For any integer n⩾1, how do you show that $\sum_{k=1}^{n} \frac{2^k}{\sqrt{k+0.5}}\leqslant2^{n+1}\sqrt{n+1}-\frac{4n^{\frac{3}{2}}}{3}$

For any integer $$n\geqslant1$$, how do you show that: $$\sum_{k=1}^{n} \frac{2^k}{\sqrt{k+0.5}}\leqslant2^{n+1}\sqrt{n+1}-\frac{4n^{\frac{3}{2}}}{3}$$

Do you need to find the formula of the partial sum of $$\sum_{k=1}^{n} \frac{2^k}{\sqrt{k+0.5}}$$ first before moving on to proof whether the inequality is true?

I could not find a suitable way of writing the formula of the partial sum, since this is a divergent series, or is there a way to do so?

• Have you tried induction? May 11, 2020 at 6:52
• Have you tried integral comparison ?
– EDX
May 11, 2020 at 7:02
• Actually, this is supposed to be a high school contest math question... Dec 6, 2020 at 0:23

Consider the following upper bound

$$\sum_{k=1}^n \frac{2^k}{\sqrt{k+1/2}} \leq \sum_{k=1}^n 2^k \leq 2^{n+1}.$$

This hints that the required inequality holds for sufficiently large $$n$$. We now show that it holds for all $$n\geq 1$$, i.e., for all $$n\geq 1$$ $$2^{n+1}\leq 2^{n+1}\sqrt{n+1}-4n^{1.5}/3.$$

First observe this holds for $$n=1$$. Then the claim follows since for all $$n\geq 2$$ $$\sqrt{n}\leq 2\cdot (\sqrt{n+1}-1) \quad \text{and}\quad 4\cdot n \leq 3\cdot 2^{n}.$$

The following manipulations prove the claim \begin{align} \sqrt{n}&\leq 2\cdot (\sqrt{n+1}-1)\\ 2^{n}\sqrt{n}&\leq 2^{n+1}\cdot (\sqrt{n+1}-1)\\ 4n^{1.5}/3&\leq 2^{n+1}\cdot (\sqrt{n+1}-1)\\ 2^{n+1}&\leq 2^{n+1}\sqrt{n+1}-4n^{1.5}/3. \end{align}