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?
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}