Evaluating the Sum of $\frac{1}{k_n}$ for $n=1,2,\cdots,1980$

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The discussion focuses on evaluating the sum of the series $\sum_{n=1}^{1980} \frac{1}{k_n}$, where $k_n$ is defined as the integer closest to $\sqrt{n}$. Participants analyze the behavior of $k_n$ as $n$ increases, noting that $k_n$ changes values at specific intervals related to perfect squares. The final evaluation of the sum reveals that it converges to a specific numerical value, demonstrating the relationship between the sum and the distribution of integers around square roots.

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Let $k_n$ denote the integer closest to $\sqrt{n}$. Evaluate the sum $\dfrac{1}{k_1}+\dfrac{1}{k_2}+\cdots+\dfrac{1}{k_{1980}}$.
 
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anemone said:
Let $k_n$ denote the integer closest to $\sqrt{n}$. Evaluate the sum $\dfrac{1}{k_1}+\dfrac{1}{k_2}+\cdots+\dfrac{1}{k_{1980}}$.

There are 2n numbers that is $n^2-(n-1)$ to $n^2 + n$ closest to n and 2n times reciprocal of n (that is 1/n) = 2
Now 1980 = 44 * 45
which is $44 ^2 + 44$ which is closest to $44^2$ and 1981 is closest to $45^2$
So the sum is 2 * 44 = 88
 
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