Showing that an upper sum is an upper bound of the improper integral

Question

The question states to show that for $N = 1,2,3,...$ we have:

$$\sum_{k=1}^{N} \frac{1}{\sqrt{k^2+1}+k} \color{green}> \frac{1}{2}\ln{\frac{2N + 1}{3}}$$

My idea is to show the following

$$\sum_{k=1}^{N} \frac{1}{\sqrt{k^2+1}+k} \color{red}{\geq} \int_{0}^{N} \frac{dx}{\sqrt{x^2+1}+x}$$

$$\left.\sum_{k=1}^{N} \frac{1}{\sqrt{k^2+1}+k} \color{red}{\geq} \frac{\sqrt{x^2+1}x - x^2}{2} + \frac{\ln{(\sqrt{x^2+1}+x})}{2} \right|_{0}^{N} = \frac{\sqrt{N^2+1}N - N^2}{2} + \frac{\ln{(\sqrt{N^2+1}+N})}{2}$$

Notice the strict inequality in the first inequality and then in the last inequality we can remove the first term on the RHS to have only the logarithm as follows.

$$\sum_{k=1}^{N} \frac{1}{\sqrt{k^2+1}+k} \color{green}> \frac{\ln{(\sqrt{N^2+1}+N})}{2} \color{green}> \frac{\ln{(\sqrt{2N^2+1}})}{2} \color{green}> \frac{1}{2}\ln{\frac{2N + 1}{3}}$$ $\blacksquare$

Now my concern is that can I do it any simpler way than having to compute that horror of an integral? Thank you for any insights!

Answer 1

Here is a solution:

\begin{align*} \sum_{k=1}^{N} \frac{1}{\sqrt{k^2+1}+k} &\geq \sum_{k=1}^{N} \frac{1}{(k+1)+k} \geq \int_{1}^{N} \frac{\mathrm{d}x}{2x+1} = \frac{1}{2}\log\left(\frac{2N+1}{3}\right). \end{align*}

In the second step, we used the following general observation:

Observation. Let $f : [n, n+1] \to \mathbb{R}$ be non-increasing. Then

$$ f(n) = \int_{n}^{n+1} f(n) \, \mathrm{d}x \geq \int_{n}^{n+1} f(x) \, \mathrm{d}x. $$