If Σ_(r=1) ^n t_r =((n(n+1)(n+2)(n+3))/8) then lim_(n→∞) Σ_(r=1) ^n (1/t


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Question Number 110173 by ajfour last updated on 27/Aug/20

$I f \underset{r = 1}{\sum^{n}} t_{r} = \frac{n (n + 1) (n + 2) (n + 3)}{8}$ $t h e n \lim_{n \to \infty} \underset{r = 1}{\sum^{n}} \frac{1}{t_{r}} = ?$
	Answered by Olaf last updated on 27/Aug/20

	$t_{n} = \underset{r = 1}{\sum^{n}} t_{r} - \underset{r = 1}{\sum^{n - 1}} t_{r}$ $t_{n} = \frac{n (n + 1) (n + 2) (n + 3)}{8} - \frac{(n - 1) n (n + 1) (n + 2)}{8}$ $t_{n} = \frac{n (n + 1) (n + 2) [n + 3 - n + 1]}{8}$ $t_{n} = \frac{n (n + 1) (n + 2)}{2}$ $t_{r} = \frac{r (r + 1) (r + 2)}{2}$ $\frac{1}{t_{r}} = \frac{2}{r (r + 1) (r + 2)} = \frac{1}{r} - \frac{2}{r + 1} + \frac{1}{r + 2}$ $\underset{r = 1}{\sum^{n}} \frac{1}{t_{r}} = \underset{r = 1}{\sum^{n}} \frac{1}{r} - 2 \underset{r = 1}{\sum^{n}} \frac{1}{r + 1} + \underset{r = 1}{\sum^{n}} \frac{1}{r + 2}$ $\underset{r = 1}{\sum^{n}} \frac{1}{t_{r}} = \underset{r = 1}{\sum^{n}} \frac{1}{r} - 2 \underset{r = 2}{\sum^{n + 1}} \frac{1}{r} + \underset{r = 3}{\sum^{n + 2}} \frac{1}{r}$ $\underset{r = 1}{\sum^{n}} \frac{1}{t_{r}} = \underset{r = 1}{\sum^{n}} \frac{1}{r} - 2 (\underset{r = 1}{\sum^{n}} \frac{1}{r} - 1 + \frac{1}{n + 1})$ $+ (\underset{r = 1}{\sum^{n}} \frac{1}{r} - 1 - \frac{1}{2} + \frac{1}{n + 1} + \frac{1}{n + 2})$ $\underset{r = 1}{\sum^{n}} \frac{1}{t_{r}} = 2 - \frac{2}{n + 1} - \frac{3}{2} + \frac{1}{n + 1} + \frac{1}{n + 2}$ $\underset{r = 1}{\sum^{n}} \frac{1}{t_{r}} = \frac{1}{2} - \frac{1}{n + 1} + \frac{1}{n + 2}$ $\underset{r = 1}{\sum^{n}} \frac{1}{t_{r}} = \frac{1}{2} - \frac{1}{(n + 1) (n + 2)}$ $\lim_{n \to \infty} \underset{r = 1}{\sum^{n}} \frac{1}{t_{r}} = \frac{1}{2}$
	Commented by ajfour last updated on 27/Aug/20

	$r i g h t a n s w e r, t h a n k s S i r .$
	Answered by Dwaipayan Shikari last updated on 27/Aug/20

	$△ \underset{r = 1}{\sum^{n}} t_{r} = \frac{n (n + 1) (n + 2) (n + 3)}{8} - \frac{(n - 1) n (n + 1) (n + 2)}{8}$ $t_{n} = \frac{n (n + 1) (n + 2)}{2}$ $\sum^{\infty} \frac{1}{t_{n}} = \sum^{\infty} \frac{2}{n (n + 1) (n + 2)} = \sum^{\infty} \frac{n + 2 - n}{n (n + 1) (n + 2)} = \sum^{\infty} \frac{1}{n} - \frac{1}{n + 1} - \sum^{\infty} \frac{1}{n + 1} - \frac{1}{n + 2}$ $= \lim_{n \to \infty} (1 - \frac{1}{n + 1}) - \lim_{n \to \infty} (\frac{1}{2} - \frac{1}{n + 2})$ $= 1 - \frac{1}{2} = \frac{1}{2}$
	Commented by ajfour last updated on 27/Aug/20

	$G r e a t w a y S i r; t h a n k s a l o t .$
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