Try this limit question..

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

Question

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

anonymous · Answer

should be 'n-> infinity'

experimentx · Answer

try multiplying by conjugate ..

turingtest · Answer

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

experimentx · Answer

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

turingtest · Answer

really?

experimentx · Answer

you missed n

turingtest · Answer

oh ic

turingtest · Answer

then you must be right, 1

anonymous · Answer

hmm should be sqrt{2}/2

turingtest · Answer

hm...

experimentx · Answer

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

experimentx · Answer

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

turingtest · Answer

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

anonymous · Answer

@experimentX correct!!

experimentx · Answer

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

anonymous · Answer

which method?

experimentx · Answer

calculating without pen and paper!! lol

anonymous · Answer

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

experimentx · Answer

ty

turingtest · Answer

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

anonymous · Answer

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

turingtest · Answer

right, I see that now thanks :)

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

I got it 1/rt2.

anonymous · Answer

The limit is
$$
\frac {\sqrt 2}2
$$

anonymous · Answer

By the method described by one of the poster.