# 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,

$$\text{Let }\lim_{x \to \infty}{\frac{1}{1+x^2} + \frac{2}{2+x^2} + \frac{3}{3+x^2}+...+\frac{x}{x + x^2}} = g(x)$$ $$g(x) < \frac{1}{1+x^2} + \frac{2}{2 + x^2}+...+\frac{x}{1+x^2}$$ $$g(x) < \frac{x(x+1)}{2(1 + x^2)}$$ $$g(x) > \frac{1}{x+x^2} + \frac{2}{x+x^2} +...+\frac{x}{x+x^2}$$ $$g(x) > \frac{x(x+1)}{2(x+x^2)} \to g(x) > \frac{1}{2}$$ $$\text{So, we get } \frac{1}{2} Applying limits, $$\lim_{x \to \infty}{\frac{x^2+x}{2(1+x^2)}} = 1/2$$ $$\text{Finally, we get } \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.

• can't use algebra of limits for infinitely many terms – J. W. Tanner Mar 5 '19 at 4:01
• May I know why I can't use it for infinite terms? @J.W.Tanner – rash Mar 5 '19 at 4:02
• math.stackexchange.com/q/92989/444015 maybe it will be useful. – Corrêa Mar 5 '19 at 4:14
• $\lim_{N \to \infty} \sum_{i=1}^N \frac 1 N = \lim_{N \to \infty} 1 = 1$ but your argument would be analogous to saying this is $0$ – J. W. Tanner Mar 5 '19 at 4:14
• 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. – Angina Seng Mar 5 '19 at 4:16

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 $$S_n=\frac{1}{2}+\frac{1}{6 n}-\frac{1}{4 n^2}+O\left(\frac{1}{n^3}\right)$$
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.