Showing a Sequence Stays Within a Distance from 0

[Wanderer]

Hello! The following question has been tripping me up a little bit:

"Show that the sequence \(\displaystyle a_n=\dfrac{n}{n^2+1}\) eventually stays within a distance \(\displaystyle \dfrac{1}{10,000}\) from 0." Here is my working:

If the sequence stays within a distance \(\displaystyle \frac{1}{10,000}\) from 0, there exists N such that \(\displaystyle \forall n>N\),

\(\displaystyle |\frac{n}{n^2+1} - 0|<\frac{1}{10,000}\)

I then solve this inequality to get that if we take N=10,000, the sequence will stay within the required distance. But I don't really feel like I've shown this, and even if I have, it's a very inelegant and clunky method that I feel like there's probably a quicker way which is eluding me, and I feel sad for being unable to see this.

Any help would be appreciated.

 

[stapel]

What value did you find for N, such that, for n >, the rational expression has a value less than 1/10,000?

 

[Wanderer]

I took N=10,000 as that value, which seems to work as \(\displaystyle a_{10,000}=0.0000999...\) I suppose since n>N, I could just have taken N as 9,999. Either way, I still don't feel satisfied that I've 'shown' anything. What I've done would hardly constitute a proof of any sort, surely?

 

[Bob Brown MSEE]

Try this

\(\displaystyle |\frac{n}{n^2+1} |<|\frac{1}{n}|<\frac{1}{10,000}\)

 

[stapel]

\(\displaystyle |\frac{n}{n^2+1} |<|\frac{1}{n}|<\frac{1}{10,000}\)

Hint to the original poster: Note that a fraction is bigger if its denominator is smaller. Since n^2 is smaller than n^2 + 1, then 1/(n^2) is bigger than 1/(n^2 + 1).

As for "proving", show the work. If your proof value is N = 10,000, then show that, for n > N, the necessary inequality is valid.