Sum?

Algebra Level 3

What is the largest integer less than or equal to ∑ n = 1 2019 1 + 1 n 2 + 1 ( n + 1 ) 2 \sum_{n = 1}^{2019}\sqrt{1+\frac{1}{n^2}+\frac{1}{(n+1)^2}}


The answer is 2019.

Junghwan Han by Junghwan Han , 19, South Korea

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2 solutions

4080399 2020 \frac{4080399}{2020}

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The i'th term of the series is 1+ 1 i ( i + 1 ) \dfrac{1}{i(i+1)} . Sum of the series upto n terms is n+1- 1 n + 1 \dfrac{1}{n+1} . The largest integer less than this is n. For n=2019, the answer is 2019.

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