0
$\begingroup$

Find the radius of convergence of the Power series $$1+z+\frac{z^2}{2^2}+\frac{z^3}{3!}+\frac{z^4}{2^4}+\frac{z^5}{5!}\cdots $$

Put the series in the form

$$\left[1+\frac{z^2}{2^2}+\frac{z^4}{2^4}+\cdots\right]+\left[z+\frac{z^3}{3!}+\frac{z^5}{5!}+\cdots\right] =\left[\sum_{n=0}^\infty\frac{z^{2n}}{2^{2n}}\right] +\sinh(z)$$ Radius of convergence of the 1st series is $4$ & since $-\infty<\sinh(z)<\infty$ so the radius of convergence og the given series is $4$.

Is my approach right??

$\endgroup$
2
$\begingroup$

Your approach is in the right direction.

The first series $\displaystyle\sum_{n=0}^\infty\frac{z^{2n}}{2^{2n}}$ has radius of convergence 2 - try the root test for example, while the second one has infinite radius of convergence. Thus their sum has radius of convergence equal to 2.

Root test for $\{a_n\}_{n\in\mathbb N}$, where $$ a_n=\left\{\begin{array}{lll} 2^{-n} & \text{if} & \text{$n$ even}, \\ 0 & \text{if} & \text{$n$ odd}, \end{array}\right. $$ is $$ \limsup \lvert a_n\rvert^{1/n}=\frac{1}{2}, $$ and hence radius of convergence$=2$.

Note that $\displaystyle\sum_{n=0}^\infty\frac{z^{2n}}{2^{2n}}=\sum_{n=0}^\infty a_nz^n$.

$\endgroup$
  • $\begingroup$ Using root test the radius of convergence becomes 4. How it is 2? $\endgroup$ – Empty Sep 19 '14 at 2:17
  • $\begingroup$ @SayantanPanja: See edit. $\endgroup$ – Yiorgos S. Smyrlis Sep 19 '14 at 7:43
  • $\begingroup$ Here a_n is 1/(2)^(2n). How it is 0 when n is odd? $\endgroup$ – Empty Sep 22 '14 at 11:07
  • $\begingroup$ @SayantanPanja: No it is not! $\endgroup$ – Yiorgos S. Smyrlis Sep 22 '14 at 11:12
  • $\begingroup$ I can not understand the construction of the sequence $\{a_{n}\}$..Please explain clearly .. $\endgroup$ – Empty Jan 14 '15 at 12:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.