Question Number 4454 by FilupSmith last updated on 29/Jan/16
$${S}=\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}+\frac{{x}−\mathrm{2}}{{x}^{\mathrm{2}} −\mathrm{2}}+\frac{{x}−\mathrm{3}}{{x}^{\mathrm{3}} +\mathrm{3}}+\frac{{x}−\mathrm{4}}{{x}^{\mathrm{4}} −\mathrm{4}}+… \\ $$$$ \\ $$$${S}=\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}−{i}}{{x}^{{i}} −\left(−\mathrm{1}\right)^{{i}} {i}} \\ $$$${D}\mathrm{oes}\:\mathrm{S}\:{limit}\:{t}\mathrm{o}\:\mathrm{a}\:\mathrm{value}\:\mathrm{for}\:\pm{x}? \\ $$
Commented by prakash jain last updated on 30/Jan/16
$$\mid{x}\mid\leqslant\mathrm{1}\Rightarrow\mathrm{does}\:\mathrm{not}\:\mathrm{converge}\:\mathrm{as}\:\underset{{i}\rightarrow\infty} {\mathrm{lim}}\:{a}_{{i}} \neq\mathrm{0} \\ $$$$\mid{x}\mid>\mathrm{1}\Rightarrow\mathrm{converges}.\:\mathrm{Ratio}\:\mathrm{test}. \\ $$
Answered by Jens last updated on 29/Jan/16
![Probably yes if ⌈x]>1. Divide top and bottom by i. Let i⇒∞ and the nominator ⇒−1 while the denominator ⇒x^i /i ⇒∞ so each term ⇒0 faster than necessary for convergence. In fact Σ_1 ^∞ ix^(−i) =−x(d/dx)((1/(x−1)))=(x/((x−1)^2 )) assuming ∣x∣>1.](https://www.tinkutara.com/question/Q4464.png)
$$\left.{Probably}\:{yes}\:{if}\:\lceil{x}\right]>\mathrm{1}.\:{Divide}\:{top}\: \\ $$$${and}\:{bottom}\:{by}\:{i}.\:{Let}\:{i}\Rightarrow\infty\:{and}\:\:{the} \\ $$$${nominator}\:\Rightarrow−\mathrm{1}\:{while}\:{the}\: \\ $$$${denominator}\:\Rightarrow{x}^{{i}} /{i}\:\Rightarrow\infty\:{so}\:{each}\: \\ $$$${term}\:\Rightarrow\mathrm{0}\:{faster}\:{than}\:{necessary}\: \\ $$$${for}\:{convergence}.\:{In}\:{fact}\:\sum_{\mathrm{1}} ^{\infty} \:{ix}^{−{i}} \\ $$$$=−{x}\frac{{d}}{{dx}}\left(\frac{\mathrm{1}}{{x}−\mathrm{1}}\right)=\frac{{x}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\:{assuming} \\ $$$$\mid{x}\mid>\mathrm{1}. \\ $$$$ \\ $$