Finding the limit of sqrt(n/(n^4+1)) as n approaches infinity

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
2 replies · 2K views
Harmony
Messages
201
Reaction score
0

Homework Statement



The nth term for a sequence is the square root of [n/ (n^4 + 1)]
Investigate whether it is convergence or divergence.

Homework Equations



Ratio test and integral test

The Attempt at a Solution



Ratio test will fail for this question, since no conclusion can be drawn if the ratio is 1. So I have to use integral test. I tried integrating the function using integration by part, bu that doesn't work. Trigonometric substitution fail as well. Any idea how to integrate the function?
 
Physics news on Phys.org
I'm assuming you want to find whether
[tex]{\sum_n^\infty \frac{n}{n^4+1}}[/tex]
converges.

Anyways, try doing a u-substitution with [tex]u=n^2[/tex].
 
Harmony is actually looking for the limit of the sequence [tex]a_n = \sqrt{ \frac{n}{n^4+1}}[/tex].

The reason foxjwill may have thought you wanted that series was because you spoke of the integral test, which is indeed a good idea - Prove the series converges, the nth term of the series must go to zero, and hence our sequence goes to zero. If you wished to take this route, as foxjwill said, u= n^2 is an easy substitution.

However much easier than any method you have tried so far is the comparison test:

[tex]a_n = \sqrt{ \frac{n}{n^4+1}} < \sqrt{ \frac{n}{n^4}}[/tex].

That should do it.