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

[anemone]

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

 

[kaliprasad]

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

Spoiler

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