Total-sum-of-this-below-infinite-series-1-1-1-2-1-2-3-1-3-4-1-n-n-1-2-0-6-0-06-0-006-6-10-n-

Question Number 91796 by zainal tanjung last updated on 03/May/20

Total sum of this below infinite series :

1) . \frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + \dots \frac{1}{n (n + 1)} = \dots

2) . 0.6 + 0.06 + 0.006 + \dots \frac{6}{10^{n}} = \dots

Commented by mathmax by abdo last updated on 04/May/20

2) S = \sum_{k = 1}^{n} \frac{6}{10^{k}} = \frac{6}{10} \sum_{k = 1}^{n} \frac{1}{10^{k - 1}} = \frac{6}{10} \sum_{p = 0}^{n - 1} {(\frac{1}{10})}^{p}

= \frac{6}{10} \times \frac{1 - {(\frac{1}{10})}^{n}}{1 - \frac{1}{10}} = \frac{6}{10} \times \frac{10}{9} (1 - \frac{1}{10^{n}}) = \frac{2}{3} (1 - \frac{1}{10^{n}})

= \frac{2}{3} - \frac{2}{3 . 10^{n}}

Answered by $@ty@m123 last updated on 03/May/20

(1) 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \dots \dots + \frac{1}{n - 1} - \frac{1}{n} + \frac{1}{n} - \frac{1}{n + 1}

= 1 - \frac{1}{n + 1} = \frac{n}{n + 1}

(2) a = 0.6, r = 0.1

∴ S_{n - 1} = \frac{a (1 - r^{n})}{1 - r}

= \frac{0.6 (1 - 0 . 1^{n})}{1 - 0.1}

= \frac{0.6 (1 - 0 . 1^{n})}{0.9}

= \frac{2 (1 - 0 . 1^{n})}{3}