# Solution 88: 1/n and Square Roots

Share
Embed
• Published on Apr 19, 2019
• Using two different methods--bounded/monotonic argument and Squeeze Theorem--we analyze the convergence of our sequence of nested radicals.
Congratulations to Gabriel N., Alon Heller, Minh Cong Nguyen, Rishav Gupta, Kevin Lu, andabata43, fmakofmako, JIN ZHI PHOONG, Florent Bréart, and Fertog for successfully solving this math challenge question! Gabriel N. was the first person to solve the question.
Your support is a heartfelt source of encouragement that propels the channel forward.
Please consider taking a second to subscribe in order to express your valuable support and receive notifications for the latest videos!
Any likes, subscriptions, comments, constructive criticisms, etc., are very much appreciated.
For more Weekly Math Challenges:
usclip.net/p/PLpoKXj-PWCbaDXYHES37_zX4O-kCWxguM

## Comments • 30

• My solution: Solve for n. n=x(n-1)/[x²(n)-1].
Take lim(n→∞) on both sides. Hence, the RHS must diverge.
This is possible in 2 cases:
Case I: x(n) converges to 1, hence making the denominator 0 and the numerator 1.
Case II: x(n-1) diverges to ∞. However, if that's the case, x(n) must also diverge to ∞, and more importantly, x(n)>x(n-1). But if that's the case, x²(n)-1>x(n-1) (since x(n) is arbitrarily big).
Hence, if {x(n)} diverges, the RHS must go to 0, not ∞.
Hence, Case I is right and Case II is wrong, and {x(n)} converges to 1.
QED.

• Arkadiy Gertsman 3 months ago

Another way of doing it:

Step 1: Show that the sequence converges as you have done
Step 2: Let lim x_n = x, and thus lim x_{n-1} = x as well since {x_{n-1}} is just the 1-tail of {x_n}. Now take the limit of both sides of the equation n(x_n)^2 - x_{n-1} - n = 0 to obtain nx^2 - x - n = 0 for all integers n. The only x which satisfies this equation is x=1, and so lim {x_n} = 1.

• Birat Acharya 4 months ago

such an elegant problem and solution was equally amazing too!

• Cienturista 4 months ago +1

Hello! I would like to know which program are you using to write on the screen, please.

• matyourin 4 months ago

I found a MUCH easier way to show this in two lines. I will write x(n) for the sequence:
n*x²(n) - x(n-1) = n is equivalent to x²(n) - x(n-1) / n = 1.
If x(n) has a limit, it has to be the same limit as x(n-1) has, lets call this limit c.
So: lim(x(n)) - lim(x(n-1))/n = 1 means: c² - 0 = 1. Since x(n) is positive, the limit c has to be 1.

• sjieh wang 4 months ago +1

wow, the second solution is stunning

• Jasper Robb 4 months ago +1

I dare you to prove that for any positive value for a, the limit as n --> infinity of n*integral (a^x)/x^n) dx from 1 to x will always approach a.

• The Religious Atheists Day ago

That doesn't even make sense. x is your integration variable, and you're also putting it as the upper bound.
I'm assuming that you meant "integral from 1 to *n*", and not x.

• Jonas Hammerich 4 months ago +2

This guy is secretly the asian version of Dr. Peyam

• Xander Gouws 4 months ago +1

Using the squeeze theorem there was so clever!

• Typo 4 months ago +2

I'm having trouble understanding as to how you proved x_(n+1)

• Chan Dan 4 months ago +1

Bro plz upload more challenges on sequences !

• pascal delcombel 4 months ago +2

I'm afraid you did change the challenge, by adding xn positive...

• Jan Von Schreibe 4 months ago

Yes I did not see this assumption in the original video

• el tapa 4 months ago +2

Yes! I was looking for the solution, very nice and elegant problem

• Zakiror Rahman 4 months ago +5

Lim.( xcos^3(x) - ln(1+ x) - sin^(-1)((x^2/2))/x^3
x-0

• Zakiror Rahman 4 months ago +2

Are you Black pen Red pen ?!?!

• Hyunwoo Park 3 months ago

He might be Korean, Steve is chinese

• Mokou Fujiwara 4 months ago

Their voices are a little bit similar, but I think they are different people.

• King Munch 4 months ago +1

I like Alon Heller’s prove of x_n

• Uttam Bhadauriya 4 months ago +5

This question is easy if you consider a multiple choice scenario ... but if you consider a subjective scenario where you have to explain your steps then it does test your knowledge of convergent sequences

• deus vult 3 months ago

Easy. Prove it's monotonic. Prove it's bounded. Use the recursive definition to find the limit. A really easy question. Compare that with some other calculous question and this one is a piece of cake.

• Benjamin Wang 4 months ago +3

At 4:00, does the weaker assumption that x_n is bounded already show that the limit is 1?

• LetsSolveMathProblems  4 months ago +2

The boundedness of {x_n} does imply that the limit of x_n / n is 0 (which in turn implies that the limit of {x_n} is 0 in our particular problem). I probably should have mentioned this in the video. Here is a short proof for the sake of completeness:

We assume that 0

• About Math 4 months ago

There is a little mistake, where you talk about the limit as n->+oo of x(n-1)/n=xεR, that's kinda wrong because if x(n-1)->o then the limit is 0/0 so we dont know where it tends.So it would be better if you wrote xεR*.Just a small parenthesis, anyway great video!

• Mokou Fujiwara 4 months ago +1

Multiple ways to prove, but actually x_n may not converge, but |x_n| surely converges.

• LetsSolveMathProblems  4 months ago +1

@Keshav Ramamurthy It depends on your strengths and weaknesses, but in general, I would recommend starting with Volume 1 since it exposes you to a variety of foundational topics; furthermore, most of the main ideas from the Intro to Counting book should be included in Volume 1, just in a more condensed form. Having said that, if you can consistently solve a lot of middle school algebra, arithmetic, and geometric problems but particularly struggle with counting problems, purchasing the Intro to Counting may not be a bad idea.

• Mokou Fujiwara 4 months ago

Revised or not, this problem is much easier than the current one lol.

• Keshav Ramamurthy 4 months ago +2

@LetsSolveMathProblems If I am a sixth-grader, would you reccomend I finish volume 1 or intro to counting for some middle school contest i have during may.
Thanks,
Keshav

• LetsSolveMathProblems  4 months ago +5

The revised version of the problem (which is presented in the video) assumes that x_n is non-negative for all n, so we know that x_n must converge.