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how do you find the values of x for which the series (x+1)^n from n=1 to infinity converges?
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from -2 to 0 yes
they gotta make it a fraction in there
guru don't you even sleep?
nvm, its ratio test right?
i thought i was asleep lol been doing algebra all day and its soooo boring
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\[\sum x^n\] converges for (-1,1) so \[\sum(x+1)^n \] just move over one. i think it is that simple
yeah I had found it, thanks for the help though guys
i am sure you can use a test if you like but there is no difference between x and x +1. if the first converges on (-1,1) the second does on (-2,0)
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