Prove that $\sum_{k=n^2+1}^{n^2+2n+1}\sqrt{k}\le 2n^2+2n+\frac{7}{6}$

Question

prove that:

$$\displaystyle\sum_{k=n^2+1}^{n^2+2n+1}\sqrt{k}\le 2n^2+2n+\dfrac{7}{6},n\ge 1$$

I find this inequality is very strong. Thank you!

such:when $n=100$,we use the mathmatic $$\displaystyle\sum_{k=10001}^{10201}\approx 20201.16666254125412794937\cdots $$

and $2n^2+2n+\dfrac{7}{6}|_{n=100}=20201.166666666666\cdots$

The sum has $2n+1$ terms each of which is no greater than $\sqrt{n^2+2n+1}=n+1$. The sum can thus be bounded by $(2n+1)(n+1)=2n^2+3n+1$ which is a little weaker than what you are asking for. — Eckhard, Mar 28 '13 at 15:02
Thank you ,Are sure? you have $2n^2+3n+1>2n^2+2n+\dfrac{7}{6}$!,but is wrong — math110, Mar 28 '13 at 15:16

score 4 · Answer 1 · edited Jul 10 '23 at 13:13

Since $x\mapsto\sqrt{x}$ $\ (x>0)$ is a concave function we have for any $m>0$ the inequality $${1\over 2}\bigl(\sqrt{m}+\sqrt{m+1}\bigr)\leq\int_m^{m+1}\sqrt{x}\ dx\ .$$ Putting $$m:=n^2+k\qquad (0\leq k\leq 2n)$$ and summing over $k$ we get $${1\over2}\sqrt{n^2} +\sum_{k=1}^{2n}\sqrt{n^2 +k}+{1\over2}\sqrt{n^2+2n+1} \leq \int_{n^2}^{(n+1)^2}\sqrt{x}\ dx={2\over3}\bigl((n+1)^3-n^3\bigr)\ .$$ Now extend the sum on the LHS to $k=2n+1$ and collect the remains. You get $$\sum_{k=1}^{2n+1}\sqrt{n^2 +k}-{1\over2}\leq 2n^2+2n+{2\over3}\ ,$$ which is equivalent with the stated inequality.

Calvin Lin · Accepted Answer · 2013-03-28T15:38:58.753

Note that your bound gets much tighter as $n$ increases, suggesting that some kind of limiting argument is involved.

With that insight, we have that

$$ \begin{align} \sum_{k = n^2 + 1}^{k=n^2+2n+1} \sqrt{k} & = \sum_{k=n^2} ^{k=n^2 + 2n} \sqrt{k} + 1 \\ & = 1 + \sum_{k=n^2} ^{k=n^2+2n} \int_k^{k+1} \sqrt{k} \, dx \\ & \leq 1 + \sum_{k=n^2} ^ {k=n^2 + 2n} \int_{k}^{k+1} \sqrt{x} \, dx \\ & = 1 + \int_{n^2} ^{(n+1)^2} \sqrt{x} \, dx\\ & = 1 + [ \frac {2}{3} k^{\frac {3}{2}} ]_{n^2} ^{(n+1)^2} \\ & = 1 + [\frac {2}{3} (3n^2 + 3n+1) ] \\ & = 2n^2 + 2n + \frac {5}{3} \end{align} $$

Now, let's consider how much we over approximated by. Looking at the inequality, we over approximated by $\int_{k}^{k+1} \sqrt{x} - \sqrt{k}$. Since the square root function is concave, we have the extra triangle on top, which tells us that we over approximated by at least $\frac {1}{2} (\sqrt{k+1}-\sqrt{k})$. Summing this up, we get $$\sum_{k=n^2}^{n^2+2n} \frac {1}{2} ( \sqrt{k+1}- \sqrt{k} ) = \frac {1}{2} ( k+1 - k) = \frac {1}{2}. $$

Hence, (back)accounting for this over-approximation, we know that

$$ \sum_{k = n^2 + 1}^{k=n^2+2n+1} \sqrt{k} \leq 2n^2 + 2n + \frac {5}{3} - \frac {1}{2} = 2n^2 + 2n + \frac {7}{6}. $$

Prove that $\sum_{k=n^2+1}^{n^2+2n+1}\sqrt{k}\le 2n^2+2n+\frac{7}{6}$

2 Answers2