show that $$1342<2\sum_{k=1}^{100}\sqrt{k}<1343$$
My idea: since $$x\in(k-1,k)\Longrightarrow \sqrt{k-1}\le\sqrt{x}\le \sqrt{k}$$ so $$\int_{k-1}^{k}\sqrt{k-1}dx\le\int_{k-1}^{k}\sqrt{x}dx\le\int_{k-1}^{k}\sqrt{k}dx$$ or $$\sqrt{k-1}\le\int_{k-1}^{k}\sqrt{x}\le\sqrt{k}$$ But I found this inequality can't solve my problem,so maybe can use other methods to solve it,
Thank you
Since $\sqrt{x}$ is a concave function, its integral over $[2,100]$ is larger than its estimate over the same interval using Trapezoid rule. As a result,
$$2 \sum_{k=1}^{100} \sqrt{k} \le 2\left[ 1 + \frac12(\sqrt{2} + \sqrt{100}) + \int_2^{100} \sqrt{x} dx \right] = 12 + \sqrt{2} + \frac43\left(100^{3/2} - 2^{3/2}\right)\\ = \frac{4036-5\sqrt{2}}{3} \approx 1342.97631 < 1343. $$ For any $k \ge 1$, $\sqrt{x}$ is concave over $[k-\frac12,k+\frac12]$ implies
$$\sqrt{k} \ge \int_{k-1/2}^{k+1/2} \sqrt{x}dx = \frac23 \left( (k+\frac12)^{3/2} - (k-\frac12)^{3/2} \right)$$ This leads to a bound in the other direction: $$2\sum_{k=1}^{100}\sqrt{k} \ge \frac43 \sum_{k=1}^{100}\left( (k+\frac12)^{3/2} - (k-\frac12)^{3/2} \right)= \frac43 \left[ \left( \frac{201}{2} \right)^{3/2} - \left( \frac12 \right)^{3/2} \right]\\ \approx 1342.87442 > 1342$$
Creative telescoping techniques often provides better bounds with respect to the pure analytic counterpart - as an example, have a look at this similar problem. In my answer I proved that $$\frac{2}{3}N\sqrt{N}\leq\sum_{k=1}^{N}\sqrt{k}\leq\frac{4N+3}{6}\sqrt{N},$$ that give a gap of $10$ between the LHS and the RHS in your problem. However, we can carry on the telescoping technique one step further, hoping to get a smaller gap. Consider that: $$(n+1)^{3/2}-n^{3/2}=\frac{3n^2+3n+1}{(n+1)^{3/2}+n^{3/2}}=\frac{3}{4}\left(\sqrt{n+1}+\sqrt{n}\right)+R(n),\tag{1}$$ where $$ 0\leq R(n)\leq\frac{1}{32\cdot n^{3/2}}.\tag{2}$$ If we define $S$ as $S=\sum_{k=1}^{100}\sqrt{k}$, $(1)$ and $(2)$ give: $$S-\frac{1}{2}-5=\sum_{k=1}^{99}\frac{\sqrt{n+1}+\sqrt{n}}{2}\leq\frac{2}{3}\sum_{k=1}^{99}\left((n+1)^{3/2}-n^{3/2}\right)=\frac{2}{3}\cdot999=666,\tag{3}$$ hence $S\leq 1343$, while: $$666-(S-11/2)\leq\frac{2}{3}\sum_{k=1}^{100}\frac{1}{32\cdot k^{3/2}}\leq\frac{\zeta(3/2)}{48}<\frac{1}{18},\tag{3}$$ hence $S\geq 1343-\frac{1}{9}$.
The following proof relies entirely on rational arithmetic.
Denote the quantity in question by $Q$, and put $S:=\sum_{k=2}^{99}\sqrt{k}$. Since the function $x\mapsto\sqrt{x}$ is concave the trapezoidal rule undershoots the integral, and we have $$666=\int_1^{100}\sqrt{x}\ dx>{1\over2}\bigl(\sqrt{1}+\sqrt{100}\bigr)+S\ ,$$ which leads to $$Q=2 (S+11)<1343\ .$$ We now approximate the same integral a second time from above with trapezoids, by drawing tangents at the points $(k,\sqrt{k})$ $\>(1\leq k\leq 100)$. The end trapezoids have width ${1\over2}$, the remaining ones have midline at $x=k$ and width $1$. In this way we get $$666<{1\over2}{1+{5\over4}\over2}+ S +{1\over2}{10+(10-{1\over40})\over 2}\ .$$ This gives $$S>{105\,671\over160}\ ,$$ from which we obtain $$Q=2(S+11)>{107\,431\over80}\doteq 1342.89\ .$$ The true value is $\doteq 1342.9259$.
