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??


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

  • $\begingroup$ 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

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