# 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.

• Is the bracket supposed to refer to the floor function? – John Lou Sep 15 '17 at 18:01
• Yes. Sorry, I forgot to specify it. – ztefelina Sep 15 '17 at 18:03
• Try it like here. – Dietrich Burde Sep 15 '17 at 18:03
• Note that $(\sqrt k+\frac1k)^2=k+\frac2{\sqrt k}+\frac1{k^2}<k+1$ for all but small $k$, hence you can use $\lfloor \sqrt k\rfloor$ for most summands – Hagen von Eitzen Sep 15 '17 at 18:04

Note that for $k\ge5$, $$\sqrt{k+1}-\sqrt{k}=\frac1{\sqrt{k+1}+\sqrt{k}}\gt\frac1k\tag{1}$$ so that $$\sqrt{k}\lt\sqrt{k}+\frac1k\lt\sqrt{k+1}\tag{2}$$ Taking the floor of $(2)$ says that $$\left\lfloor\sqrt{k}\right\rfloor\le\left\lfloor\sqrt{k}+\frac1k\right\rfloor\le\left\lfloor\sqrt{k+1}\right\rfloor\tag{3}$$ The only time that $\left\lfloor\sqrt{k}\right\rfloor\lt\left\lfloor\sqrt{k+1}\right\rfloor$ is when $k+1$ is a perfect square, which, in light of $(2)$ means that $\left\lfloor\sqrt{k}+\frac1k\right\rfloor\lt\left\lfloor\sqrt{k+1}\right\rfloor$. Thus, we get that for $k\ge5$ $$\left\lfloor\sqrt{k}+\frac1k\right\rfloor=\left\lfloor\sqrt{k}\right\rfloor\tag{4}$$ We can also verify $(4)$ for the case $k=4$.

Thus, \begin{align} \sum_{k=1}^{100}\left\lfloor\sqrt{k}+\frac1k\right\rfloor &=\overbrace{2+1+2+10}^{k=1,2,3,100}+\sum_{k=4}^{99}\left\lfloor\sqrt{k}\right\rfloor\\ &=15+\sum_{j=2}^9j\left((j+1)^2-j^2\right)\quad\text{floor times number of terms with that floor}\\ &=15+\sum_{j=2}^9\left(2j^2+j\right)\\ &=15+\sum_{j=2}^9\left(4\binom{j}{2}+3\binom{j}{1}\right)\\ &=15+\left[4\binom{j+1}{3}+3\binom{j+1}{2}\right]_1^9\\[3pt] &=15+4\binom{10}{3}+3\binom{10}{2}-4\binom{2}{3}-3\binom{2}{2}\\[12pt] &=15+480+135-0-3\\[18pt] &=627 \end{align}

$f(k)=\left \lfloor \sqrt{k}+\frac{1}{k} \right \rfloor$. 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 $$\sum_\limits{k=1}^{100}\left \lfloor \sqrt{k}+\frac{1}{k} \right \rfloor=622+5=\color{red}{627}$$

• (+1) I didn't see your answer before I posted mine, but yours is essentially the same, so I deleted it. – robjohn Sep 15 '17 at 19:55
• It is a situation that has happened to me many times but mainly because my writing in English requires more time than I would use to write in either Spanish or French. I do however think dear friend that you should have posted your post anyway. Regards. – Piquito Sep 15 '17 at 21:29
• By the way I have had a type with $\sqrt{k}+\dfrac{1}{k}=n$ without the symbol of floor function. I don't edit because this would imply a new publication. I record it here. – Piquito Sep 15 '17 at 21:36
• I will undelete it because there is a bit of extra justification included in it. – robjohn Sep 15 '17 at 21:37
• Very well, you are right. – Piquito Sep 15 '17 at 21:39