Direct comparison

Now, while $k>0$ , $\begin{aligned}k-1<k+1\implies\frac{k-1}{k(k+1)}&<\frac1k\\ \implies\sqrt{\frac{k-1}{k(k+1)}}&<\frac1{\sqrt k}\\ \implies\sum_{k=1}^n\sqrt{\frac{k-1}{k(k+1)}}&<\sum_{k=1}^n\frac1{\sqrt k}\qquad\qquad(1)\end{aligned}$ Now, we can rearrange the sum of $\frac1{\sqrt k}$ to get something that looks like a Riemann sum for the function $f(x)=\frac1{\sqrt x}$ : $\sum_{k=1}^n\frac1{\sqrt k}=\sqrt n\left(\frac1n\sum_{k=1}^n\sqrt{\frac nk}\right)\approx\sqrt n\int_0^1\frac1{\sqrt x}dx$ Since $k$ begins at $1$ and ends at $n$ , we can see that this Riemann sum is using the right end of each subinterval for the height. Then, since $\frac1{\sqrt x}$ is a decreasing function, this must be a lower Riemann sum, and hence: $\sum_{k=1}^n\frac1{\sqrt k}<\sqrt n\int_0^1\frac1{\sqrt x}dx=\sqrt n\big[2\sqrt x\big]_0^1=2\sqrt n$ $\implies\sum_{k=1}^n\frac1{\sqrt k}<2\sqrt n\qquad\qquad(2)$ From $(1)$ and $(2)$ , we conclude: $\boxed{\sum_{k=1}^n\sqrt{\frac{k-1}{k(k+1)}}<2\sqrt n\text{ for all positive integer values of }n}$

For any positive integer $n$ , we have $\begin{aligned} n^3 + 6n^2 + 8n + 4 \; > \; 0 & \Rightarrow \; n^4 + 4n^3 + 8n^2 + 8n + 4 \; > \; n^2(n+1)(n+2) \; \Rightarrow \; n^2 + 2n + 2 \; > \; n\sqrt{(n+1)(n+2)} \\ & \Rightarrow \; 4n^2 + 11n + 8 \; > \; 4n^2 + 8n + 8 \; > \; 4n\sqrt{(n+1)(n+2)} \; \Rightarrow \; 4(n+1)(n+2) \; > \; n + 4n\sqrt{(n+1)(n+2)} \\ & \Rightarrow \; 4 \; > \; \frac{n}{(n+1)(n+2)} + \frac{4n}{\sqrt{(n+1)(n+2)}} \; \Rightarrow \; 4(n+1) \; > \; 4n + \frac{n}{(n+1)(n+2)} + \frac{4n}{\sqrt{(n+1)(n+2)}} \\ & \Rightarrow \; 2\sqrt{n+1} \; > \; 2\sqrt{n} + \sqrt{\frac{n}{(n+1)(n+2)}} \; \Rightarrow \; \sqrt{\frac{n}{(n+1)(n+2)}} \; < \; 2\big(\sqrt{n+1} - \sqrt{n}\big) \end{aligned}$ Note that the inequality $\sum_{m=1}^n \sqrt{\frac{m-1}{m(m+1)}} \; < \, 2\sqrt{n} \hspace{3cm} (\star)$ is true when $n=1$ . Since $\sum_{m=1}^{n+1} \sqrt{\frac{m-1}{m(m+1)}} \; = \; \sum_{m=1}^n \sqrt{\frac{m-1}{m(m+1)}} + \sqrt{\frac{n}{(n+1)(n+2)}} \; < \; \sum_{m=1}^n \sqrt{\frac{m-1}{m(m+1)}} + 2\big(\sqrt{n+1}-\sqrt{n}\big)$ the truth of inequality $(\star)$ for all positive integers $n$ follows by induction.

Let $f(n) = \sum_{i=1}^{n} \sqrt{\frac{i-1}{i(i+1)}}$ denote the left-hand side and $g(n) = 2\sqrt{n}$ , the right-hand side of the inequality.

Claim: for all positive integral $n, f(n) < g(n)$ .

Proof by induction:

True for the base case $n=1$ , as $f(1) = 0 < g(1) = 2$ .

Now for the induction step. Assume $f(n-1) < g(n-1)$ . Now we wish to see that $f(n) < g(n)$ . It suffices to show $f(n) - f(n-1) < g(n) - g(n-1)$ , as that would show that the right-hand side is growing faster than the left-hand side.

Note $f(n) - f(n-1) = \sqrt{\frac{n-1}{n(n+1)}}$ . And we can use the mean value theorem to bound $g(n) - g(n-1)$ .

The mean value theorem tells us that $g(n) - g(n-1) = g'(x_0)$ , for a value of $x_0 \in (n-1,n)$ . Now $g'(x) = \sqrt{\frac{1}{x}}$ , and $g'(x)$ is a decreasing function.

Thus $g'(x_0) > \sqrt{\frac{1}{n}}$ .

In conclusion, $f(n) - f(n-1) = \sqrt{\frac{n-1}{n(n+1)}} < \sqrt{\frac{1}{n}} < g(n) - g(n-1)$ . The inequality in the middle is obvious, since $\frac{n-1}{n+1} < 1$ .

According to the first mathematical induction, the problem is an actually easy one.

The only thing we need to do is just to substract $2\sqrt{k+1}$ from $2\sqrt{k}+\sqrt{\frac{k}{(k+1)(k+2)}}$ , and judge whether the outcome is negative absolutely.

The next step where we can employ a direct and violent way is to let $f(x)=2\sqrt{x}+\sqrt{\frac{x}{(x+1)(x+2)}}-2\sqrt{x+1}$

Then we have $\lim_{x \to +\infty}f(x)=0$ and $f'(x)=\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{x+1}}+(\sqrt{\frac{x}{(x+1)(x+2)}})'>0$ . Thus we come to the conclusion that $f(x)<0$ when $x>0$ , which means $f(n)=2\sqrt{n}+\sqrt{\frac{n}{(n+1)(n+2)}}-2\sqrt{n+1}<0$ for every positive integer $n$ .

Then the problem is solved with the validity of the first mathematical induction.