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Question 1

Question 2

Решите задачу:

$2)\; \; \Sigma \frac{1}{n(n+1)}=\Sigma (\frac{1}{n}-\frac{1}{n+1})\\\\S_{n}=(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+...+(\frac{1}{n-1}-\frac{1}{n})+(\frac{1}{n}-\frac{1}{n+1})=\\\\=1-\frac{1}{n+1}=\frac{n}{n+1}\\\\lim_{n\to \infty}S_{n}=lim\frac{n}{n+1}=1\; \; \to \; \; S=1$

$1)\; \; \Sigma \frac{1}{n^2+6n+8}=\Sigma \frac{1}{(n+2)(n+4)}=\Sigma (\frac{1/2}{n+2}-\frac{1/2}{n+4})=\Sigma \, \frac{1}{2}(\frac{1}{n+2}-\frac{1}{n+4})\\\\S_{n}=\frac{1}{2}\cdot (\frac{1}{3}-\frac{1}{5})+(\frac{1}{4}-\frac{1}{6})+(\frac{1}{5}-\frac{1}{7})+(\frac{1}{6}-\frac{1}{8})+...+\\\\+(\frac{1}{n-1}-\frac{1}{n+1})+(\frac{1}{n}-\frac{1}{n+2})+(\frac{1}{n+1}-\frac{1}{n+3})+(\frac{1}{n+1}-\frac{1}{n+4})=\\\\=\frac{1}{2}\cdot (\frac{1}{3}+\frac{1}{4}-\frac{1}{n+3}-\frac{1}{n+4})=$

$=\frac{1}{2}\cdot (\frac{7}{12}-\frac{2n+7}{(n+3)(n+4)})\\\\S=lim_{n\to \infty}S_{n}=lim_{n\to \infty}\frac{1}{2}(\frac{7}{12}+\cdot \frac{2n+7}{(n+3)(n+4)})=\frac{1}{2}\cdot \frac{7}{12}+0=\frac{7}{24}$

$3)\; \; \Sigma \frac{1}{3^{n-1}}=\frac{1}{3^0}+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...\; \; -\; \; geom.\; progressiya\\\\S=\frac{b_1}{1-q}=\frac{1}{1-\frac{1}{3}}=\frac{3}{2}$

Question 3

Спасибо за старание !