If-r-1-n-t-r-n-n-1-n-2-n-3-8-then-lim-n-r-1-n-1-t-r-

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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