*Proof verification* If a rational sequence $(y_n)$ converges to non-zero $y$, then $(y_n^{-1})$ converges to...











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For $epsilon > 0$, there exists $Nin mathbb{N}$ such that $|y_n - y| < epsilon$ for $ngeq N$. Removing the absolute value, adding $y$, then "inverting" the inequalities gives us $(y + epsilon)^{-1} < y_n^{-1} < (y - epsilon)^{-1}$.



We want $(y - epsilon)^{-1} = epsilon_0 + y^{-1}$ for $epsilon_0 > 0$. If we solve for $epsilon$ , we get $epsilon = y(1 - (epsilon_0y + 1)^{-1})$. I observed that, given a small enough $epsilon_0>0$, $epsilon$ is positive regardless of if $y$ is positive or negative. Then we can find a positive $epsilon$, and a corresponding index $N_0 in mathbb{N}$, for an arbitrarily small $epsilon_0$.



I would proceed by defining a third epsilon, $epsilon^* = max{|(y+ epsilon)^{-1} - y^{-1}|,epsilon_0}$, observing that since the former expression tends to zero as $epsilon_0$ tends to zero, $epsilon^*$ can be made arbitrarily small. Then $|y_n^{-1} - y^{-1}|<epsilon^*$ for $ngeq N_0$.






real-analysis sequences-and-series convergence







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    Note that this follows immediately from the fact that $xmapsto 1/x$ is a continuous map $mathbb Rsetminus{0}tomathbb Rsetminus{0}$
    – MPW
    Nov 22 at 22:54















up vote
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For $epsilon > 0$, there exists $Nin mathbb{N}$ such that $|y_n - y| < epsilon$ for $ngeq N$. Removing the absolute value, adding $y$, then "inverting" the inequalities gives us $(y + epsilon)^{-1} < y_n^{-1} < (y - epsilon)^{-1}$.



We want $(y - epsilon)^{-1} = epsilon_0 + y^{-1}$ for $epsilon_0 > 0$. If we solve for $epsilon$ , we get $epsilon = y(1 - (epsilon_0y + 1)^{-1})$. I observed that, given a small enough $epsilon_0>0$, $epsilon$ is positive regardless of if $y$ is positive or negative. Then we can find a positive $epsilon$, and a corresponding index $N_0 in mathbb{N}$, for an arbitrarily small $epsilon_0$.



I would proceed by defining a third epsilon, $epsilon^* = max{|(y+ epsilon)^{-1} - y^{-1}|,epsilon_0}$, observing that since the former expression tends to zero as $epsilon_0$ tends to zero, $epsilon^*$ can be made arbitrarily small. Then $|y_n^{-1} - y^{-1}|<epsilon^*$ for $ngeq N_0$.






real-analysis sequences-and-series convergence







share|cite|improve this question




















  • 2




    Note that this follows immediately from the fact that $xmapsto 1/x$ is a continuous map $mathbb Rsetminus{0}tomathbb Rsetminus{0}$
    – MPW
    Nov 22 at 22:54













up vote
1
down vote

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up vote
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For $epsilon > 0$, there exists $Nin mathbb{N}$ such that $|y_n - y| < epsilon$ for $ngeq N$. Removing the absolute value, adding $y$, then "inverting" the inequalities gives us $(y + epsilon)^{-1} < y_n^{-1} < (y - epsilon)^{-1}$.



We want $(y - epsilon)^{-1} = epsilon_0 + y^{-1}$ for $epsilon_0 > 0$. If we solve for $epsilon$ , we get $epsilon = y(1 - (epsilon_0y + 1)^{-1})$. I observed that, given a small enough $epsilon_0>0$, $epsilon$ is positive regardless of if $y$ is positive or negative. Then we can find a positive $epsilon$, and a corresponding index $N_0 in mathbb{N}$, for an arbitrarily small $epsilon_0$.



I would proceed by defining a third epsilon, $epsilon^* = max{|(y+ epsilon)^{-1} - y^{-1}|,epsilon_0}$, observing that since the former expression tends to zero as $epsilon_0$ tends to zero, $epsilon^*$ can be made arbitrarily small. Then $|y_n^{-1} - y^{-1}|<epsilon^*$ for $ngeq N_0$.






real-analysis sequences-and-series convergence







share|cite|improve this question















For $epsilon > 0$, there exists $Nin mathbb{N}$ such that $|y_n - y| < epsilon$ for $ngeq N$. Removing the absolute value, adding $y$, then "inverting" the inequalities gives us $(y + epsilon)^{-1} < y_n^{-1} < (y - epsilon)^{-1}$.



We want $(y - epsilon)^{-1} = epsilon_0 + y^{-1}$ for $epsilon_0 > 0$. If we solve for $epsilon$ , we get $epsilon = y(1 - (epsilon_0y + 1)^{-1})$. I observed that, given a small enough $epsilon_0>0$, $epsilon$ is positive regardless of if $y$ is positive or negative. Then we can find a positive $epsilon$, and a corresponding index $N_0 in mathbb{N}$, for an arbitrarily small $epsilon_0$.



I would proceed by defining a third epsilon, $epsilon^* = max{|(y+ epsilon)^{-1} - y^{-1}|,epsilon_0}$, observing that since the former expression tends to zero as $epsilon_0$ tends to zero, $epsilon^*$ can be made arbitrarily small. Then $|y_n^{-1} - y^{-1}|<epsilon^*$ for $ngeq N_0$.






real-analysis sequences-and-series convergence




real-analysis sequences-and-series convergence






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edited Nov 22 at 22:57









José Carlos Santos

146k22116215




146k22116215










asked Nov 22 at 22:42









hiroshin

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685








  • 2




    Note that this follows immediately from the fact that $xmapsto 1/x$ is a continuous map $mathbb Rsetminus{0}tomathbb Rsetminus{0}$
    – MPW
    Nov 22 at 22:54














  • 2




    Note that this follows immediately from the fact that $xmapsto 1/x$ is a continuous map $mathbb Rsetminus{0}tomathbb Rsetminus{0}$
    – MPW
    Nov 22 at 22:54








2




2




Note that this follows immediately from the fact that $xmapsto 1/x$ is a continuous map $mathbb Rsetminus{0}tomathbb Rsetminus{0}$
– MPW
Nov 22 at 22:54




Note that this follows immediately from the fact that $xmapsto 1/x$ is a continuous map $mathbb Rsetminus{0}tomathbb Rsetminus{0}$
– MPW
Nov 22 at 22:54










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This is wrong. You cannot define a $varepsilon^*$. You need to work with any $varepsilon>0$.



First of all, note that there is a $N_1inmathbb N$ such that$$ngeqslant N_1implieslvert y-y_nrvert<frac{lvert yrvert}2implieslvert y_nrvert>frac{lvert yrvert}2.$$ But then, if $ngeqslant N_1$,$$leftlvertfrac1y-frac1{y_n}rightrvert=frac{lvert y-y_nrvert}{lvert yrvertrvert y_nrvert}leqslantfrac2{lvert yrvert^2}lvert y-y_nrvert.$$Now, take $N_2inmathbb N$ such that $ngeqslant N_2implieslvert y-y_nrvert<dfrac{varepsilonlvert yrvert^2}2.$ Then$$ngeqslantmax{N_1,N_2}impliesleftlvertfrac1y-frac1{y_n}rightrvert<varepsilon.$$






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  • Your proof is very clear, thank you! Regarding $epsilon^*$, if given how I've "assigned" the new epsilon I can find an index $N$ such that $|y_n^{-1}- y^{-1}|<epsilon^*$ for $ngeq N$ and for however small an $epsilon^*$, won't that satisfy the "for every $epsilon > 0$" condition for convergence?
    – hiroshin
    Nov 23 at 7:00








  • 1




    Yes, that would work. To be more precise, if you define $varepsilon^*$ as a function of $varepsilon$ in such a way that $varepsilonto0iffvarepsilon^*to0$, that would work.
    – José Carlos Santos
    Nov 23 at 7:13













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up vote
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This is wrong. You cannot define a $varepsilon^*$. You need to work with any $varepsilon>0$.



First of all, note that there is a $N_1inmathbb N$ such that$$ngeqslant N_1implieslvert y-y_nrvert<frac{lvert yrvert}2implieslvert y_nrvert>frac{lvert yrvert}2.$$ But then, if $ngeqslant N_1$,$$leftlvertfrac1y-frac1{y_n}rightrvert=frac{lvert y-y_nrvert}{lvert yrvertrvert y_nrvert}leqslantfrac2{lvert yrvert^2}lvert y-y_nrvert.$$Now, take $N_2inmathbb N$ such that $ngeqslant N_2implieslvert y-y_nrvert<dfrac{varepsilonlvert yrvert^2}2.$ Then$$ngeqslantmax{N_1,N_2}impliesleftlvertfrac1y-frac1{y_n}rightrvert<varepsilon.$$






share|cite|improve this answer























  • Your proof is very clear, thank you! Regarding $epsilon^*$, if given how I've "assigned" the new epsilon I can find an index $N$ such that $|y_n^{-1}- y^{-1}|<epsilon^*$ for $ngeq N$ and for however small an $epsilon^*$, won't that satisfy the "for every $epsilon > 0$" condition for convergence?
    – hiroshin
    Nov 23 at 7:00








  • 1




    Yes, that would work. To be more precise, if you define $varepsilon^*$ as a function of $varepsilon$ in such a way that $varepsilonto0iffvarepsilon^*to0$, that would work.
    – José Carlos Santos
    Nov 23 at 7:13

















up vote
3
down vote



accepted










This is wrong. You cannot define a $varepsilon^*$. You need to work with any $varepsilon>0$.



First of all, note that there is a $N_1inmathbb N$ such that$$ngeqslant N_1implieslvert y-y_nrvert<frac{lvert yrvert}2implieslvert y_nrvert>frac{lvert yrvert}2.$$ But then, if $ngeqslant N_1$,$$leftlvertfrac1y-frac1{y_n}rightrvert=frac{lvert y-y_nrvert}{lvert yrvertrvert y_nrvert}leqslantfrac2{lvert yrvert^2}lvert y-y_nrvert.$$Now, take $N_2inmathbb N$ such that $ngeqslant N_2implieslvert y-y_nrvert<dfrac{varepsilonlvert yrvert^2}2.$ Then$$ngeqslantmax{N_1,N_2}impliesleftlvertfrac1y-frac1{y_n}rightrvert<varepsilon.$$






share|cite|improve this answer























  • Your proof is very clear, thank you! Regarding $epsilon^*$, if given how I've "assigned" the new epsilon I can find an index $N$ such that $|y_n^{-1}- y^{-1}|<epsilon^*$ for $ngeq N$ and for however small an $epsilon^*$, won't that satisfy the "for every $epsilon > 0$" condition for convergence?
    – hiroshin
    Nov 23 at 7:00








  • 1




    Yes, that would work. To be more precise, if you define $varepsilon^*$ as a function of $varepsilon$ in such a way that $varepsilonto0iffvarepsilon^*to0$, that would work.
    – José Carlos Santos
    Nov 23 at 7:13















up vote
3
down vote



accepted







up vote
3
down vote



accepted






This is wrong. You cannot define a $varepsilon^*$. You need to work with any $varepsilon>0$.



First of all, note that there is a $N_1inmathbb N$ such that$$ngeqslant N_1implieslvert y-y_nrvert<frac{lvert yrvert}2implieslvert y_nrvert>frac{lvert yrvert}2.$$ But then, if $ngeqslant N_1$,$$leftlvertfrac1y-frac1{y_n}rightrvert=frac{lvert y-y_nrvert}{lvert yrvertrvert y_nrvert}leqslantfrac2{lvert yrvert^2}lvert y-y_nrvert.$$Now, take $N_2inmathbb N$ such that $ngeqslant N_2implieslvert y-y_nrvert<dfrac{varepsilonlvert yrvert^2}2.$ Then$$ngeqslantmax{N_1,N_2}impliesleftlvertfrac1y-frac1{y_n}rightrvert<varepsilon.$$






share|cite|improve this answer














This is wrong. You cannot define a $varepsilon^*$. You need to work with any $varepsilon>0$.



First of all, note that there is a $N_1inmathbb N$ such that$$ngeqslant N_1implieslvert y-y_nrvert<frac{lvert yrvert}2implieslvert y_nrvert>frac{lvert yrvert}2.$$ But then, if $ngeqslant N_1$,$$leftlvertfrac1y-frac1{y_n}rightrvert=frac{lvert y-y_nrvert}{lvert yrvertrvert y_nrvert}leqslantfrac2{lvert yrvert^2}lvert y-y_nrvert.$$Now, take $N_2inmathbb N$ such that $ngeqslant N_2implieslvert y-y_nrvert<dfrac{varepsilonlvert yrvert^2}2.$ Then$$ngeqslantmax{N_1,N_2}impliesleftlvertfrac1y-frac1{y_n}rightrvert<varepsilon.$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 22 at 23:11

























answered Nov 22 at 22:51









José Carlos Santos

146k22116215




146k22116215












  • Your proof is very clear, thank you! Regarding $epsilon^*$, if given how I've "assigned" the new epsilon I can find an index $N$ such that $|y_n^{-1}- y^{-1}|<epsilon^*$ for $ngeq N$ and for however small an $epsilon^*$, won't that satisfy the "for every $epsilon > 0$" condition for convergence?
    – hiroshin
    Nov 23 at 7:00








  • 1




    Yes, that would work. To be more precise, if you define $varepsilon^*$ as a function of $varepsilon$ in such a way that $varepsilonto0iffvarepsilon^*to0$, that would work.
    – José Carlos Santos
    Nov 23 at 7:13




















  • Your proof is very clear, thank you! Regarding $epsilon^*$, if given how I've "assigned" the new epsilon I can find an index $N$ such that $|y_n^{-1}- y^{-1}|<epsilon^*$ for $ngeq N$ and for however small an $epsilon^*$, won't that satisfy the "for every $epsilon > 0$" condition for convergence?
    – hiroshin
    Nov 23 at 7:00








  • 1




    Yes, that would work. To be more precise, if you define $varepsilon^*$ as a function of $varepsilon$ in such a way that $varepsilonto0iffvarepsilon^*to0$, that would work.
    – José Carlos Santos
    Nov 23 at 7:13


















Your proof is very clear, thank you! Regarding $epsilon^*$, if given how I've "assigned" the new epsilon I can find an index $N$ such that $|y_n^{-1}- y^{-1}|<epsilon^*$ for $ngeq N$ and for however small an $epsilon^*$, won't that satisfy the "for every $epsilon > 0$" condition for convergence?
– hiroshin
Nov 23 at 7:00






Your proof is very clear, thank you! Regarding $epsilon^*$, if given how I've "assigned" the new epsilon I can find an index $N$ such that $|y_n^{-1}- y^{-1}|<epsilon^*$ for $ngeq N$ and for however small an $epsilon^*$, won't that satisfy the "for every $epsilon > 0$" condition for convergence?
– hiroshin
Nov 23 at 7:00






1




1




Yes, that would work. To be more precise, if you define $varepsilon^*$ as a function of $varepsilon$ in such a way that $varepsilonto0iffvarepsilon^*to0$, that would work.
– José Carlos Santos
Nov 23 at 7:13






Yes, that would work. To be more precise, if you define $varepsilon^*$ as a function of $varepsilon$ in such a way that $varepsilonto0iffvarepsilon^*to0$, that would work.
– José Carlos Santos
Nov 23 at 7:13




















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