For $n\in Z^{+}$ and $n>2$ prove that$$\frac{n(n-1)}{2}\le \sum_{k=1}^{n^2}\{\sqrt{k}\}\le \frac{n^2-1}{2},$$where $0\le k=[k]+\{k\}.$
I check that $$3<\sum_{k=1}^{9}\{\sqrt{k}\}=\sum_{k=1}^{9}(\sqrt{k}-[\sqrt{k}])=\sum_{k=1}^{9}\sqrt{k}-\sum_{k=1}^{9}[\sqrt{k}]<4,$$ where $$\sum_{k=1}^{9}\sqrt{k}\approx 19.306$$$$\sum_{k=1}^{9}[\sqrt{k}]=1+1+1+2+2+2+2+2+3=16$$ But I can not prove it in general case. Can you help me?
If $0\le a \le 2k$, then
$$\sqrt{k^2 + a} - k = \frac{a}{k + \sqrt{k^2+a}}$$
which means $\{\sqrt{k^2+a}\} = \frac{a}{k + \sqrt{k^2+a}}$, therefore
$$\sum_{m=1}^{n^2} \{\sqrt{m}\} = \sum_{l=1}^{n-1} \sum_{k=0}^{2 l} \frac{k}{l + \sqrt{l^2+k}} $$
Now you can estimate $ \frac{k}{2l} \ge \frac{k}{l + \sqrt{l^2+k}} \ge \frac{k}{2l+1}$, which gives you the bounds.
Proof.
By your work, $n=3$ $$3\le \sum_{k=1}^{9}\{\sqrt{k}\}\le 4.$$ The proof for RHS.$$\sum_{k=1}^{n^2}\{\sqrt{k}\}\le \frac{n^2-1}{2}.\tag {*}$$
Assuming that $n=m\ge 3.$ It gives$$\sum_{k=1}^{m^2}\{\sqrt{k}\}\le \frac{m^2-1}{2}.$$ We'll prove $(*)$ is true for $n=m+1.$
Notice that$$\sum_{k=1}^{(m+1)^2}\{\sqrt{k}\}=\sum_{k=1}^{m^2}\{\sqrt{k}\}+\sum_{k=m^2+1}^{(m+1)^2}\{\sqrt{k}\}\le \frac{m^2-1}{2}+\sum_{k=m^2+1}^{(m+1)^2}\{\sqrt{k}\}.$$ Thus, it suffices to prove$$\frac{m^2-1}{2}+\sum_{k=m^2+1}^{(m+1)^2}\{\sqrt{k}\}\le \frac{(m+1)^2-1}{2}.$$Or$$\sum_{k=m^2+1}^{(m+1)^2}\{\sqrt{k}\}\le \frac{2m+1}{2}.$$ Since $\{\sqrt{(m+1)^2}\}=\{m+1\}=0,$ it's$$\{\sqrt{m^2+1}\}+\{\sqrt{m^2+2}\}+...+\{\sqrt{m^2+2m}\}\le\frac{2m+1}{2} .$$ Also, it's obvious that$$m<\sqrt{m^2+1}<\sqrt{m^2+2}<...<\sqrt{m^2+2m}<m+1.$$
Thus, $[\sqrt{m^2+t}]=m,\forall 1\le t\le 2m.$ It means$$\{\sqrt{m^2+t}\}=\sqrt{m^2+t}-m,\forall 1\le t\le 2m.$$ Also, we may use$$\sqrt{m^2+t}-m=\sqrt{m^2+2\cdot\frac{t}{2m}\cdot m+\frac{t^2}{4m^2}-\frac{t^2}{4m^2}}-m< \frac{t}{2m},\forall 1\le t\le 2m.$$ Hence, $$\{\sqrt{m^2+1}\}+\{\sqrt{m^2+2}\}+...+\{\sqrt{m^2+2m}\}\le \frac{\sum_{t=1}^{2m}t}{2m}=\frac{2m+1}{2} .$$
The proof for LHS.$$\sum_{k=1}^{n^2}\{\sqrt{k}\}\ge \frac{n(n-1)}{2}.\tag {**}$$
Assuming that $n=m\ge 3.$ It gives$$\sum_{k=1}^{m^2}\{\sqrt{k}\}\le \frac{m(m-1)}{2}.$$ We'll prove $(*)$ is true for $n=m+1.$
Notice that$$\sum_{k=1}^{(m+1)^2}\{\sqrt{k}\}=\sum_{k=1}^{m^2}\{\sqrt{k}\}+\sum_{k=m^2+1}^{(m+1)^2}\{\sqrt{k}\}\ge \frac{m(m-1)}{2}+\sum_{k=m^2+1}^{(m+1)^2}\{\sqrt{k}\}.$$ Thus, it suffices to prove$$\frac{m(m-1)}{2}+\sum_{k=m^2+1}^{(m+1)^2}\{\sqrt{k}\}\ge \frac{m(m+1)}{2}.$$Or$$\{\sqrt{m^2+1}\}+\{\sqrt{m^2+2}\}+...+\{\sqrt{m^2+2m}\}\ge m .$$ Also, it's obvious that$$m<\sqrt{m^2+1}<\sqrt{m^2+2}<...<\sqrt{m^2+2m}<m+1.$$
Thus, $[\sqrt{m^2+t}]=m,\forall 1\le t\le 2m.$ It means$$\{\sqrt{m^2+t}\}=\sqrt{m^2+t}-m=\frac{t}{\sqrt{m^2+t}+m}\ge \frac{t}{\sqrt{m^2+2m+1}+m}=\frac{t}{1+2m} ,\forall 1\le t\le 2m.$$ Hence, $$\{\sqrt{m^2+1}\}+\{\sqrt{m^2+2}\}+...+\{\sqrt{m^2+2m}\}\ge \frac{\sum_{t=1}^{2m}t}{1+2m}= m .$$ We're done!
