Technical question regarding showing sqrt(n+1) - sqrt(n) converges to 0

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SUMMARY

The sequence defined by sn = √(n+1) - √n converges to zero as n approaches infinity. The limit can be shown by rationalizing the numerator, resulting in sn = 1/(√(n+1) + √n). Both bounds sn < 1/(2√n) and sn < 1/n^(1/2) are valid, but the former provides a tighter upper bound. This demonstrates that sn converges to zero more effectively, aligning with mathematical rigor preferred in academic settings.

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Homework Statement



show the sequence sn= (n+1)1/2 - n1/2 converges to zero

Homework Equations





The Attempt at a Solution



I don't have that much of a problem showing the limit goes to zero, rationalize the numerator (or whatever it's called) to get (n+1)1/2 - n1/2 = 1/((n+1)1/2 + n1/2). My question is that I show this goes to zero because sn<1/n(1/2) which goes to zero, but my professor provides a solution where he writes sn<1/2(n1/2). I don't understand why the 2 is there. Is saying that sn<1/(n1/2) insufficient or not true?

Thanks
 
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both are true, and sufficient to show it converges to zero, the 2nd is just a little tighter

how about this, as
(n+1)^{1/2} &gt; n^{1/2}
then
\frac{1}{s_n} = n^{1/2} + (n+1)^{1/2} &gt; n^{1/2} + n^{1/2} = 2n^{1/2}
then inverting
s_n = \frac{1}{n^{1/2} + (n+1)^{1/2}} &lt; \frac{1}{n^{1/2} + n^{1/2}} = \frac{1}{2n^{1/2}}
 
\sqrt{n+1}+\sqrt{n}&gt; 2 \sqrt{n}.
taking the inverse on both sides yields
\frac{1}{\sqrt{n+1}+\sqrt{n}} &lt; \frac{1}{2 \sqrt{n}}
your professor is just using a smaller upper bound for s_n. professors like to use bounds that are as small as possible:smile:
 

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