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The alternating series `sum_(n=2) ^ oo (-1)^n b_n` is convergent if the following two conditions are satisfied:
1) `b_(n+1) <= b_n` for all n
2) `lim_(n - > oo) b_n = 0`
In the given series, `b_n = (n - 1)/(n + 1)` , and the second condition is not satisfied, because
`lim_(n - > oo)(n-1)/(n+1) = lim_(n - > oo) (1 - 1/n)(1 + 1/n)= 1`
So we cannot applied the test above to prove convergence. Let's apply the Test for Divergence to prove that the given series is divergent:
if `lim_(n - > oo) a_n` does not exest OR exists but is not equal to 0, then the series `sum_(n=1) ^ oo a_n` is divergent.
Consider the limit of nth term of the given series:
`lim_(n - > oo) (-1)^n (n - 1)/(n+1)`
We have shown above that the limit of `(n-1)/(n+1)` is 1. That means the limit of the `(-1)^n (n-1)/(n+1)` is +1 when n is even and it is -1 when n is odd. Thus, the limit does not exists.
The given series is divergent by the Test for Divergence.
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