Answer to Question #155550 in Real Analysis for Gaurav Dabas

Question

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

Answer 1

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

$\sum_{n=1}^{\infty} (-1)^{n+1} {1 \over n^3} =1 - {1 \over2^3}+{1 \over3^3}-{1 \over4^3}+{1 \over5^3} ...$

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

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

$\int_{1}^{∞} f(x) dx$

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

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

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

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

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

$={1 \over 3}$

$\because$ the series is absolutely convergent