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How to find the convergence radius of $$\sum_{k=1}^{\infty}\frac{(-5)^k}{(k+1)!}\left(\frac x2\right)^{2k-1}$$

I guess I should use $R = \lim_{n\rightarrow\infty} \left| \frac{c_n}{c_{n+1}} \right|$, but I have trouble differentiating between $c_k$ - the $k^{th}$ complex coefficient, and $x$ - the complex variable.

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    You guessed right. I tend to use $\frac{c_{n+1}}{c_n}$ – 2017-02-28
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    If you want to deal with every case in one stroke, simply define $$a_k=\frac{(-5)^k}{(k+1)!}\left(\frac x2\right)^{2k-1}$$ and consider $$r_k=\left|\frac{a_{k+1}}{a_k}\right|$$ In the present case, you are left with $$r_k=\frac{25|x|^2}{4(k+3)(k+2)}$$ hence $$r_k\to0$$ for every $x$ hence the radius of convergence is $$R=\infty$$ More generally, you will obtain $$r_k\to0$$ for every $|x|$$r_k\to\infty$$ for every $|x|>R$, which will allow to determine $R$. – 2017-02-28
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    @Did. I probably made a mistake somewhere since I get $r_k=\frac{5 |x|^2}{4 (k+2)}$. The problem is that I don't know where. – 2017-02-28
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    @ClaudeLeibovici My bad, somehow $x^{2k-1}$ must have influenced my view of $(-5)^k$ and transformed it into $(-5)^{2k-1}$... or whichever reason. Thanks for the correction, your $r_k$ is the right one. – 2017-02-28
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    @Did. You are very welcome, for sure ! – 2017-02-28

2 Answers 2

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Hint: First change the summation index: $$\sum_{k=1}^{\infty}\frac{(-5)^k}{(k+1)!}\left(\frac x2\right)^{2k-1}=\sum_{k=0}^{\infty}\frac{(-5)^{k+1}}{(k+2)!}\left(\frac x2\right)^{2k+1}$$

Secondly, split $x/2$ from the sum and isolate the power of $x$: $$\frac{x}{2}\sum_{k=0}^{\infty}\frac{(-5)^{k+1}}{2^{2k}(k+2)!}x^{2k}.$$

Now substitute: $x^2 = u$ and calculate the radius of convergence for: $$\sum_{k=0}^{\infty}\frac{(-5)^{k+1}}{2^{2k}(k+2)!}u^k.$$

After having evaluated the radius of convergence for $u$ (e.g. by ratio rule) take the square root and you are done. Note, that the $x/2$ does not change the radius of convergence, that is why it is left out (see comment).

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    Why can we split x/2 from the sum? – 2017-02-28
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    It does not change the radius of convergence. In non-academic examples, the radius of convergence is always dictated by the coefficients for $k \to \infty$, hence the higher order powers. Another way of thinking about it is thinking of $x/2$ as a power series, which has obviously an infinitely large radius of convergence multiplied by a power series, which has a radius of convergence of $R$ (which needs to be determined). The radius of convergence of the product will be the smallest of both radii, hence $R$. – 2017-02-28
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Every time you increase $k$, you multiply the current term by $-5\left(\dfrac x2\right)^2$ and divide it by $k+2$. As soon as $k+2$ exceeds $\dfrac{5x^2}4$, the absolute series becomes bounded by a geometric series of common ratio less than $1$, and converges.