the series ((-1)^n+1)(x-2)^n)/n .... the interval of convergence is 1

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the series ((-1)^n+1)(x-2)^n)/n .... the interval of convergence is 1

Mathematics
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another end point one, this time i do it with paper before typing
\[|x-2|<1\]
ok got it

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now we try replacing x by 1 btw you have done all the hard work, the rest should be the easy part
at x=1 i think it is \[\sum_{?}^{?} (-1)^{n+1}(-1)^{n}\div n\] ???
ok very smilar to the last one
but this time we don't do anything stupid \[(-1)^{n+1}(-1)^n=(-1)^{2n+1}=-1\]
on account of \[2n+1\] is odd.
so does not converge at x = 1
what do you want to bet it does converge at x = 3?
.ah! um a lot? haha.
yeah well we can just about visualize it because it is the same as the last one. this time you get \[(-1)^{n+1}\times 1=(-1)^{n+1}\]so we are in good shape
thank you! another question...
this one alternates in other words, whereas the other one does not
keep posting
you got zarkon watching so if i mess up he will step in wish he had stepped in to the last one so i wouldn't have made a fool out of myself

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