Get inspired by the success stories of our students in IIT JAM MS, ISI  MStat, CMI MSc DS.  Learn More 

A Telescopic Sequence| ISI MStat 2018 PSB Problem 2

This is a beautiful problem from ISI MStat 2018 problem 2, which uses the cute little ideas of telescopic sum and partial fractions.

Problem

Let \{x_{n}\}_{n \geq 1} be a sequence defined by x_{1}=1 and

    \[x_{n+1}=\left(x_{n}^{3}+\frac{1}{n(n+1)(n+2)}\right)^{1 / 3}, \quad n \geq 1\]


Show that \{x_{n}\}_{n \geq 1} converges and find its limit.

Prerequisities

  • Telescopic Sum \sum_{i = 1}^{\infty} \left(\underbrace{\frac1{i+1} -\frac1i}\right) =  \lim_{n \to \infty} \frac1n - 1 = -1

Solution

x_{n+1} = (x_{n}^{3}+\frac{1}{n(n+1)(n+2)})^{1 / 3} \Rightarrow {x_{n+1}}^3 = x_{n}^{3}+\frac{1}{n(n+1)(n+2)}

\Rightarrow {x_{n+1}}^3 -  x_{n}^{3} = \frac{1}{i(i+1)(i+2)}; x_1 = 1.

\Rightarrow \sum_{i  = 1}^{n-1} {x_{i+1}}^3 - x_{i}^{3} = \sum_{i = 1}^{n-1} \frac{1}{n(n+1)(n+2)} ; x_1 = 1.

x_{n}^{3} - x_{1}^{3} = \sum_{i = 1}^{n-1} \frac{1}{i(i+1)(i+2)} = \sum_{i = 1}^{n-1} -\frac12\left(\underbrace{\frac1{i+1} -\frac1i}\right)+\frac12\left(\underbrace{\frac1{i+2}-\frac1{i+1}}\right)

\lim_{n \to \infty} (x_{n}^{3} - x_{1}^{3}) = \sum_{i = 1}^{\infty} \frac{1}{i(i+1)(i+2)} = \sum_{i = 1}^{\infty} -\frac12\left(\underbrace{\frac1{i+1} -\frac1i}\right)+\frac12\left(\underbrace{\frac1{i+2}-\frac1{i+1}}\right) = \frac14

\lim_{n \to \infty} x_{n}^{3} = \frac54 \Rightarrow  \lim_{n \to \infty} x_{n} = ({\frac54})^\frac13.

This is a beautiful problem from ISI MStat 2018 problem 2, which uses the cute little ideas of telescopic sum and partial fractions.

Problem

Let \{x_{n}\}_{n \geq 1} be a sequence defined by x_{1}=1 and

    \[x_{n+1}=\left(x_{n}^{3}+\frac{1}{n(n+1)(n+2)}\right)^{1 / 3}, \quad n \geq 1\]


Show that \{x_{n}\}_{n \geq 1} converges and find its limit.

Prerequisities

  • Telescopic Sum \sum_{i = 1}^{\infty} \left(\underbrace{\frac1{i+1} -\frac1i}\right) =  \lim_{n \to \infty} \frac1n - 1 = -1

Solution

x_{n+1} = (x_{n}^{3}+\frac{1}{n(n+1)(n+2)})^{1 / 3} \Rightarrow {x_{n+1}}^3 = x_{n}^{3}+\frac{1}{n(n+1)(n+2)}

\Rightarrow {x_{n+1}}^3 -  x_{n}^{3} = \frac{1}{i(i+1)(i+2)}; x_1 = 1.

\Rightarrow \sum_{i  = 1}^{n-1} {x_{i+1}}^3 - x_{i}^{3} = \sum_{i = 1}^{n-1} \frac{1}{n(n+1)(n+2)} ; x_1 = 1.

x_{n}^{3} - x_{1}^{3} = \sum_{i = 1}^{n-1} \frac{1}{i(i+1)(i+2)} = \sum_{i = 1}^{n-1} -\frac12\left(\underbrace{\frac1{i+1} -\frac1i}\right)+\frac12\left(\underbrace{\frac1{i+2}-\frac1{i+1}}\right)

\lim_{n \to \infty} (x_{n}^{3} - x_{1}^{3}) = \sum_{i = 1}^{\infty} \frac{1}{i(i+1)(i+2)} = \sum_{i = 1}^{\infty} -\frac12\left(\underbrace{\frac1{i+1} -\frac1i}\right)+\frac12\left(\underbrace{\frac1{i+2}-\frac1{i+1}}\right) = \frac14

\lim_{n \to \infty} x_{n}^{3} = \frac54 \Rightarrow  \lim_{n \to \infty} x_{n} = ({\frac54})^\frac13.

Leave a Reply

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.

2 comments on “A Telescopic Sequence| ISI MStat 2018 PSB Problem 2”

Knowledge Partner

Cheenta is a knowledge partner of Aditya Birla Education Academy
Cheenta

Cheenta Academy

Aditya Birla Education Academy

Aditya Birla Education Academy

Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.
JOIN TRIAL
support@cheenta.com
Menu
Trial
Whatsapp
ISI Entrance Solutions
ISI CMI Self Paced
rockethighlight