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We have just started learning this and I do not really understand it fully.

Given: $$ \sum_{n=1}^{\infty} {(\pi^n + n + 1) \cdot x^n}$$

We should check what is the radius of convergence $R$ and what happens in the far ends.

So what I did so far is find $R$:

$$ R = \lim_{n \to \infty} { \left\lvert\frac{c_n}{c_{n+1}} \right\rvert} = \lim_{n \to \infty} {\frac{\pi^n + n + 1}{\pi^{n+1} + n + 2}} = 1 = R$$

What does that exactly mean? That the power sum converges in the area $(-1,1)$ ? And how to check what happens in the far ends?

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The limit is not $1$. Try again. –  Pedro Tamaroff Apr 25 '13 at 19:03
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Here is a not so rigorous approach: The Radius of Convergence can be seen as a sort of a Domain of a function. As long as you pick values of x that are within the Interval of Convergence (which is determined by the ROC)then your series will converge to the same value as your original function of which you have set up its powerseries. Conversely, if you take a value of x that falls outside yout IOC, then the powerseries is going to behave divergent and thus it won't produce the y value from the function (if it produces anything at all). –  imranfat Apr 25 '13 at 19:05
    
And the limit is indeed not 1, you forgot something with the pi –  imranfat Apr 25 '13 at 19:09

2 Answers 2

up vote 1 down vote accepted

The radius of convergence is: $$R = \lim_n \frac{\pi^n + n + 1}{\pi^{n+1} + (n+1) + 1} = \lim_n \frac{\pi^n(1 + \frac{n+1}{\pi^n})}{\pi^{n+1}(1 + \frac{n+2}{\pi^{n+1}})} = \frac{1}{\pi}.$$

It is clear or do you need further explanaitions? :D

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Thanks alot :)) –  TheNotMe Apr 25 '13 at 19:14
    
Wait, you said it should include the x-part. Your limit doesn't. Why? –  TheNotMe Apr 25 '13 at 21:05
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I was writing something and then changed my mind, it's wrong an meaningless, of course! –  user01123581321345589144... Apr 25 '13 at 21:14

$$(\pi^n + n + 1) x^n\sim_\infty \pi^n x^n$$ so $$R=\frac{1}{\pi}$$

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To be honest? I understood nothing. –  TheNotMe Apr 25 '13 at 19:04
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@TheNotMe this means that the series $\sum (\pi^n+n+1)x^n$ and $\sum \pi^n x^n$ have the same radius of convergence since $\lim_\infty \frac{(\pi^n+n+1)}{\pi^n}=1$ –  Sami Ben Romdhane Apr 25 '13 at 19:09

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