A-sequence-is-given-by-3-2-2-3-5-4-4-5-Write-down-the-general-term-of-the-sequence-and-find-its-limit-

Question Number 4445 by Rasheed Soomro last updated on 28/Jan/16

A sequence is given by

\frac{3}{2}, \frac{2}{3}, \frac{5}{4}, \frac{4}{5}, \dots .

Write down the general term of the

sequence and find its limit .

Commented by 123456 last updated on 28/Jan/16

c_{n} = \frac{a_{n}}{b_{n}}

a_{n} = {\begin{cases} 3, n = 1 \\ a_{n - 1} - 1, n > 1, n \equiv 0 (\mod 2) \\ a_{n - 1} + 3, n > 1, n \equiv 1 (\mod 2) \end{cases}

b_{n} = n + 1

\frac{3}{2}, \frac{2}{3}, \frac{5}{4}, \frac{4}{5}, \frac{7}{6}, \frac{6}{7}, \frac{9}{8}, \frac{8}{9}, \frac{11}{10}, \frac{10}{11}, \dots

Commented by prakash jain last updated on 29/Jan/16

n^{t h} term = \frac{n + 1 + {(- 1)}^{n + 1}}{n + 1}

a_{1} = \frac{1 + 1 + 1}{2} = \frac{3}{2}

a_{2} = \frac{2 + 1 - 1}{3} = \frac{2}{3}

a_{3} = \frac{3 + 1 + 1}{4} = \frac{5}{4}

a_{4} = \frac{4 + 1 - 1}{5} = \frac{4}{5}

Answered by prakash jain last updated on 29/Jan/16

a_{n} = \frac{n + 1 + {(- 1)}^{n + 1}}{n + 1} = 1 + \frac{{(- 1)}^{n + 1}}{n + 1}

\lim_{n \to \infty} a_{n} = 1