Simplify the general term of the sum: sqrt(1 + (1)/(n2) + (1)/((n+1)2))

Step 1: Simplify the general term of the sum.
Let the general term be $T_n = \sqrt{1 + \frac{1}{n^2} + \frac{1}{(n+1)^2}}$.
The expression inside the square root is:

$$1 + \frac{1}{n^2} + \frac{1}{(n+1)^2} = \frac{n^2(n+1)^2 + (n+1)^2 + n^2}{n^2(n+1)^2}$$ $$= \frac{n^2(n^2+2n+1) + (n^2+2n+1) + n^2}{n^2(n+1)^2}$$ $$= \frac{n^4+2n^3+n^2 + n^2+2n+1 + n^2}{n^2(n+1)^2}$$ $$= \frac{n^4+2n^3+3n^2+2n+1}{n^2(n+1)^2}$$

The numerator can be factored as $(n^2+n+1)^2$.

$$= \frac{(n^2+n+1)^2}{n^2(n+1)^2}$$

Now, substitute this back into the expression for $T_n$:

$$T_n = \sqrt{\frac{(n^2+n+1)^2}{n^2(n+1)^2}}$$ $$T_n = \frac{n^2+n+1}{n(n+1)}$$

We can rewrite this expression:

$$T_n = \frac{n(n+1)+1}{n(n+1)}$$ $$T_n = \frac{n(n+1)}{n(n+1)} + \frac{1}{n(n+1)}$$ $$T_n = 1 + \frac{1}{n(n+1)}$$

Using partial fraction decomposition for $\frac{1}{n(n+1)}$:

$$\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$$

So, the simplified general term is:

$$T_n = \boxed{1 + \frac{1}{n} - \frac{1}{n+1}}$$