What is the interval of convergence of #sum_1^oo ((-1)^n(x-2)^n )/ (n+1) #?

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Answer 1 · 2015-12-13T14:37:14

The interval of convergence is #(1, 3]#

If #abs(x-2) < 1# then this sum converges faster than a geometric series with common ratio #x-2#.

If #x <= 1# then all of the terms are positive and greater than #1/(n+1)#, so the sum diverges.

If #x > 3# then for large enough #n# we find #(x-2) > (n+2)/(n+1)#, so the sum diverges faster than a geometric series with common ratio #< -1#.

If #x = 3# then the sum is:

#sum_(n=1)^oo (-1)^n/(n+1)#

which converges.

So the interval of convergence is #(1, 3]#