question? $\endgroup$
    – user557493
    Sep 8 '18 at 9:48
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    $\begingroup$ @Bell I would say so! $\endgroup$
    – MSDG
    Sep 8 '18 at 9:50
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    $\begingroup$ @Bell If $\alpha = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|$ is not equal to $0$, then the radius of convergence $R = \frac{1}{\alpha}$ is finite. $\endgroup$
    – MSDG
    Sep 8 '18 at 9:52
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As you've noted, $$|z|^2 \cdot \lim_{k \to \infty} \left|\frac{1 - k}{(k + 2)(k + 1)}\right| = 0$$ for all $k$, regardless of the value of $z$. The ratio test tells us that, when this limiting ratio is strictly less than $1$, the series converges absolutely.

The logic behind this involves comparison to a geometric series. We know that a series with common ratio $r$ such that $|r| < 1$ converges. If we have a limiting ratio of $L < 1$, then the gaps between terms will eventually become less than gaps between terms of a geometric series with common ratio $\frac{L + 1}{2} < 1$. Such a geometric series converges, so the original series must converge too.

In this case, we have limiting ratio is $0$, which is an extreme case. Not only will it converge, it will converge quickly!

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