Summation is inequal

Algebra Level 4

∑ r = 1 n 4 r + 1 ( 4 r − 1 ) 2 ( 4 r + 3 ) 2 > 1 75 \large \sum _{ r=1 }^{ n }{ \frac { 4r+1 }{ { ( 4r - 1 ) }^{ 2 }{ ( 4r + 3 ) }^{ 2 } } > \frac { 1 }{ 75 } }

Find the minimum possible value of natural number n n such that the inequality above is fulfilled.


The answer is 4.

Priyanshu Mishra by Priyanshu Mishra , 20, India

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1 solution

Firstly to do, we simplify the sum so we can do multiply and dividing easier ∑ r = 1 n 4 r + 1 ( 4 r − 1 ) 2 ( 4 r + 3 ) 2 = 1 8 ( ∑ r = 1 n 1 ( 4 r − 1 ) 2 − 1 ( 4 r + 3 ) 2 ) = 1 8 ( 1 9 − 1 ( 4 n + 3 ) 2 ) By using Telescoping Series \begin{aligned} \large \sum_{r=1}^{n} \frac{4r+1}{(4r-1)^2 (4r+3)^2} & \large = \frac{1}{8} {\left( \sum_{r=1}^{n} \frac{1}{(4r-1)^2} - \frac{1}{(4r+3)^2} \right)} \\ & \large = \frac{1}{8} {\left(\frac{1}{9} - \frac{1}{(4n+3)^2}\right)} & \small \color{#3D99F6}{\text{By using Telescoping Series}}\\ \end{aligned}

So, now that we have simplify the sum, we can go back to the inequality

1 8 ( 1 9 − 1 ( 4 n + 3 ) 2 ) > 1 75 1 9 − 1 ( 4 n + 3 ) 2 > 8 75 225 > ( 4 n + 3 ) 2 ± 15 > 4 n + 3 n > 3 Note that 0 > n is not acceptable \begin{aligned} \large \frac{1}{8} {\left(\frac{1}{9} - \frac{1}{(4n+3)^2}\right)} & \large > \frac{1}{75} \\ \large \frac{1}{9} - \frac{1}{(4n+3)^2} & \large > \frac{8}{75} \\ \large 225 & \large > (4n+3)^2 \\ \large \pm 15 & > \large 4n + 3 \\ \large n & \large > 3 & \small \color{#D61F06}{\text{Note that }\ 0 > n\ \text{is not acceptable} }\\ \end{aligned}

Therefore, the minimal value for n n is 4 \color{#D61F06}{\boxed{4}}

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