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I'm having some trouble understanding why the following power series interval of convergences is equal to 0.

$$\sum\limits_{n=0}^{\infty} \frac{n!x^n}{100^n}$$

According to my calculation, my answer is equal to $-100 < x < 100$ since I end up with: $|x/100| < 1$

I did it with the ratio test.

Can somebody explain to me why it equal to $0$? Thank you

Dave
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ATrashInTheWorld
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1 Answers1

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Let $a_n:=\frac{n!x^n}{100^n}$. Then for all $x\neq 0$ we have $$\left|\frac{a_{n+1}}{a_n}\right|=\frac{(n+1)|x|}{100}\xrightarrow{n\to\infty}\infty.$$ Factorials beat powers!

Dave
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  • You don't need $x$ to determine the radius of convergence – Bernard Dec 09 '19 at 23:35
  • OK, so all that multiplied by infinity equal to infinity, then 1/ifinity = 0 right? – ATrashInTheWorld Dec 09 '19 at 23:36
  • I'm not sure why you're taking reciprocal. The sum $\sum_{n=0}^\infty a_n$ diverges for $x\neq 0$ by the ratio test (seeing my above computation). See here: https://en.wikipedia.org/wiki/Ratio_test – Dave Dec 09 '19 at 23:38
  • @Bernard I was just using the ratio test on the series $\sum_{n=0}^\infty a_n$. – Dave Dec 09 '19 at 23:40
  • School homework restrictions, we only learned the ration test so far. I know there are many other tests, but it is the only one we learned. Else, I just want to thank you for making me realising it – ATrashInTheWorld Dec 09 '19 at 23:46
  • But I am using the ratio test. – Dave Dec 09 '19 at 23:47
  • Yes, but it is wall known that for power series it is enough to apply this test on the coefficients (same for the root test). – Bernard Dec 09 '19 at 23:47