If $f(n)$ be the integer closest to $\sqrt{n}.$ then value of $ \sum^{2016}_{k=1}\frac{1}{f(k)}$ If $f(n)$ be the integer closest to $\sqrt{n}.$ then value of $\displaystyle \sum^{2016}_{k=1}\frac{1}{f(k)}$
could some help me with ths, thanks

Taking a cue from sequence A000194 in the OEIS, we have $$S = \sum_{k=1 }^{2016} \frac{1}{f (k)}$$ $$ = \frac {2}{1} + \frac {4}{2} + \frac {6}{3} + \cdots + \frac {2 \lfloor \sqrt {2016} \rfloor}{\lfloor \sqrt {2016} \rfloor} + \frac {36}{\lceil \sqrt {2016} \rceil}\,\,(\text{ why? })$$ $$=2+2+2\cdots +2  +\frac {36}{45} $$ where $2$ is added $44$ times giving an answer of $\boxed {88.8} $.
Hope it helps. 
