Find the largest Integer, without programming!

Algebra Level 5

1 2 + 3 + 1 4 + 15 + … + 1 2 n + 4 n 2 − 1 < 2013 2 \frac1{\sqrt{2+\sqrt3}} + \frac1{\sqrt{4+\sqrt{15}}} + \ldots + \frac1{\sqrt{2n+\sqrt{4n^2-1}}} < 2013 \sqrt2

Find the largest positive integer n n such that it satisfies the inequality above.


The answer is 8108363.

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Satyajit Mohanty by Satyajit Mohanty , 24, India

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3 solutions

Chew-Seong Cheong
Jul 12, 2015

Let S = L H S S = LHS , then we have:

S = ∑ k = 1 n 1 2 k + 4 k 2 − 1 = ∑ k = 1 n 1 ( k + 1 2 ) + ( k − 1 2 ) + 2 ( k + 1 2 ) ( k − 1 2 ) = ∑ k = 1 n 1 ( k + 1 2 + k − 1 2 ) 2 = ∑ k = 1 n 1 k + 1 2 + k − 1 2 = ∑ k = 1 n k + 1 2 − k − 1 2 k + 1 2 + k − 1 2 = ∑ k = 1 n ( k + 1 2 − k − 1 2 ) = ∑ k = 1 n k + 1 2 − ∑ k = 1 n k − 1 2 = ∑ k = 1 n k + 1 2 − 1 2 − ∑ k = 1 n − 1 k + 1 2 = n + 1 2 − 1 2 \begin{aligned} S & = \sum_{k=1}^n {\frac{1}{\sqrt{2k+\sqrt{4k^2-1}}}} \\ & = \sum_{k=1}^n {\frac{1}{\sqrt{(k+\frac{1}{2} ) + (k - \frac{1}{2}) + 2\sqrt{(k+\frac{1}{2})(k-\frac{1}{2})}}}} \\ & = \sum_{k=1}^n {\frac{1}{\sqrt{\left(\sqrt{k+\frac{1}{2}} + \sqrt{k - \frac{1}{2}}\right)^2}}} \\ & = \sum_{k=1}^n {\frac{1}{\sqrt{k+\frac{1}{2}} + \sqrt{k - \frac{1}{2}}}} \\ & = \sum_{k=1}^n {\frac{\sqrt{k+\frac{1}{2}} - \sqrt{k-\frac{1}{2}}}{k+\frac{1}{2} + k - \frac{1}{2}}} \\ & = \sum_{k=1}^n {\left(\sqrt{k+\frac{1}{2}} - \sqrt{k-\frac{1}{2}} \right)} \\ & = \sum_{k=1}^n \sqrt{k+\frac{1}{2}} - \sum_{k=1}^n \sqrt{k-\frac{1}{2}} \\ & = \sum_{k=1}^n \sqrt{k+\frac{1}{2}} - \sqrt{\frac{1}{2}} - \sum_{k=1}^{n-1} \sqrt{k+\frac{1}{2}} \\ & = \sqrt{n+\frac{1}{2}} - \sqrt{\frac{1}{2}} \end{aligned}

⇒ S = n + 1 2 − 1 2 < 2013 2 ⇒ n + 1 2 < 2013 2 + 1 2 n + 1 2 < 201 3 2 ˙ 2 + 2 ˙ 2013 2 ˙ 1 2 + 1 2 n < 2 ˙ 2013 ( 2013 + 1 ) = 8108364 ⇒ n = 8108363 \begin{aligned} \Rightarrow S = \sqrt{n+\frac{1}{2}} - \sqrt{\frac{1}{2}} & < 2013\sqrt{2} \\ \Rightarrow \sqrt{n+\frac{1}{2}} & < 2013\sqrt{2} + \sqrt{\frac{1}{2}} \\ n + \frac{1}{2} & < 2013^2 \dot{} 2 + 2 \dot{} 2013 \sqrt{2} \dot{} \sqrt{\frac{1}{2}} + \frac{1}{2} \\ n & < 2\dot{} 2013 (2013+1) = 8108364 \\ \Rightarrow n & = \boxed{8108363} \end{aligned}

7 Helpful 0 Interesting 1 Brilliant 0 Confused

Moderator note:

Simple standard approach with telescoping sums.

How can one tell that the LHS allows for a nice telescoping sum?

Overrated, seriously!

Kartik Sharma - 5 years, 11 months ago

Though, I have solved many questions like this where a bizarre looking expression telescopes to a simple number, but, everytime I have to write at least three to four terms from beginning and end so as to recognise the fashion in which term cancels. So, I want to know, is there any technique by which we can ascertain in advance that a series having the general term as difference of two terms telescopes to a certain number and if it is then to what number it telescopes ? Any response will be greatly appreciated !

Aditya Sky - 5 years, 6 months ago
Satyajit Mohanty
Jul 11, 2015

2 Helpful 0 Interesting 0 Brilliant 0 Confused
Ravi Dwivedi
Jul 11, 2015

S = 1 2 + 3 + 1 4 + 15 + . . . + 1 2 n + 4 n 2 − 1 S=\frac{1}{\sqrt{2+\sqrt{3}}} + \frac{1}{\sqrt{4+\sqrt{15}}} +...+\frac{1}{\sqrt{2n+\sqrt{4n^2-1}}}\\ S = 2 4 + 2 3 + 2 8 + 2 15 + . . . + 2 4 n + 2 4 n 2 − 1 S=\frac{\sqrt{2}}{\sqrt{4+2\sqrt{3}}} + \frac{\sqrt{2}}{\sqrt{8+2\sqrt{15}}} +...+\frac{\sqrt{2}}{\sqrt{4n+2\sqrt{4n^2-1}}}\\ (Multiplying numerator and denomiantor by 2 \sqrt{2} in prev step) S = 2 1 + 3 + 2 3 + 5 + . . . + 2 2 n − 1 + 2 n + 1 S=\frac{\sqrt{2}}{1+\sqrt{3}} + \frac{\sqrt{2}}{\sqrt{3}+\sqrt{5}}+...+\frac{\sqrt{2}}{\sqrt{2n-1}+\sqrt{2n+1}}\\ S = 2 2 ( ( 3 − 1 ) + ( 5 − 3 ) + ( 7 − 5 ) + . . . + ( 2 n + 1 − 2 n − 1 ) ) S=\frac{\sqrt{2}}{2}((\sqrt{3}-1)+(\sqrt{5}-\sqrt{3})+(\sqrt{7}-\sqrt{5})+...+(\sqrt{2n+1}-\sqrt{2n-1}))\\ We obtained the last step by taking 2 \sqrt{2} common and rationalizing all the terms.

We proceed by noticing that it is a telescopic series, so all terms cancel except 2 n + 1 \sqrt{2n+1} and − 1 -1

S = 1 2 ( 2 n + 1 − 1 ) < 2013 2 (Given) S=\frac{1}{\sqrt{2}}(\sqrt{2n+1}-1)<2013\sqrt{2} \text{(Given)} 2 n + 1 − 1 < 4026 \sqrt{2n+1}-1 < 4026 2 n + 1 < 4027 \sqrt{2n+1}<4027 2 n + 1 < 16216729 2n+1<16216729 2 n < 16216728 2n<16216728 n < 8108364 n<8108364

Answer is 8108363 \boxed{8108363}

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Simple standard approach.

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