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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.

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    $\begingroup$ can't use algebra of limits for infinitely many terms $\endgroup$ – J. W. Tanner Mar 5 '19 at 4:01
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    $\begingroup$ May I know why I can't use it for infinite terms? @J.W.Tanner $\endgroup$ – rash Mar 5 '19 at 4:02
  • $\begingroup$ math.stackexchange.com/q/92989/444015 maybe it will be useful. $\endgroup$ – Corrêa Mar 5 '19 at 4:14
  • $\begingroup$ $\lim_{N \to \infty} \sum_{i=1}^N \frac 1 N = \lim_{N \to \infty} 1 = 1$ but your argument would be analogous to saying this is $0$ $\endgroup$ – J. W. Tanner Mar 5 '19 at 4:14
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    $\begingroup$ Your first argument is analogous to saying that $1/n+\cdots+1/n$ (with $n$ terms) tends to zero as $n\to\infty$ as $1/n\to0$ etc. $\endgroup$ – Lord Shark the Unknown Mar 5 '19 at 4:16
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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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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.

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