Question

Try this limit question..

lim(x-> infinity) [sqrt{1+2+3+.....+n} - sqrt{1+2+3+...+(n-1)}

Answer 1

should be 'n-> infinity'

Answer 2

try multiplying by conjugate ..

Answer 3

\[\lim_{n\to\infty}\sqrt{1+2+3+...+n}-\sqrt{1+2+3+...+(n-1)}\]\[=\lim_{n\to\infty}\frac1{\sqrt{1+2+3+...+n}+\sqrt{1+2+3+...+(n-1)}}=0\]I think..

Answer 4

sorry ... wrong ... should be 1 ... i ignored root on denominator.

Answer 5

really?

Answer 6

you missed n

Answer 7

oh ic

Answer 8

then you must be right, 1

Answer 9

hmm should be sqrt{2}/2

Answer 10

hm...

Answer 11

lol ... we ignored ... 1's at the bottom!!

Answer 12

n(n+1)/2 and n(n-1)/2

Answer 13

\[\lim_{n\to\infty}\sqrt{1+2+3+...+n}-\sqrt{1+2+3+...+(n-1)}\]\[=\lim_{n\to\infty}\frac n{\sqrt{1+2+3+...+n}+\sqrt{1+2+3+...+(n-1)}}\]I don't see it yet :(

Answer 14

@experimentX correct!!

Answer 15

i guess i can't use the method my friend described ... try to solve everything in head!!

Answer 16

which method?

Answer 17

calculating without pen and paper!! lol

Answer 18

well actually using n(n+1)/2 that method can solve it. :) well done.

Answer 19

ty

Answer 20

ok I see what you are saying now, but I would have never thought to use that formula here
nice one, thanks for showing me @experimentX

Answer 21

@TuringTest your method also works. Just need to simplify the denominator using the formula.

Answer 22

right, I see that now thanks :)

Answer 23

use Sn=n(n+1)/2, this works.

Answer 24

[(n+1/2)*(1+n)]-[(1+(n-1)/2)*(1+(n-1)] ???

Answer 25

I got it 1/rt2.

Answer 26

The limit is
\[
\frac {\sqrt 2}2
\]

Answer 27

By the method described by one of the poster.