On inf and sup of a sequence












1














My question is following



Let $(x_{n})$ be a positive sequence and satisfies the inequalities



$frac{1}{K+x_{n+1}}leq x_{n}leq frac{1}{e^{-v}+x_{n+1}}$



where $K>1$ and $v>0$. How to find inf $x_{n}$ and sup $x_{n}$?



My try: Iterating the given inequalities I have found upper and lower bounds for $x_{n}$ as follow



$$
frac{sqrt{K^{2}+4Ke^{v}}-K}{2Ke^{v}}leq x_{n} leq frac{sqrt{K^{2}+4Ke^{v}}-K}{2}.
$$



But I am not so sure whether they are inf and sup?










share|cite|improve this question
























  • May I ask how the iteration is done? Edit: I suppose it's "Solving homogeneous linear recurrence relations with constant coefficients", right? Explained also in Wikipedia.
    – Abdullah UYU
    Nov 20 '18 at 13:02












  • Do you know something about $x_0$?
    – Keen-ameteur
    Nov 20 '18 at 13:04










  • @AbdullahUYU. Using given inequalities we can show $x_{n}leq frac{1}{e^{-v}+frac{1}{K+x_{n+2}}} $. Then we can continue this inequaity for infinite level
    – Alex
    Nov 20 '18 at 13:06












  • @Keen-ameteur I know only $x_{0}in (0,1)$
    – Alex
    Nov 20 '18 at 13:08
















1














My question is following



Let $(x_{n})$ be a positive sequence and satisfies the inequalities



$frac{1}{K+x_{n+1}}leq x_{n}leq frac{1}{e^{-v}+x_{n+1}}$



where $K>1$ and $v>0$. How to find inf $x_{n}$ and sup $x_{n}$?



My try: Iterating the given inequalities I have found upper and lower bounds for $x_{n}$ as follow



$$
frac{sqrt{K^{2}+4Ke^{v}}-K}{2Ke^{v}}leq x_{n} leq frac{sqrt{K^{2}+4Ke^{v}}-K}{2}.
$$



But I am not so sure whether they are inf and sup?










share|cite|improve this question
























  • May I ask how the iteration is done? Edit: I suppose it's "Solving homogeneous linear recurrence relations with constant coefficients", right? Explained also in Wikipedia.
    – Abdullah UYU
    Nov 20 '18 at 13:02












  • Do you know something about $x_0$?
    – Keen-ameteur
    Nov 20 '18 at 13:04










  • @AbdullahUYU. Using given inequalities we can show $x_{n}leq frac{1}{e^{-v}+frac{1}{K+x_{n+2}}} $. Then we can continue this inequaity for infinite level
    – Alex
    Nov 20 '18 at 13:06












  • @Keen-ameteur I know only $x_{0}in (0,1)$
    – Alex
    Nov 20 '18 at 13:08














1












1








1







My question is following



Let $(x_{n})$ be a positive sequence and satisfies the inequalities



$frac{1}{K+x_{n+1}}leq x_{n}leq frac{1}{e^{-v}+x_{n+1}}$



where $K>1$ and $v>0$. How to find inf $x_{n}$ and sup $x_{n}$?



My try: Iterating the given inequalities I have found upper and lower bounds for $x_{n}$ as follow



$$
frac{sqrt{K^{2}+4Ke^{v}}-K}{2Ke^{v}}leq x_{n} leq frac{sqrt{K^{2}+4Ke^{v}}-K}{2}.
$$



But I am not so sure whether they are inf and sup?










share|cite|improve this question















My question is following



Let $(x_{n})$ be a positive sequence and satisfies the inequalities



$frac{1}{K+x_{n+1}}leq x_{n}leq frac{1}{e^{-v}+x_{n+1}}$



where $K>1$ and $v>0$. How to find inf $x_{n}$ and sup $x_{n}$?



My try: Iterating the given inequalities I have found upper and lower bounds for $x_{n}$ as follow



$$
frac{sqrt{K^{2}+4Ke^{v}}-K}{2Ke^{v}}leq x_{n} leq frac{sqrt{K^{2}+4Ke^{v}}-K}{2}.
$$



But I am not so sure whether they are inf and sup?







real-analysis sequences-and-series inequality






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 20 '18 at 13:00

























asked Nov 20 '18 at 12:24









Alex

112




112












  • May I ask how the iteration is done? Edit: I suppose it's "Solving homogeneous linear recurrence relations with constant coefficients", right? Explained also in Wikipedia.
    – Abdullah UYU
    Nov 20 '18 at 13:02












  • Do you know something about $x_0$?
    – Keen-ameteur
    Nov 20 '18 at 13:04










  • @AbdullahUYU. Using given inequalities we can show $x_{n}leq frac{1}{e^{-v}+frac{1}{K+x_{n+2}}} $. Then we can continue this inequaity for infinite level
    – Alex
    Nov 20 '18 at 13:06












  • @Keen-ameteur I know only $x_{0}in (0,1)$
    – Alex
    Nov 20 '18 at 13:08


















  • May I ask how the iteration is done? Edit: I suppose it's "Solving homogeneous linear recurrence relations with constant coefficients", right? Explained also in Wikipedia.
    – Abdullah UYU
    Nov 20 '18 at 13:02












  • Do you know something about $x_0$?
    – Keen-ameteur
    Nov 20 '18 at 13:04










  • @AbdullahUYU. Using given inequalities we can show $x_{n}leq frac{1}{e^{-v}+frac{1}{K+x_{n+2}}} $. Then we can continue this inequaity for infinite level
    – Alex
    Nov 20 '18 at 13:06












  • @Keen-ameteur I know only $x_{0}in (0,1)$
    – Alex
    Nov 20 '18 at 13:08
















May I ask how the iteration is done? Edit: I suppose it's "Solving homogeneous linear recurrence relations with constant coefficients", right? Explained also in Wikipedia.
– Abdullah UYU
Nov 20 '18 at 13:02






May I ask how the iteration is done? Edit: I suppose it's "Solving homogeneous linear recurrence relations with constant coefficients", right? Explained also in Wikipedia.
– Abdullah UYU
Nov 20 '18 at 13:02














Do you know something about $x_0$?
– Keen-ameteur
Nov 20 '18 at 13:04




Do you know something about $x_0$?
– Keen-ameteur
Nov 20 '18 at 13:04












@AbdullahUYU. Using given inequalities we can show $x_{n}leq frac{1}{e^{-v}+frac{1}{K+x_{n+2}}} $. Then we can continue this inequaity for infinite level
– Alex
Nov 20 '18 at 13:06






@AbdullahUYU. Using given inequalities we can show $x_{n}leq frac{1}{e^{-v}+frac{1}{K+x_{n+2}}} $. Then we can continue this inequaity for infinite level
– Alex
Nov 20 '18 at 13:06














@Keen-ameteur I know only $x_{0}in (0,1)$
– Alex
Nov 20 '18 at 13:08




@Keen-ameteur I know only $x_{0}in (0,1)$
– Alex
Nov 20 '18 at 13:08















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