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I need to prove that $(n+1)^{1/2} - n^{1/2}$ is null. I know there's a similar question already posted but I need to prove this stating rules etc.

What I've done already is use difference of 2 squares to get to:

$1/[(n+1)^{1/2} + n^{1/2}]$ which I'm sure is near the end. Do I need to use the Squeeze Rule?

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    Multiplying by its 'conjugate' (n+1)^1/2+n^1/2 gives 1. but then we get to divide by the 'conjugate' which clearly goes to infinity and thus we get 0.2016-11-05

2 Answers 2

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We have $a_n=\sqrt{n+1}-\sqrt{n}$, and we want to show that $\lim a_n=0$.

Let see, what is $\sqrt{n+1}-\sqrt{n}$:

$\sqrt{n+1}-\sqrt{n}=\frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}+\sqrt{n}}=\frac{n+1-n}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{\sqrt{n+1}+\sqrt{n}}$

So, when $n\to\infty$, we get $\frac{1}{\sqrt{n+1}+\sqrt{n}}\to 0$.

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According to the definition of a null sequence, if E (Epsilon) > 0 then there exists a number X such that absolute value of |an| < E for all n>X Here an = (n+1)^1/2 - ( n)^1/2 Let Epsilon E > 0 then |(n+1)^1/2 - ( n)^1/2| < E (n+1)^1/2 - ( n)^1/2 < E (n+1)^1/2 < E + (n)^1/2
squaring both sides gives n+1 < E^2 +2E(n)^1/2 +n n-n +1 < E^2 + 2E(n)^1/2 -2E (n)^1/2 n>(2/E^2 -2)^2 so according to the def of null seq |an| < E for all n>X here X =(2/E^2 -2)^2 so n>X proved it is a null seq

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    Please write your answer clearly, use the math formatting. Your answer is unclear.2016-11-05