Problem

Determine whether the sequence converges or diver…

01:05

Problem 33 Hard Difficulty

Determine whether the sequence converges or diverges. If it converges, find the limit.

$ a_n = \frac {n^2}{\sqrt {n^3 + 4n}} $

Name: determine whether the sequence converges or diverges if it converges find the limit a_n frac n2sqrt
Uploaded: 2018-02-02
Duration: 2 min 49 s
Channel: Numerade
Description: Determine whether the sequence converges or diverges. If it converges, find the limit. $ a_n = \frac {n^2}{\sqrt {n^3 + 4n}} $

Answer

Diverges

Topics

Sequences

Series

Calculus: Early Transcendentals

Chapter 11

Infinite Sequences and Series

Section 1

Sequences

Discussion

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Video Transcript

okay. We need to look at the Limited and goes to infinity of a m. Figure out whether or not that limit exists and whether or not it's fine. So this is limit as n goes to infinity of in squared over square root of in cubed plus foreign. This problem we do a similar trick to what we've done before As far as we look in the denominator, we looked to see what term in the denominator is going to infinity the fastest except here instead of dividing top and bottom by that term will need to factor out something. So we're gonna need to factor out this in cubed that we have happening here. So factor that out in the denominator and the square root function is multiplication. So we just gotta pull that outside like that. And then we have one plus for over and squared here. So now we can just rewrite this a little bit. Mhm in squared square root of n cubed is in to the 3/2. Mhm. Mhm. Okay, and now we can simplify this part a bit. So this is gonna be in to the Tu minus 3/2 so into the one half. And then we're still dividing by square root of one plus four over n squared. Okay, so hopefully that's legible. Okay, so now if we do that limit on top, divided by limit on bottom, which is something that we're allowed to do as long as we don't get indeterminate form, then we'll see that we're going to get infinity divided by one, which is just gonna be infinity. Yeah, right. So we're allowed to limit on top over limit on bottom as long as we don't get indeterminate form. Indeterminate form would be like infinity over infinity or something divided by zero. If we try doing that here, we're just gonna get infinity over one which is not considered indeterminate form. So that's fine. Okay, So the limit turns out to be infinity, so the limit exists, but it's not finite. So the sequence is still said to diverge

Gabriel R.

Topics

Sequences

Series

Calculus: Early Transcendentals

Chapter 11

Infinite Sequences and Series

Section 1

Sequences

Problem