Question

See below for question. Take the limit as x goes to infinity of...

Answer 1

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

Answer 2

by x do you mean n?

Answer 3

there's a formula that you can use to cancel out \[\frac{ 1 }{ n }\]
and \[\frac{ 1 }{ n^2 }\]
and then take the limit as n -> ∞ but not x.

Answer 4

When there's i i think the formula is:
\[\frac{ n(n+1) }{ 2 }\]
I THINK.

Answer 5

I'm terrible at infinite series lol, good luck.

Answer 6

whoops my bad

\[\lim_{n \rightarrow \infty } \frac{ 2 }{ n } (n+2\frac{ n(n+1) }{ 2 })\]
\[\lim_{n \rightarrow \infty} 2(1+\frac{ 2n^2+2n }{ 2n })\]
\[\lim_{n \rightarrow \infty} 2(1+\frac{ 2n^2 }{ 2n }\ + \frac{ 2n }{ 2n })\]
so then it's 2(1+1+2n)
am i doing something wrong?

Answer 7

Distribute. Wait, is this still a series question or a limits question?

Answer 8

i'm not sure. i think it's both? well it's just limits to infinity.

Answer 9

Haha, ok. Well, bu the looks of it, it seems like they already gave you the formula's for "i".
But when i distribute everythhing, I get 2n+4.

Answer 10

\[\lim_{n \rightarrow ∞} 2n+4\]

Answer 11

Which makes sense, b/c your series expantion @ n = ∞ is 2n+4.

Answer 12

ohh. but the answer is 4...

Answer 13

Isn't the limit as n-> of 2n+4 = 4?

Answer 14

i thought that because there's a "n" there that there'd be no limit because it goes to infinity..?

Answer 15

I'm imagining somehow taking l'hopitals rule of this to get 4 but it really doesn't make any sense, but it "kinda" looks like 4. just distribute the 2/n... =/

Answer 16

Idk. I get ∞ as the answer.

Answer 17

and i think it's 4n+4 but same same
what's the l'hopitals rule?

Answer 18

And I just did it, twice. I get 2n+4.

Answer 19

oh yeah i just wrote 2n instead of n
but even if it's 2n+4, the limit is infinity...?

Answer 20

yes

Answer 21

but the answer is 4...
did i do something wrong in my work?

Answer 22

ohh i think i see what i did wrong.
i left out a "n"

Answer 23

An easy way would be to split them out and work on each separately:
lim sum (1+2i/n)(2/n)
= lim sum [(2/n) + 4i/n^2]
= lim sum 2/n + lim sum 4i/n^2
= lim 2/n sum(1) + lim (4/n^2) sum(i)
= lim (2/n)*n + lim (4/n^2)(n)(n+1)/2
= 2 + 2 lim (n)(n+1)/n^2
= 2+2
= 4

Answer 24

lol