Question Number 4587 by FilupSmith last updated on 09/Feb/16
$${L}=\underset{{i}\rightarrow\infty} {\mathrm{lim}}\:\frac{\left(−\mathrm{1}\right)^{{i}+\mathrm{1}} {i}}{{i}+\mathrm{1}} \\ $$$${L}=? \\ $$
Commented by Yozzii last updated on 14/Feb/16
$${L}\:{does}\:{not}\:{exist}.\:{While}\:\mid{L}\mid=\underset{{i}\rightarrow\infty} {\mathrm{lim}}\frac{{i}}{{i}+\mathrm{1}} \\ $$$$\mid{L}\mid=\underset{{i}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{{i}}}=\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\infty}}=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{0}}=\mathrm{1} \\ $$$${this}\:{indicates}\:{that}\:{for}\:{very}\:{large}\:{i} \\ $$$${the}\:{terms}\:{of}\:\left\{\frac{{i}}{\mathrm{1}+{i}}\right\}_{{i}=\mathrm{1}} ^{\infty} \:{converge}\:{to}\:\mathrm{1}.\: \\ $$$${However},\:\left\{\left(−\mathrm{1}\right)^{{i}+\mathrm{1}} \frac{{i}}{\mathrm{1}+{i}}\right\}_{{i}=\mathrm{1}} ^{\infty} \:{alternates} \\ $$$${periodically}\:{between}\:\mathrm{1}\:{and}\:−\mathrm{1}\:{for}\: \\ $$$${very}\:{large}\:{i}\:{according}\:{to}\:{the}\:{parity}\:{of} \\ $$$${i}.\:{Since}\:{this}\:{sequence}\:{behaves}\:{in}\:{an} \\ $$$${alternating}\:{manner}\:{ceaselessly}\:{between} \\ $$$$\mathrm{1}\:{and}\:−\mathrm{1}\:\left({with}\:{period}\:\mathrm{2}\right)\:{as}\:{i}\rightarrow\infty, \\ $$$${where}\:{if}\:{L}\:{exists}\:{it}\:{takes}\:{one}\:{value}, \\ $$$${the}\:{limit}\:{L}\:{of}\:{the}\:{sequence}\:\left\{\frac{\left(−\mathrm{1}\right)^{{i}+\mathrm{1}} {i}}{{i}+\mathrm{1}}\right\}_{{i}=\mathrm{1}} ^{\infty} {does}\:{not} \\ $$$${exist}. \\ $$$$ \\ $$$$ \\ $$