Inequality problem: $\sum_{i=1}^{n}\sqrt{i}\le (n+1)\sqrt{n+1}-1\quad\forall n\in\mathbb{N}$

Question

Hey guys have some problems with this one. Hope you can help me with solving it. I'm happy for just some tipps.

$$\displaystyle\sum_{i=1}^{n} \sqrt{i} \leq (n+1) \sqrt{n+1}-1 \quad\forall\ n\in \mathbb{N}$$

Answer 1

You may prove first that for any $k\geq 1$

$$ \sqrt{k} \leq \frac{2}{3}\left((k+1)^{3/2}-k^{3/2}\right) \tag{1}$$ holds by Lagrange's theorem: $$\frac{2}{3}\left((k+1)^{3/2}-k^{3/2}\right) = \sqrt{\xi},\quad \xi\in(k,k+1). $$

Then by summing both sides of $(1)$ over $k=1,2,\ldots,n$ we get the better bound $$ \sum_{k=1}^{n}\sqrt{k}\leq \color{red}{\frac{2}{3}\left((n+1)^{3/2}-1\right)}\tag{2}$$ through a telescopic sum. With a similar approach, we also have: $$ \sum_{k=1}^{n}\sqrt{k}\geq \frac{2}{3}\left(\left(n+\frac{1}{2}\right)^{3/2}-\frac{1}{2\sqrt{2}}\right).$$

Answer 2

Note that $\sqrt{i}\leq\sqrt{n+1}$ for $i=1,\dots,n$. Therefore $$\sum_{i=1}^{n} \sqrt{i}\leq n\cdot \sqrt{n+1}.$$ Since $\sqrt{n+1}\geq 1$, $$\sum_{i=1}^{n} \sqrt{i}\leq n\cdot \sqrt{n+1}+\sqrt{n+1}-1=(n+1)\cdot \sqrt{n+1}-1.$$

Answer 3

$$\sqrt{\lfloor t}\rfloor\le\sqrt t,$$

then integrating in unit intervals,

$$\sum_{i=1}^{n}\sqrt i\le\sum_{i=1}^{n}\int_{t=i}^{i+1}\sqrt tdt=\int_1^{n+1}\sqrt tdt=\frac23((n+1)\sqrt{n+1}-1)\le(n+1)\sqrt{n+1}-1.$$

You actually get a tighter bound.

Answer 4

Another take on it, a-la Cauchy-Schwarz:

\begin{align} \left(\sum_{k=1}^n\sqrt{k}\right) & \le \sqrt{ \left(\sum_{k=1}^n k\right)}\times \sqrt{ n}\\ &=\frac{ \sqrt{n+1} \times n }{\sqrt{2}}\\&=\frac{ \sqrt{n+1}(n+1) - \sqrt{n+1}}{\sqrt{2}}\\ &\le \frac{ \sqrt{n+1}(n+1)}{\sqrt{2}}-1. \end{align}

Note:

1. The inequality in the first line is an equality if and only if $n=1$.
2. The inequality in the last line is an equality if and only if $n=1$.
3. For all $n$, the expression on the last line is strictly smaller than $\sqrt{n+1}(n+1)-1$.