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$\sum k!(x+3)^k$

Ok, I've tried and I'm a bit stuck... The sum is something like:

$1+(x+3)+2(x+3)^2$

So $|\dfrac{x+3}{1}|<1 \Rightarrow -4<x<-2$

The answer in the book says the radius is $0$ and it will only converge at $x=-3$

However, my working suggests the radius is $1$..? Have I made a mistake somewhere?

Jim
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    It looks as if you were trying to use the Ratio Test, and got the ratio wrong, it is $(k+1)|x+3|$. – André Nicolas Jan 19 '15 at 20:24
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    Why did someone downvote this? He tried some things, showed his work and asked for help... I don't understand all the haters out there – jameselmore Jan 19 '15 at 20:25
  • @AndréNicolas Can I not work out the first few terms of the sequence and do the ratio test on them? What do I then do with |(k+1)(x+3)|<1? I thought you had to take the limit as k->infinity... I'm a bit confused now – Jim Jan 19 '15 at 20:37
  • @Alex I'm confused because the answer says the radius of convergence is - and it only converges at x=-3... so is the answer a typo? – Jim Jan 19 '15 at 20:42
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    @Jun: The "ratio" is $\frac{(k+1)!|x+3|^{k+1}}{k!|x+3|^k}$ which simplifies to $(k+1)|x+3|$. If $|x+3|\ne 0$, this has infinite limit as $k\to\infty$. So if $|x+3|\ne 0$ the series diverges. Of course it converges if $x=-3$. – André Nicolas Jan 19 '15 at 20:51

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