Question

Find a formula for the sum of n terms. Use the formula to find the limit as n → [infinity]. lim n → [infinity] n 2 + i n 8 n i = 1

Answer 1

Complete Question

Find a formula for the sum of n terms.   $\sum\limits_{i=1}^n ( 8 + \frac{i}{n} )(\frac{2}{n} )$

Use the formula to find the limit as $n \to \infty$

 

Answer:

   $K_n = \frac{n + 73 }{n}$

  $\lim_{n \to \infty} K_n = 1$

Step-by-step explanation:

     So let assume that

                  $K_n = \sum\limits_{i=1}^n ( 8 + \frac{i}{n} )(\frac{2}{n} )$

=>             $K_n = \sum\limits_{i=1}^n ( \frac{16}{n} + \frac{2i}{n^2} )$

=>              $K_n = \frac{2}{n} \sum\limits_{i=1}^n (8) + \frac{2}{n^2} \sum\limits_{i=1}^n(i)$

Generally  

         $\sum\limits_{i=1}^n (k) = \frac{1}{2} n (n + 1)$

So  

      $\sum\limits_{i=1}^n (8) = \frac{1}{2} * 8* (8 + 1)$

      $\sum\limits_{i=1}^n (8) = 36$

$K_n = \frac{2}{n} \sum\limits_{i=1}^n (8) + \frac{2}{n^2} \sum\limits_{i=1}^n(i)$  

and  

  $\sum\limits_{i=1}^n (i) = \frac{1}{2} n (n + 1)$

  Therefore

         $K_n = \frac{72}{n} + \frac{2}{n^2} * \frac{1}{2} n (n + 1 )$

         $K_n = \frac{72}{n} + \frac{1}{n} (n + 1 )$

         $K_n = \frac{72}{n} + 1 + \frac{1}{n}$

        $K_n = \frac{72 + 1 + n }{n}$

        $K_n = \frac{n + 73 }{n}$

Now  $\lim_{n \to \infty} K_n = \lim_{n \to \infty} [\frac{n + 73 }{n} ]$

=>     $\lim_{n \to \infty} [\frac{n + 73 }{n} ] = \lim_{n \to \infty} [\frac{n}{n} + \frac{73 }{n} ]$

=>     $\lim_{n \to \infty} [\frac{n + 73 }{n} ] = \lim_{n \to \infty} [1 + \frac{73 }{n} ]$

=>     $\lim_{n \to \infty} [\frac{n + 73 }{n} ] = \lim_{n \to \infty} [1 ] + \lim_{n \to \infty} [\frac{73 }{n} ]$

=>    $\lim_{n \to \infty} [\frac{n + 73 }{n} ] = 1 + 0$

Therefore

      $\lim_{n \to \infty} K_n = 1$