Okay, where does it end?

Algebra Level 4

$\large\displaystyle Y= \left[ \left(1+\frac{1}{n}\right) \left(1+\frac{2}{n}\right) \left(1+\frac{3}{n}\right) \cdots \left(1+\frac{n}{n}\right) \right]^{\frac{1}{n}}$

Find the value of $\lceil Y \rceil$ , where $n$ is a positive integer.

Submit your answer as 123 if you think that the answer depends on the value of $n$

Notation : $\lceil \cdot \rceil$ denotes the ceiling function .

This problem is a part of the set All-Zebra

The answer is 2.

3 solutions

Chew-Seong Cheong
May 8, 2016

Using the AM-GM inequality, we have:

$\begin{aligned} Y & \le \frac{1}{n} \sum_{k=1}^n \left(1+\frac{k}{n}\right) \\ & \le \frac{1}{n} \sum_{k=1}^n 1 + \frac{1}{n^2} \sum_{k=1}^n k \\ & \le \frac{n}{n} + \frac{n(n+1)}{2n^2} \\ & \le \frac{3}{2} + \frac{1}{2\color{#3D99F6}{n}^2} \quad \quad \small \color{#3D99F6}{\text{As } \frac{3}{2} + \frac{1}{2n^2} \text{ is maximum when }n=1 \text{ and }\frac{3}{2} + \frac{1}{2n^2}\le 2} \\ \implies Y & \le 2 \end{aligned}$

Since each factor $\left(1 +\dfrac {k} {n} \right) >1$ , their product $Y>1$ .

Therefore $1 <Y\le 2 \implies \lceil Y \rceil =\boxed {2}$

Elegant :) !!

Abhay Tiwari - 5 years, 1 month ago

Yes, but can there be an alternative approach for solving problems which state that n is a positive integers by substituting n = 1,2,3,... For example
when n=1, then $\lceil Y\rceil = 2$
when n=2, then $\lceil Y\rceil = 2$
when n=3, then $\lceil Y\rceil = 2$
Thoughts?

Ashish Menon - 5 years, 1 month ago

But it does not prove that for all infinitely many values of $n$ it equals 2.

Nihar Mahajan - 5 years, 1 month ago

@Nihar Mahajan – Yes it is but then the question itself becomes wrong if it the answer is equal for some n but not for others right? :thinkingface:

Ashish Menon - 5 years, 1 month ago

@Ashish Menon – Its a more like proof based question rather than answer-computing.Proving that would improve your skills rather than taking advantage of the brilliant's system of answer-accepting ;)

Nihar Mahajan - 5 years, 1 month ago

@Nihar Mahajan – Accepted :nomouth:

Ashish Menon - 5 years, 1 month ago

@Nihar Mahajan – I think the question should include "this should be true for all n. If you think this is not possible, enter 123".

Aryan Gaikwad - 5 years, 1 month ago

@Nihar Mahajan – It never equals to 2, but always stays between 1 and 2.

Abhay Tiwari - 5 years, 1 month ago

@Abhay Tiwari – I obviously know that.But $\lceil Y \rceil$ remains 2 right?

Nihar Mahajan - 5 years, 1 month ago

@Nihar Mahajan – Yes, it will $always$ remain 2.

Abhay Tiwari - 5 years, 1 month ago

@Abhay Tiwari – What about when n=1?

Ashish Menon - 5 years, 1 month ago

@Ashish Menon – Ohk, it equal to 2, forgot to mention that. :P

Abhay Tiwari - 5 years, 1 month ago

@Abhay Tiwari – Nvm, i agree with nihar though it is better to prove it for all values than that method, btw it is valid still in JEE right?

Ashish Menon - 5 years, 1 month ago

@Ashish Menon – Yep, you can use it as a "shortcut" or "time-saving" method in JEE. (Though such a question has a very low probability to be asked in JEE).However, its not mathematically complete.

Nihar Mahajan - 5 years, 1 month ago

@Nihar Mahajan – Congo on 2000 followers

Ashish Menon - 5 years, 1 month ago

Ashish, there is one more solution at the last, maybe that is the answer to all the questions you seek. ;), he has done simply by logic plus by conventional method.

Abhay Tiwari - 5 years, 1 month ago

@Abhay Tiwari – That was a small typo plz correct it thanks! :))

Ashish Menon - 5 years, 1 month ago

Awesome solution from you as always @Chew-Seong Cheong

Sudhir Aripirala - 5 years, 1 month ago

展豪張
May 8, 2016

On one hand, $Y>((1+\dfrac 1n)(1+\dfrac 1n)\cdots(1+\dfrac 1n))^{\frac 1n}=1+\dfrac 1n > 1$
On the other hand, $Y<((1+\dfrac nn)(1+\dfrac nn)\cdots(1+\dfrac nn))^{\frac 1n}=1+\dfrac nn = 2$
Concluded, $1<Y<2$
$\therefore \lceil Y \rceil = 2$

Great, can you find the exact value?

Abhay Tiwari - 5 years, 1 month ago

Exact value? Do you mean when $n \to \infty$ ?

展豪張 - 5 years, 1 month ago

Yes. When $n \to \infty$

Abhay Tiwari - 5 years, 1 month ago

@Abhay Tiwari – $\;\;\;\;\ln\lim_{n\to\infty} ((1+\dfrac 1n)(1+\dfrac 2n)\cdots(1+\dfrac nn))^{\frac 1n}$
$=\lim_{n\to\infty}\dfrac 1n (\ln (1+\dfrac 1n)+\ln (1+\dfrac 2n)+\cdots \ln (1+\dfrac nn))$
$=\displaystyle \int_0^1 \ln(1+x)\mathrm dx$
$=\displaystyle(1+x)\ln(1+x)|_0^1 - \int_0^1 \mathrm dx$
$=2\ln 2 -1$
$\displaystyle\therefore \lim_{n\to\infty} Y=e^{2\ln 2-1}=\dfrac 4e$

展豪張 - 5 years, 1 month ago

@展豪張 – Great :(+1): :)

Abhay Tiwari - 5 years, 1 month ago

@Abhay Tiwari – At first I thought your question is asking this, but I soon find out it's actually asking $\lceil Y\rceil$ . =D
:like: (+1)

展豪張 - 5 years, 1 month ago

@展豪張 – :D, still the answer remains same, Lol.

Abhay Tiwari - 5 years, 1 month ago

Abhay, it was me who change the wording of the problem. I hope it is okay for you.

Chew-Seong Cheong - 5 years, 1 month ago

Sir, it looks great now. :)

Abhay Tiwari - 5 years, 1 month ago

A P
May 9, 2016

Without loss of generality, we can assume n is equal to 1. Plugging it in, we find that Y is 2.

Okay, where does it end?

The answer is 2.

3 solutions

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