Question #212285

show that∑n=0∞ (-1)^(n+1)(5)/(7n+2)` is conditionally convergent 


1
Expert's answer
2021-07-16T02:38:08-0400

Consider the series


n=0(1)n+17n+2=n=017n+2\displaystyle\sum_{n=0}^{\infin}|\dfrac{(-1)^{n+1}}{7n+2}|=\displaystyle\sum_{n=0}^{\infin}\dfrac{1}{7n+2}

Use the Integral Test

Let f(n)=an=17n+2,n=0,1,2,...f(n)=a_n=\dfrac{1}{7n+2}, n=0,1,2,...


017x+2dx=limt0t17x+2dx\displaystyle\int_{0}^{\infin}\dfrac{1}{7x+2}dx=\lim\limits_{t\to\infin}\displaystyle\int_{0}^{t}\dfrac{1}{7x+2}dx

=limt17[ln7x+2]t0==\lim\limits_{t\to\infin}\dfrac{1}{7}[\ln|7x+2|]\begin{matrix} t \\ 0 \end{matrix}=\infin

Since this improper integral is divergent, the series n=017n+2\displaystyle\sum_{n=0}^{\infin}\dfrac{1}{7n+2} diverges by the Integral Test.

Consider the alternating series


n=0(1)n+17n+2\displaystyle\sum_{n=0}^{\infin}\dfrac{(-1)^{n+1}}{7n+2}bn=17n+2>0,n=0,1,2,...b_n=\dfrac{1}{7n+2}>0, n=0,1,2,...

bn+1=17(n+1)+2<17n+2=bn,n=0,1,2,...b_{n+1}=\dfrac{1}{7(n+1)+2}<\dfrac{1}{7n+2}=b_n, n=0, 1, 2, ...

limnbn=limn17n+2=0\lim\limits_{n\to \infin}b_n=\lim\limits_{n\to \infin}\dfrac{1}{7n+2}=0

Then the series n=0(1)n+17n+2\displaystyle\sum_{n=0}^{\infin}\dfrac{(-1)^{n+1}}{7n+2} is convergent by the Alternating Series Test.


Therefore the series n=0(1)n+17n+2\displaystyle\sum_{n=0}^{\infin}\dfrac{(-1)^{n+1}}{7n+2} is conditionally convergent.



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