Confusion in this algebraic limit approaching infinity 
Question : Evaluate $\lim_{x \to \infty}{\frac{1}{1+x^2} + \frac{2}{2+x^2} + \frac{3}{3+x^2}+...+\frac{x}{x + x^2}}$.

My working:
$$\lim_{x \to \infty}{\frac{1}{1+x^2} + \frac{2}{2+x^2} + \frac{3}{3+x^2}+...+\frac{x}{x + x^2}}$$
Dividing and transforming each fraction by $x^2$in the numerator and denominator.
$$=\lim_{x \to \infty}{\frac{\frac{1}{x^2}}{\frac{1}{x^2} +1} +...+\frac{\frac{1}{x}}{\frac{1}{x} + 1}}$$
Using algebra of limits we get,
$$0+0+...+0 = 0$$
But when solving this using Sandwich Theorem I get,
$$\text{Let }\lim_{x \to \infty}{\frac{1}{1+x^2} + \frac{2}{2+x^2} + \frac{3}{3+x^2}+...+\frac{x}{x + x^2}} = g(x)$$
$$g(x) < \frac{1}{1+x^2} + \frac{2}{2 + x^2}+...+\frac{x}{1+x^2}$$
$$g(x) < \frac{x(x+1)}{2(1 + x^2)}$$
$$g(x) > \frac{1}{x+x^2} + \frac{2}{x+x^2} +...+\frac{x}{x+x^2}$$
$$g(x) > \frac{x(x+1)}{2(x+x^2)} \to g(x) > \frac{1}{2} $$
$$\text{So, we get } \frac{1}{2}<g(x)<\frac{x(x+1)}{2(1 + x^2)}$$
Applying limits,
$$\lim_{x \to \infty}{\frac{x^2+x}{2(1+x^2)}} = 1/2$$
$$\text{Finally, we get } \frac{1}{2}<g(x)<\frac{1}{2}$$
By Sandwich theorem, $\lim_{x \to \infty}{\frac{1}{1+x^2} + \frac{2}{2+x^2} + \frac{3}{3+x^2}+...+\frac{x}{x + x^2}} =\frac{1}{2}$
So, my question is Why am I getting two different answers when worked out differently.
If there is any error or misconception in my working, please correct me.

Two different answers with different methods in solving limits.
here you know the two important things are there one is $x\rightarrow\infty$ and and also the terms are variying as 1,2,3.....
you mean by individualy taking them as constant even when they are going to $\infty$
for example this below
similar misleading question by your point of view
here also the sum might tend to something different.
but the $k$ as a variable makes difference.
