<s>[INST] 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.
 [/INST] I may be totally wrong but looking at the top formula, it seems that you are rather considering 
$$S_n=\sum_{i=1}^n \frac i{i+n^2}=n \left(n H_{n^2}-n H_{n^2+n}+1\right)$$ where appear harmonic numbers.
Using the asymptotics
$$H_p=\gamma +\log \left({p}\right)+\frac{1}{2 p}-\frac{1}{12
   p^2}+O\left(\frac{1}{p^4}\right)$$ Replacing and continuing with Taylor expansions for large value of $n$, you get
$$S_n=\frac{1}{2}+\frac{1}{6 n}-\frac{1}{4 n^2}+O\left(\frac{1}{n^3}\right)$$
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