Question

Show that Sigma infinity n=1 ( -1)^n+1 5÷7n+2 is conditionally convergent

Answer 1

Solution:

$\sum _{n=1}^{\infty \:}\left(-1\right)^{n+1}\frac{5}{7n+2} \\=5\cdot \sum _{n=1}^{\infty \:}\left(-1\right)^{n+1}\frac{1}{7n+2} \\=5\cdot \sum _{n=1}^{\infty \:}\left(-1\right)^n\left(-1\right)\frac{1}{7n+2} \\=5\left(-\sum _{n=1}^{\infty \:}\left(-1\right)^n\frac{1}{7n+2}\right) \\=5\left(-\mathrm{converges}\right)\ \ [\text{using Alternating series test}] \\=\mathrm{converges}$

Here, $a_n=\dfrac1{7n+1}$ which is positive and decreasing sequence.

Also, $\lim_{n\rightarrow \infty}a_n=\lim_{n\rightarrow \infty}\dfrac1{7n+1}=0$

Thus, by alternating series test $\sum _{n=1}^{\infty \:}\left(-1\right)^{n}a_n\ or\ \sum _{n=1}^{\infty \:}\left(-1\right)^{n+1}a_n$ converge.