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have to calculate the ratio of the serie in the title. So using the ratio test criteria I find that $\frac{9^{n+1}}{9^{n}}=9$ and so that $R=\frac{1}{9}$. My professor's result is $\frac{1}{3}$ however. I think that's for that $z^{2n}$ term, but I can't figure out. How can I see that the ray is $\frac{1}{3}$? Thankyou!

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    $\begingroup$ The ratio of consecutive terms gives a $z^2$ not a $z$. $\endgroup$ – Paul Feb 10 '16 at 9:59
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$$\sum_{n=0}^\infty9^nz^{2n}=\sum_{n=0}^\infty(9z^2)^n$$ will converge iff

$$\left|9z^2\right|<1\iff |z|^2<\dfrac19\iff |z|<\dfrac13$$

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Yes the answer is indeed $1/3$.because while finding the roc you not knowingly have taken $z^2=t $.and thats the reason after calculating $1/9$ you have to take its root

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The ratio test you used is wrong. Let $(a_n)$ be a sequence of $n$th coefficients of given series. List a few terms, then: $$ (a_n):1,0,9,0,81,0,\cdots $$ Therefore, $\lim_{n\to\infty} \frac{a_{n+1}}{a_n}$ does not exist. In this situation, the root test using $\lim$ is also unavailable, because $\lim_{n\to\infty} \sqrt[n]{a_n}$ doesn't exist also. Instead, you can use the convergence of geometric series like lab bhattacharjee's, or root test using $\limsup$, usually learned in basic real analysis course. Since $$\limsup_{n\to\infty} \sqrt[n]{|a_n z^n|}=3|z|,\text{ (why?)}$$ given series converges when $|z|<\frac{1}{3}$.

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