Determine if sequence converges, If it converges what does it converge to?

Question

anonymous · Answer

$$\left\{ n-\sqrt{n^2+5n} \right\}\ _{n=0}^{\infty} $$

perl · Answer

try multiplying by teh conjugate

anonymous · Answer

sorry, n=1

anonymous · Answer

I don't know what that means...

anonymous · Answer

$$\lim_{n \rightarrow \infty}n-\sqrt{n^2+5n}$$

anonymous · Answer

is infinity-infinity

perl · Answer

multiply by ( n + sqrt( n^2  + 5n) / ( n + sqrt(n^2 + 5n) )

anonymous · Answer

oh, yea we did that before, I got stuck later down

anonymous · Answer

$$\lim_{n \rightarrow \infty} \frac{ 5n }{ n+\sqrt{n^2+5n} }$$

anonymous · Answer

$$\lim_{n \rightarrow \infty}\frac{ 5n }{ n+\sqrt{n^2(1+\frac{ 5 }{ n })} }$$

anonymous · Answer

$$\lim_{n \rightarrow \infty}\frac{ 5 }{ n \sqrt{1+\frac{ 5 }{ n }} }$$

anonymous · Answer

and then I get stuck

anonymous · Answer

$$n-\sqrt{n^{2} + 5n}*\frac{ n+\sqrt{n^{2}+5n} }{ n+\sqrt{n^{2}+5n} }= \frac{ n^{2} -(n^{2}+5n)}{ n+\sqrt{n^{2}+5n} }= \frac{ -5n }{ n+\sqrt{n^{2}+5n} }$$
What you can do from there is multiply top and bottom by 1/sqrt(n^2). Because n is going to infinity, sqrt(n^2) is positive and we don't have to worry about any absolute value issues. Once you do that, you should be fine to take your limit to all terms and get your answer.