Find-S-n-and-S-of-the-series-1-1-2-3-4-1-2-3-4-5-1-3-4-5-6-

Question Number 75301 by vishalbhardwaj last updated on 09/Dec/19

Find S_{n} and S_{\infty} of the series : \frac{1}{1.2.3.4} + \frac{1}{2.3.4.5} + \frac{1}{3.4.5.6} + . . .

Commented by vishalbhardwaj last updated on 09/Dec/19

please solve this

Commented by vishalbhardwaj last updated on 09/Dec/19

please solve this

Commented by mind is power last updated on 09/Dec/19

= \sum_{k ⩾ 1} \frac{1}{k (k + 1) (k + 2) (k + 3)}

= \frac{1}{2} Σ {\frac{1}{k (k + 3)} - \frac{1}{(k + 1) (k + 2)}}

= \frac{1}{6} Σ \frac{(k + 3) - k}{k (k + 3)} - \frac{1}{2} Σ \frac{(k + 2) - (k + 1)}{(k + 1) (k + 2)}

= \frac{1}{6} Σ (\frac{1}{k} - \frac{1}{k + 3}) - \frac{1}{2} Σ (\frac{1}{k + 1} - \frac{1}{k + 2})

S_{N} = \underset{k = 1}{\sum^{N}} (\frac{1}{k (k + 1) (k + 2) (k + 3})

= \frac{1}{6} \underset{k = 1}{\sum^{1 N}} (\frac{1}{k} - \frac{1}{k + 3}) - \frac{1}{2} \underset{k = 1}{\sum^{N}} (\frac{1}{k + 1} - \frac{1}{k + 2})

= \frac{1}{6} (\underset{k = 1}{\sum^{N}} \frac{1}{k} - \underset{k = 4}{\sum^{N + 3}} \frac{1}{k}) - \frac{1}{2} (\underset{k = 1}{\sum^{N}} \frac{1}{k + 1} - \underset{k = 2}{\sum^{N + 1}} \frac{1}{k + 1})

= \frac{1}{6} (1 + \frac{1}{2} + \frac{1}{3} - \frac{1}{N + 1} - \frac{1}{N + 2} - \frac{1}{N + 3}) - \frac{1}{2} (\frac{1}{2} - \frac{1}{N + 2})

Double subscripts: use braces to clarify

Commented by peter frank last updated on 09/Dec/19

t h a n k y o u