Mathematics OpenStudy (anonymous):

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

should be 'n-> infinity' OpenStudy (experimentx):

try multiplying by conjugate .. OpenStudy (turingtest):

$\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.. OpenStudy (experimentx):

sorry ... wrong ... should be 1 ... i ignored root on denominator. OpenStudy (turingtest):

really? OpenStudy (experimentx):

you missed n OpenStudy (turingtest):

oh ic OpenStudy (turingtest):

then you must be right, 1 OpenStudy (anonymous):

hmm should be sqrt{2}/2 OpenStudy (turingtest):

hm... OpenStudy (experimentx):

lol ... we ignored ... 1's at the bottom!! OpenStudy (experimentx):

n(n+1)/2 and n(n-1)/2 OpenStudy (turingtest):

$\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 :( OpenStudy (anonymous):

@experimentX correct!! OpenStudy (experimentx):

i guess i can't use the method my friend described ... try to solve everything in head!! OpenStudy (anonymous):

which method? OpenStudy (experimentx):

calculating without pen and paper!! lol OpenStudy (anonymous):

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

ty OpenStudy (turingtest):

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 OpenStudy (anonymous):

@TuringTest your method also works. Just need to simplify the denominator using the formula. OpenStudy (turingtest):

right, I see that now thanks :) OpenStudy (anonymous):

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

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

I got it 1/rt2. OpenStudy (anonymous):

The limit is $\frac {\sqrt 2}2$ OpenStudy (anonymous):

By the method described by one of the poster.

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