$\sum_{k=3}^{n} \sqrt{1+\dfrac{1}{(k-1)^2}+\dfrac{1}{k^2} }$

$\sum_{k=3}^{n} \sqrt{\dfrac{((k-1)^2+1)k^2+(k-1)^2}{k^2(k-1)^2}}$

$\sum_{k=3}^{n} \sqrt{\dfrac{k^4-2k^3+3k^2-2k+1}{k^2(k-1)^2}}$

$\sum_{k=3}^{n} \sqrt{\dfrac{(k^2-k+1)^2}{k^2(k-1)^2}}$

$\sum_{k=3}^{n} \dfrac{(k^2-k+1)}{k(k-1)}$

$\sum_{k=3}^{n} 1+\dfrac{1}{k(k-1)}$

$n-3+\sum_{k=3}^{n} \dfrac{1}{k-1}-\dfrac{1}{k}$ $n-3+ \dfrac{1}{2}-\dfrac{1}{n}$ then

${\displaystyle\lim_{n\to\infty}\dfrac{\ln(n-\dfrac{5}{2}-\dfrac{1}{n})}{n}}$ = $\dfrac{\infty}{\infty}$

Using L'Hopital rule ${\displaystyle\lim_{n\to\infty}\dfrac{1+\dfrac{1}{n^2}}{n-\dfrac{5}{2}-\dfrac{1}{n}}}$ = $\dfrac{1}{\infty}= 0$

For some $n \in \mathbb{N}$ , as $n­ \rightarrow \infty$ , we would have:

$\displaystyle L=\lim_{n \to \infty} \frac{ln(x_n)}{n} \approx \lim_{n \to \infty} \frac {ln(x_{n+1})}{(n+1)}$

Expressing $x_{n+1}$ in terms of $x_n$ , follows:

$\displaystyle x_{n+1}=x_n+\sqrt{1+\frac{1}{n^2}+\frac{1}{(n+1)^2}}$

Handling terms and plugging one on another, we get:

$\displaystyle \lim_{n \to \infty} \ n.ln(x_n)+ln(x_n) \approx \lim_{n \to \infty} \ n.ln(x_{n+1})\implies$

$\displaystyle \implies L=\lim_{n \to \infty} \frac{ln(x_n)}{n} \approx \lim_{n \to \infty} ln\left(\frac{x_{n+1}}{x_n}\right)\implies$

$\displaystyle \implies L \approx \lim_{n \to \infty} ln\left(\frac{x_n+\sqrt{1+\frac{1}{n^2}+\frac{1}{(n+1)^2}}}{x_n}\right)$

One could easily use D'Alembert criterion and prove that $\{x_n\}$ diverges (i. e., the sequence diverges). Thus, $\lim_{n \to \infty} x_n=\infty$ . Then, L-limit becomes:

$\displaystyle \implies L \approx \lim_{n \to \infty} ln\left(1+\frac{\sqrt{1+\frac{1}{n^2}+\frac{1}{(n+1)^2}}}{x_n}\right) \implies$

$\displaystyle \implies L \approx \lim_{n \to \infty} ln\left(1+\frac{\sqrt{1+0+0}}{\infty}\right) \implies \boxed{L=0}$

$\blacksquare$

Only difficulty is to remove square root over kth term.this can be done as follows

for less complexity take lcm and take out (k-1) out of square root.

deviding numerator again by k^2 we get kth term as 1+1/k-1+1/k .

find xn by opening summation and apply limits