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}"
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