I have been trying to get the radius of convergence of this expression $\sum_{k=1} ^{\infty} \frac{2^k*z^{2k}}{k^2+k}$ by using the ratio test. However I keep getting $\frac{1}{2}$ and the answers say the solution is $\frac{1}{\sqrt{2}}$, could anyone explain me why they get this answer? Thanks!
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The series has $z^2$ everywhere, so it converges for $|z^2| < 1/2$. – KCd Mar 26 '16 at 13:13
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Thanks! THat las comment makes a lot of sense! – urpi Mar 26 '16 at 13:52
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Observe that in order to be this series to be considered a power series, we have to take the general sequence in accord to the "power", which is $\;k\;$ of $\;z^2\;$ , not of $\;z\;$ .
You can apply Cauchy-Haddamard's formula for the whole thing instead of only for the coefficients in order to avoid confussions:
$$\sqrt[k]{\frac{2^k|z|^{2k}}{k^2+k}}=\frac{2|z|^2}{\sqrt[k]{k^2+k}}\xrightarrow[k\to\infty]{}2|z|^2<1\iff|z|<\frac1{\sqrt2}$$
DonAntonio
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Thanks! But isn't the Cauchy-Haddamard formula of the form $R=\frac{1}{limsup |a_k|^{1/k}}$? – urpi Mar 26 '16 at 13:54
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$$\sum\limits_{k=1}^{\infty} \frac{2^k z^{2k}}{k^2+k}$$ Using the ratio test, we have $$\lim\limits_{n\to\infty}\left|\frac{\frac{2^{n+1} z^{2n+2}}{(n+1)^2+n+1}}{\frac{2^n z^{2n}}{n^2+n}}\right|$$ $$=\lim\limits_{n\to\infty}\left|\frac{2^{n+1} z^{2n+2}n(n+1)}{(n+1)(n+2)2^n z^{2n}}\right|$$ $$=\lim\limits_{n\to\infty}\left|\frac{2^{n}2 z^{2n}z^{2}n}{(n+2)2^n z^{2n}}\right|$$ $$=\lim\limits_{n\to\infty}\left|\frac{2n z^{2}}{n+2}\right|$$ $$=2 \lim\limits_{n\to\infty}\left|\frac{z^2}{1+\frac2n}\right|$$ $$=2 \left|z^2\right|$$ Therefore $$2 \left|z^2\right|\lt 1$$ $$\left|z\right|^2\lt \frac12$$ $$\left|z\right|\lt \frac{1}{\sqrt 2}$$
k170
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I thought the ratio test only used the coefficients of the series not the $z$'s too. – urpi Mar 26 '16 at 13:53
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@urpi, $z$ is indeed part of the series and the convergence of the series is dependent on the value of $z$. – k170 Mar 26 '16 at 18:56