Question #155550

show that the series 1 - 1/2³ + 1/3³ - 1/4³ + 1/5³ - .... is absolutely convergent


1
Expert's answer
2021-01-18T15:55:37-0500

show that the series 1 - 1/2³ + 1/3³ - 1/4³ + 1/5³ - .... is absolutely convergent


n=1(1)n+11n3=1123+133143+153...\sum_{n=1}^{\infty} (-1)^{n+1} {1 \over n^3} =1 - {1 \over2^3}+{1 \over3^3}-{1 \over4^3}+{1 \over5^3} ...


n=1(1)n+11n3=n=11n3|\sum_{n=1}^{\infty} (-1)^{n+1} {1 \over n^3}| =\sum_{n=1}^{\infty} {1 \over n^3}


we test if it is convergent using the integral test, in order to use the Integral Test the series terms MUST eventually be decreasing and positive, which is so with the given series


1f(x)dx\int_{1}^{∞} f(x) dx


f(x)=1x3f(x) = {1 \over x^3}


limt1t1x3dx=limt13x21t\lim_{t\to\infty} \int_{1}^{t}{1 \over x^3} dx = \lim_{t\to\infty} -{1 \over 3x^2}|_1^t


=limt13t2+13(1)2=\lim_{t\to\infty} -{1 \over 3t^2} + {1 \over 3(1)^2}


=1+13= -{1 \over\infty} + {1 \over 3}


=0+13= -0 + {1 \over 3}


=13={1 \over 3}


\because the series is absolutely convergent






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