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

should be 'n-> infinity'

try multiplying by conjugate ..

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

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

really?

you missed n

oh ic

then you must be right, 1

hmm should be sqrt{2}/2

hm...

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

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

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

@experimentX correct!!

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

which method?

calculating without pen and paper!! lol

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

ty

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

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

right, I see that now thanks :)

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

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

I got it 1/rt2.

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

By the method described by one of the poster.

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