Real Analysis - provinig a sequence converges to some value

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Let $a_n$ where $n in mathbb N$ be a sequence of rational numbers converging to $a$. Suppose $a neq 0$, for $k = 1, 2, ...$ let
$$b_k=begincases 0 & textif;a_k=0\\frac1a_k &textif;a_k neq 0endcases$$
Prove that $b_n$ converges to $frac1a$.




I was studying real analysis and got stuck on this problem.
Can you help me solve this problem or give me some hints?



Thanks



edit: is it possible to solve this in terms of Cauchy Sequence?






real-analysis







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  • Tips: to make MathJax works, use the dollar symbols to wrap up your math expressions. E.g. $mathbb N$ vs mathbb N.
    – xbh
    1 hour ago














up vote
2
down vote

favorite













Let $a_n$ where $n in mathbb N$ be a sequence of rational numbers converging to $a$. Suppose $a neq 0$, for $k = 1, 2, ...$ let
$$b_k=begincases 0 & textif;a_k=0\\frac1a_k &textif;a_k neq 0endcases$$
Prove that $b_n$ converges to $frac1a$.




I was studying real analysis and got stuck on this problem.
Can you help me solve this problem or give me some hints?



Thanks



edit: is it possible to solve this in terms of Cauchy Sequence?






real-analysis







share|cite|improve this question









New contributor




TUC is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.



















  • Tips: to make MathJax works, use the dollar symbols to wrap up your math expressions. E.g. $mathbb N$ vs mathbb N.
    – xbh
    1 hour ago












up vote
2
down vote

favorite









up vote
2
down vote

favorite












Let $a_n$ where $n in mathbb N$ be a sequence of rational numbers converging to $a$. Suppose $a neq 0$, for $k = 1, 2, ...$ let
$$b_k=begincases 0 & textif;a_k=0\\frac1a_k &textif;a_k neq 0endcases$$
Prove that $b_n$ converges to $frac1a$.




I was studying real analysis and got stuck on this problem.
Can you help me solve this problem or give me some hints?



Thanks



edit: is it possible to solve this in terms of Cauchy Sequence?






real-analysis







share|cite|improve this question









New contributor




TUC is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












Let $a_n$ where $n in mathbb N$ be a sequence of rational numbers converging to $a$. Suppose $a neq 0$, for $k = 1, 2, ...$ let
$$b_k=begincases 0 & textif;a_k=0\\frac1a_k &textif;a_k neq 0endcases$$
Prove that $b_n$ converges to $frac1a$.




I was studying real analysis and got stuck on this problem.
Can you help me solve this problem or give me some hints?



Thanks



edit: is it possible to solve this in terms of Cauchy Sequence?






real-analysis




real-analysis






share|cite|improve this question









New contributor




TUC is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









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TUC is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




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edited 1 hour ago









Chinnapparaj R

2,275520




2,275520






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asked 1 hour ago









TUC

133




133




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TUC is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






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Check out our Code of Conduct.











  • Tips: to make MathJax works, use the dollar symbols to wrap up your math expressions. E.g. $mathbb N$ vs mathbb N.
    – xbh
    1 hour ago
















  • Tips: to make MathJax works, use the dollar symbols to wrap up your math expressions. E.g. $mathbb N$ vs mathbb N.
    – xbh
    1 hour ago















Tips: to make MathJax works, use the dollar symbols to wrap up your math expressions. E.g. $mathbb N$ vs mathbb N.
– xbh
1 hour ago




Tips: to make MathJax works, use the dollar symbols to wrap up your math expressions. E.g. $mathbb N$ vs mathbb N.
– xbh
1 hour ago










2 Answers
2






active

oldest

votes

















up vote
2
down vote



accepted










It is not necessary to use Cauchy sequences.



Let $varepsilon = fraca2 > 0$ (because $a neq 0)$. The sequence $(a_k)$ converges to $a$, so there exists $N in mathbbN$, such that for all $k geq N$, $a-varepsilon < a_k < a +varepsilon$. By definition of $varepsilon$, you get $a_k neq 0$ for all $k geq N$.



So, for all $k geq N$, $b_k = frac1a_k$. Taking the limit, you get immediately that $(b_k)$ converges to $frac1a$.






share|cite|improve this answer




















  • So $lim_nto infty$ $(b_n)$ $_n in mathbb N$ = $frac1lim_nto infty (a_n)$ = $frac1a$ ?
    – TUC
    13 mins ago


















up vote
2
down vote













Hint



Because $anot=0$ (let's say $a>0$), show that $exists n_0inmathbbN: forall ngeq n_0quad a_n>0$






share|cite|improve this answer




















  • I already know that $a_n$ is a Cauchy sequence. Do I also have to show that $b_n$ is a Cauchy sequence as well?
    – TUC
    55 mins ago










  • I think that the definition of convergence is enough
    – giannispapav
    51 mins ago










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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote



accepted










It is not necessary to use Cauchy sequences.



Let $varepsilon = fraca2 > 0$ (because $a neq 0)$. The sequence $(a_k)$ converges to $a$, so there exists $N in mathbbN$, such that for all $k geq N$, $a-varepsilon < a_k < a +varepsilon$. By definition of $varepsilon$, you get $a_k neq 0$ for all $k geq N$.



So, for all $k geq N$, $b_k = frac1a_k$. Taking the limit, you get immediately that $(b_k)$ converges to $frac1a$.






share|cite|improve this answer




















  • So $lim_nto infty$ $(b_n)$ $_n in mathbb N$ = $frac1lim_nto infty (a_n)$ = $frac1a$ ?
    – TUC
    13 mins ago















up vote
2
down vote



accepted










It is not necessary to use Cauchy sequences.



Let $varepsilon = fraca2 > 0$ (because $a neq 0)$. The sequence $(a_k)$ converges to $a$, so there exists $N in mathbbN$, such that for all $k geq N$, $a-varepsilon < a_k < a +varepsilon$. By definition of $varepsilon$, you get $a_k neq 0$ for all $k geq N$.



So, for all $k geq N$, $b_k = frac1a_k$. Taking the limit, you get immediately that $(b_k)$ converges to $frac1a$.






share|cite|improve this answer




















  • So $lim_nto infty$ $(b_n)$ $_n in mathbb N$ = $frac1lim_nto infty (a_n)$ = $frac1a$ ?
    – TUC
    13 mins ago













up vote
2
down vote



accepted







up vote
2
down vote



accepted






It is not necessary to use Cauchy sequences.



Let $varepsilon = fraca2 > 0$ (because $a neq 0)$. The sequence $(a_k)$ converges to $a$, so there exists $N in mathbbN$, such that for all $k geq N$, $a-varepsilon < a_k < a +varepsilon$. By definition of $varepsilon$, you get $a_k neq 0$ for all $k geq N$.



So, for all $k geq N$, $b_k = frac1a_k$. Taking the limit, you get immediately that $(b_k)$ converges to $frac1a$.






share|cite|improve this answer












It is not necessary to use Cauchy sequences.



Let $varepsilon = fraca2 > 0$ (because $a neq 0)$. The sequence $(a_k)$ converges to $a$, so there exists $N in mathbbN$, such that for all $k geq N$, $a-varepsilon < a_k < a +varepsilon$. By definition of $varepsilon$, you get $a_k neq 0$ for all $k geq N$.



So, for all $k geq N$, $b_k = frac1a_k$. Taking the limit, you get immediately that $(b_k)$ converges to $frac1a$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 52 mins ago









TheSilverDoe

87011




87011











  • So $lim_nto infty$ $(b_n)$ $_n in mathbb N$ = $frac1lim_nto infty (a_n)$ = $frac1a$ ?
    – TUC
    13 mins ago

















  • So $lim_nto infty$ $(b_n)$ $_n in mathbb N$ = $frac1lim_nto infty (a_n)$ = $frac1a$ ?
    – TUC
    13 mins ago
















So $lim_nto infty$ $(b_n)$ $_n in mathbb N$ = $frac1lim_nto infty (a_n)$ = $frac1a$ ?
– TUC
13 mins ago





So $lim_nto infty$ $(b_n)$ $_n in mathbb N$ = $frac1lim_nto infty (a_n)$ = $frac1a$ ?
– TUC
13 mins ago











up vote
2
down vote













Hint



Because $anot=0$ (let's say $a>0$), show that $exists n_0inmathbbN: forall ngeq n_0quad a_n>0$






share|cite|improve this answer




















  • I already know that $a_n$ is a Cauchy sequence. Do I also have to show that $b_n$ is a Cauchy sequence as well?
    – TUC
    55 mins ago










  • I think that the definition of convergence is enough
    – giannispapav
    51 mins ago














up vote
2
down vote













Hint



Because $anot=0$ (let's say $a>0$), show that $exists n_0inmathbbN: forall ngeq n_0quad a_n>0$






share|cite|improve this answer




















  • I already know that $a_n$ is a Cauchy sequence. Do I also have to show that $b_n$ is a Cauchy sequence as well?
    – TUC
    55 mins ago










  • I think that the definition of convergence is enough
    – giannispapav
    51 mins ago












up vote
2
down vote










up vote
2
down vote









Hint



Because $anot=0$ (let's say $a>0$), show that $exists n_0inmathbbN: forall ngeq n_0quad a_n>0$






share|cite|improve this answer












Hint



Because $anot=0$ (let's say $a>0$), show that $exists n_0inmathbbN: forall ngeq n_0quad a_n>0$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 1 hour ago









giannispapav

1,414323




1,414323











  • I already know that $a_n$ is a Cauchy sequence. Do I also have to show that $b_n$ is a Cauchy sequence as well?
    – TUC
    55 mins ago










  • I think that the definition of convergence is enough
    – giannispapav
    51 mins ago
















  • I already know that $a_n$ is a Cauchy sequence. Do I also have to show that $b_n$ is a Cauchy sequence as well?
    – TUC
    55 mins ago










  • I think that the definition of convergence is enough
    – giannispapav
    51 mins ago















I already know that $a_n$ is a Cauchy sequence. Do I also have to show that $b_n$ is a Cauchy sequence as well?
– TUC
55 mins ago




I already know that $a_n$ is a Cauchy sequence. Do I also have to show that $b_n$ is a Cauchy sequence as well?
– TUC
55 mins ago












I think that the definition of convergence is enough
– giannispapav
51 mins ago




I think that the definition of convergence is enough
– giannispapav
51 mins ago










TUC is a new contributor. Be nice, and check out our Code of Conduct.









 

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