# Solution 88: 1/n and Square Roots

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**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.

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The Religious AtheistsDay agoMy 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 Gertsman3 months agoAnother 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 Acharya4 months agosuch an elegant problem and solution was equally amazing too!

Cienturista4 months ago^{+1}Hello! I would like to know which program are you using to write on the screen, please.

matyourin4 months agoI 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 wang4 months ago^{+1}wow, the second solution is stunning

Jasper Robb4 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 AtheistsDay agoThat 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 Hammerich4 months ago^{+2}This guy is secretly the asian version of Dr. Peyam

Xander Gouws4 months ago^{+1}Using the squeeze theorem there was so clever!

Typo4 months ago^{+2}I'm having trouble understanding as to how you proved x_(n+1)

Chan Dan4 months ago^{+1}Bro plz upload more challenges on sequences !

pascal delcombel4 months ago^{+2}I'm afraid you did change the challenge, by adding xn positive...

Jan Von Schreibe4 months agoYes I did not see this assumption in the original video

el tapa4 months ago^{+2}Yes! I was looking for the solution, very nice and elegant problem

Zakiror Rahman4 months ago^{+5}Lim.( xcos^3(x) - ln(1+ x) - sin^(-1)((x^2/2))/x^3

x-0

Zakiror Rahman4 months ago^{+2}Are you Black pen Red pen ?!?!

Hyunwoo Park3 months agoHe might be Korean, Steve is chinese

Mokou Fujiwara4 months agoTheir voices are a little bit similar, but I think they are different people.

King Munch4 months ago^{+1}I like Alon Heller’s prove of x_n

Uttam Bhadauriya4 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 vult3 months agoEasy. 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 Wang4 months ago^{+3}At 4:00, does the weaker assumption that x_n is bounded already show that the limit is 1?

LetsSolveMathProblems4 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 Math4 months agoThere 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 Fujiwara4 months ago^{+1}Multiple ways to prove, but actually x_n may not converge, but |x_n| surely converges.

LetsSolveMathProblems4 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 Fujiwara4 months agoRevised or not, this problem is much easier than the current one lol.

Keshav Ramamurthy4 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

LetsSolveMathProblems4 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.