Question

show that ⠈ ‘ (-1)^(n+1)à —5à ·(7n+2) is condintionally convergent

Answer 1

Since, given term is unclear, we assume it as $\sum _{n=0}^{\infty \:}\frac{\left(-1\right)^{n+1}\cdot \:5}{7n+2}$

Solution:

$\sum _{n=0}^{\infty \:}\frac{\left(-1\right)^{n+1}\cdot \:5}{7n+2}$

$=5\cdot \sum _{n=0}^{\infty \:}\frac{\left(-1\right)^{n+1}}{7n+2}$

$=5\cdot \sum _{n=0}^{\infty \:}\frac{\left(-1\right)^n\left(-1\right)^1}{7n+2}$

$=5\left(-1\right)\cdot \sum _{n=0}^{\infty \:}\frac{\left(-1\right)^n}{7n+2}$

Apply alternating series test

(a) To check $\lim_{n\rightarrow \infty} b_n=0$

$\lim_{n\rightarrow \infty} \dfrac{1}{7n+2}=\dfrac1{\infty}=0$

(b) And to check $b_{n+1}\le b_n$

$b_n=\dfrac1{7n+2}, b_{n+1}=\dfrac1{7(n+1)+2}=\dfrac1{7n+9}$

Clearly, $\dfrac1{7n+9}\le \dfrac1{7n+2}$

Thus, $b_{n+1}\le b_n$

Hence, by using alternating series test

$\sum _{n=0}^{\infty \:}\frac{\left(-1\right)^{n+1}\cdot \:5}{7n+2}$ $=5\left(-1\right)\mathrm{converges}$

$=\mathrm{converges}$