Question

find a formula for the sum of n terms. Use the formula to find the limit as n->infinity

limit as n goes to infinity of the sum when i = 1 (1+i/n)(2/n)

Answer 1

The equation looks more so like this

\[\lim_{n \rightarrow \infty} \sum_{i=1}^{n} (1+\frac{ i }{ n })(\frac{ 2 }{ n })\]

Answer 2

I got 3

Answer 3

I know that is the answer but I am having trouble starting it. What could I take out to get just 1+ i on the inside?

Answer 4

you can take out (2/n) since it's *i* that you're incrementing on

Answer 5

so I would have (1+i)(4/n^2) ?

Answer 6

\[(2/n)\sum_{i=1}^{n}(1+i/n) = (2/n) (\sum_{i=1}^{n}1 + \sum_{i=1}^{n}i)\]

Answer 7

the first sum is just n
the second sum is (n/2)(n+1)

Answer 8

ahhhh yes thanks!

Answer 9

now you just need to multiply them all together and take the limit. Answer should be 3

Answer 10

Yea i checked back of the book for final forumula and limit so, anyways thanks all I needed was the starting step! :)

Answer 11

opps the second sum should be
\[(1/n)\sum_{i=1}^{n}i\]

Answer 12

right now i got \[\lim_{n \rightarrow \infty} \frac{ 2 }{ n } (n) + \frac{ 1 }{ n } (\frac{ n^2+n }{ 2 }\]
and i multiplied the outsides of the 2 to get
\[\lim_{n \rightarrow \infty} \frac{ 2n }{ n } + \frac{ n^2+n }{ 2 }\] is that correct?

Answer 13

nope, you probably messed your algebra somewhere

Answer 14

there is no n^2 term

Answer 15

the sum of i though is \[\frac{ n(n+1) }{ 2 }\] is it not?

Answer 16

yes but there is also (1/n) in front of the sum. I believe I corrected mistake above

Answer 17

ahh so that neutralizes the n in front

Answer 18

yep

Answer 19

so if I simplify I end up with \[\lim_{n \rightarrow \infty} 2 + \frac{ n+1 }{ }\] right?

Answer 20

woops the n+1 is over 2

Answer 21

still no. Did add the second sum to the first sum?

Answer 22

you should have
(2/n) [n + (n+1)/2]

Answer 23

ok then the end is \[\lim_{n \rightarrow \infty} \frac{ 1 }{ n } (\frac{ 3n+1 }{ 1 })\] and the limit is 3

Answer 24

bingo

Answer 25

thanks!