<s>[INST] Calculate $\sum_\limits{k=1}^{100}\left \lfloor \sqrt{k}+\frac{1}{k} \right \rfloor$ Determine: $\sum_\limits{k=1}^{100}\left \lfloor \sqrt{k}+\frac{1}{k} \right \rfloor$.
It's well-known that $x-1<\lfloor x \rfloor \leq x, \forall x \in \mathbb{R}$, but this didn't help me at all.
I computed it and the sum equals 627, but I found no useful properties.
 [/INST] $f(k)=\left \lfloor \sqrt{k}+\frac{1}{k} \right \rfloor$. We have $f(1)=2,\space f(2)=1,\space f(3)=2$ so $$\sum_\limits{k=1}^{100}\left \lfloor \sqrt{k}+\frac{1}{k} \right \rfloor=5+\sum_\limits{k=4}^{100}\left \lfloor \sqrt{k}+\frac{1}{k} \right \rfloor$$
Now, between $n^2$ and $(n+1)^2-1$ one has $\sqrt{k}+\dfrac{1}{k}=n$ from which
$$\sum_\limits{k=4}^{100}\left \lfloor \sqrt{k}+\frac{1}{k} \right \rfloor=\sum_\limits{k=2}^{9}n(2n+1)+10=622$$
Thus $$\sum_\limits{k=1}^{100}\left \lfloor \sqrt{k}+\frac{1}{k} \right \rfloor=622+5=\color{red}{627}$$
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