Thursday, 7 January 2016

limits - Find the convergence radius of series




I need to find the convergence radius of the following series:
n=0cosin×zn
Since
cosin=coshn=ein+ein2
therefore, using D'Alembert's property I find the limit as
lim
Therefore the radius should be 1. But the answer on the book is {\frac{1}{e}}.
What's wrong with my solution?


Answer



You have two errors here. The first is that

\cosh n = \frac{e^n + e^{-n}}{2}
not what you wrote (which is \cos n. To remember: \cosh is unbounded on \mathbb{R}, so you have "true" exponentials, while \cos is bounded, so you have e^{i x} which is bounded). So you need to compute
\lim_{n\to\infty} \frac{e^{n+1}+e^{-(n+1)}}{e^{n}+e^{-n}} \tag{1} and not \lim_{n\to\infty} \frac{e^{i(n+1)}+e^{-i(n+1)}}{e^{in}+e^{-in}} (where you also made a mistake, the second: this limit does not exist, anyway!).



So let's compute (1):
\frac{e^{n+1}+e^{-(n+1)}}{e^{n}+e^{-n}} = \frac{e^n(e+e^{-(2n+1)})}{e^n(1+e^{-2n})} = = \frac{e+e^{-(2n+1)}}{1+e^{-2n}} \xrightarrow[n\to\infty]{} \frac{e+0}{1+0} = e\,.
Therefore,
\lim_{n\to\infty} \frac{a_{n+1}}{a_n} = z\cdot e \tag{2}
which explains why the radius is \frac{1}{e}.


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