Definition of convergence

  • #1
1,462
44

Homework Statement


Show that ##\displaystyle \lim_{n\to \infty} \sqrt{n}(\sqrt{n+1}-\sqrt{n}) = \frac{1}{2}##

Homework Equations




The Attempt at a Solution


We see that ##\displaystyle \sqrt{n}(\sqrt{n+1}-\sqrt{n}) - \frac{1}{2} = \frac{\sqrt{n}}{\sqrt{n+1}+\sqrt{n}} - \frac{1}{2} < \frac{\sqrt{n}}{2\sqrt{n}} - \frac{1}{2} = \frac{1}{2} - \frac{1}{2} = 0##. But I don't think this can be right... What am I doing wrong?
 

Answers and Replies

  • #2
875
54
##\frac{\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}##
Try dividing the top and bottom of this fraction by something, then take the limit as ##n\rightarrow \infty##.
 
  • #3
14,411
11,722

Homework Statement


Show that ##\displaystyle \lim_{n\to \infty} \sqrt{n}(\sqrt{n+1}-\sqrt{n}) = \frac{1}{2}##

Homework Equations




The Attempt at a Solution


We see that ##\displaystyle \sqrt{n}(\sqrt{n+1}-\sqrt{n}) - \frac{1}{2} = \frac{\sqrt{n}}{\sqrt{n+1}+\sqrt{n}} - \frac{1}{2} < \frac{\sqrt{n}}{2\sqrt{n}} - \frac{1}{2} = \frac{1}{2} - \frac{1}{2} = 0##. But I don't think this can be right... What am I doing wrong?
Nothing, except that the step of subtracting ##\dfrac{1}{2}## is unnecessary. You can directly go to
$$
\sqrt{n}(\sqrt{n+1}-\sqrt{n})=\dfrac{\sqrt{n}(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}+\sqrt{n}}=\dfrac{\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}< \dfrac{\sqrt{n}}{2\sqrt{n}}=\dfrac{1}{2}
$$
Now the limit is definitely smaller than a half. What a bout the lower bound?
 
  • #4
1,462
44
Nothing, except that the step of subtracting ##\dfrac{1}{2}## is unnecessary. You can directly go to
$$
\sqrt{n}(\sqrt{n+1}-\sqrt{n})=\dfrac{\sqrt{n}(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}+\sqrt{n}}=\dfrac{\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}< \dfrac{\sqrt{n}}{2\sqrt{n}}=\dfrac{1}{2}
$$
Now the limit is definitely smaller than a half. What a bout the lower bound?
Well a lower bound would be 0. But I am trying to do this using the definition of convergence. I'm doing the scratchwork to find the ##N## and can associate with every ##\epsilon##. Like normally in the end we get like a difference between the sequence and the limit as ##1/n < \epsilon##, and so we select ##N## s.t. ##1/N < \epsilon##, and by the archimdedian principle this exists. However, in this case I just get that the difference is 0...
 
  • #5
14,411
11,722
Well a lower bound would be 0. But I am trying to do this using the definition of convergence. I'm doing the scratchwork to find the ##N## and can associate with every ##\epsilon##. Like normally in the end we get like a difference between the sequence and the limit as ##1/n < \epsilon##, and so we select ##N## s.t. ##1/N < \epsilon##, and by the archimdedian principle this exists. However, in this case I just get that the difference is 0...
You need to prove something like ##\frac{1}{2}-\frac{c}{n} \leq a_n## so that the gap is closing down with increasing ##n##.
However, the easier method is to follow @Eclair_de_XII 's suggestion and simply cancel out ##\sqrt{n}## and then use the rules given for limit arithmetic. My question only refers to the way you have chosen, in which case you need a lower bound.
 

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