Question

18 задание ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,

Answer 1

$1)\Sigma \frac{7^{n}-1}{5^{n}(n+1)!}\\\\lim_{n\to \infty}\frac{a_{n+1}}{a_{n}}=lim_{n\to \infty }\frac{7^{n+1}-1}{5^{n+1}(n+2)!}\cdot \frac{5^{n}(n+1)!}{7^{n}-1}=\\\\=lim\frac{7^{n}\cdot 7\cdot (1-\frac{1}{7^{n+1}})\cdot 5^{n}(n+1)!}{5^{n}\cdot 5\cdot (n+1)!(n+2)\cdot 7^{n}(1-\frac{1}{7^{n}})}=\frac{7}{5}lim_{n\to \infty}\frac{1-\frac{1}{7^{n+1}}}{(n+2)(1-\frac{1}{7^{n}})}=\frac{7}{5}\cdot \frac{1-0}{\infty \cdot (1-0)}=\\\\=\frac{7}{\infty}=0<1\; \; \to \; \; sxoditsya$

$2)\Sigma a_{n}=\Sigma \frac{14}{49n^2-70n-24}\\\\\Sigma b_{n}=\Sigma \frac{1}{n^2}(sxoditsya)\\\\Sravnenie:lim_{n\to \infty}\frac{a_{n}}{b_{n}}=lim\frac{14n^2}{49n^2-70n-24}=\frac{14}{49}\ne 1\to oba sxodyatsya$

$3)\int _1^{+\infty}\frac{14}{49x^2-70x-24}dx=lim_{A\to +\infty}\int _1^{A}\frac{14\, dx}{(7x-5)^2-49}=\\\\=14*lim_{A\to +\infty}\frac{1}{2*7}ln|\frac{7x-5-7}{7x-5+7}|\, |_1^{A}=lim_{A\to +\infty}(ln|\frac{7A-12}{7A+2}|-ln/\frac{-5}{9}/)=\\\\=ln1-ln\frac{5}{9}=ln\frac{9}{5},sxoditsya$