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 $(1, 3]$

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