Question

Determine if the sequence converges, or diverges. Award Given To Best Answer!
An = Sqrt(n)/(1 + sqrt(n))

Answer 1

Write it as a fraction with denominator 1, and then multiply top and bottom by sqrt(n+1) + sqrt(n). This gives:

[(n+1) - n] / [ sqrt(n+1) + sqrt(n)] = 1/ [ sqrt(n+1) + sqrt(n)]

Now as n gets large the denominator gets large, so the limit as n-> infinity is zero.

Answer 2

Is that ok?

Answer 3

Yeah, but that doesn't make full sense..Can you explain it more? Or use the equation tab so I can understand it?

Answer 4

the limit is 1 as n approaches infinity

Answer 5

I love how I just had the whole thing typed out and then I hit the space bar and it deletes everything. fml.

Answer 6

Sorry. :P Lol that happens a lot to me too. :P

Answer 7

https://www.wolframalpha.com/input/?i=lim+n-%3E+inf+Sqrt%28n%29%2F%281+%2B+sqrt%28n%29%29+

Answer 8

\[\lim_{n \rightarrow \infty } \frac{ \sqrt{n} }{ 1+\sqrt{n} }\]
Here, we will divide by the n term in the DENOMINATOR (only the denominator) with the highest power, which would be the square root of n.
\[\lim_{n \rightarrow \infty } \frac{ \frac{ \sqrt{n} }{ \sqrt{n} } }{ \frac{ 1 }{ \sqrt{n} } + \frac{ \sqrt{n} }{ \sqrt{n} } }\]
So as n approaches infinity, the 1/sqrt n will get infinitely small, so it's safe to say it would be zero.
\[\lim_{n \rightarrow \infty} \frac{ 1 }{ 0+1 } = \frac{ 1 }{ 1 } = 1\]
It would converge towards 1.

Answer 9

@sourwing I can't view the step-by-step solution. :/

Answer 10

There's your step by step, sir.

Answer 11

Thank you so much @halorazer :D You're a lifesaver! :)

Answer 12

Pft, I know. :3

Answer 13

yes, dividing both top and bottom by sqrt(n) is the key