sum_(n=1)^oo ((-1)^(n+1))/(n+1)
Take note that an alternating series
sum a_n = sum (-1)^(n+1) b_n
is convergent if the following conditions are satisfied.
(i) b_n is decreasing, and
(ii) lim_(n->oo) b_n=0 .
In the given alternating series, the bn is:
b_n = 1/(n+1)
Then, check if the values of bn decrease as n increases by 1.
n=1 , b_n = 1/2
n=2 , b_n=1/3
n=3 , b_n=1/4
n=4 , b_n=1/5
So bn is decreasing.
Also, take the limit of bn as n approaches infinity.
lim_(n->oo) b_n = lim_(n->oo) 1/(n+1) = 0
Since the result is zero, the second condition is satisfied too.
Therefore, by Alternating Series Test, the given series is convergent.
Saturday, June 15, 2019
sum_(n=1)^oo (-1)^(n+1)/(n+1) Determine the convergence or divergence of the series.
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