Pointwise convergence of $frac{n}{xn+1}$












1












$begingroup$



For each $n ∈ Bbb N$, let $f_n : (0, 1) to Bbb R$ be defined as`$$f_n(x) = frac{n}{xn+1}$$
Prove $(f_n )^{∞}_{n=1}$ converges pointwise on $(0,1)$




That is, the $f_n$'s go to zero



I tried finding the limit as n approaches infinity of $frac{n}{nx+1}$, but that is equal to $frac{1}{x}$, and given that $x ∈ (0,1)$, this seems contrary to the idea that the sequence converges pointwise. What am I overlooking?






real-analysis sequences-and-series uniform-convergence







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$endgroup$












  • $begingroup$
    You have a sequence, not a series.
    $endgroup$
    – egreg
    Dec 5 '18 at 11:04
















1












$begingroup$



For each $n ∈ Bbb N$, let $f_n : (0, 1) to Bbb R$ be defined as`$$f_n(x) = frac{n}{xn+1}$$
Prove $(f_n )^{∞}_{n=1}$ converges pointwise on $(0,1)$




That is, the $f_n$'s go to zero



I tried finding the limit as n approaches infinity of $frac{n}{nx+1}$, but that is equal to $frac{1}{x}$, and given that $x ∈ (0,1)$, this seems contrary to the idea that the sequence converges pointwise. What am I overlooking?






real-analysis sequences-and-series uniform-convergence







share|cite|improve this question











$endgroup$












  • $begingroup$
    You have a sequence, not a series.
    $endgroup$
    – egreg
    Dec 5 '18 at 11:04














1












1








1





$begingroup$



For each $n ∈ Bbb N$, let $f_n : (0, 1) to Bbb R$ be defined as`$$f_n(x) = frac{n}{xn+1}$$
Prove $(f_n )^{∞}_{n=1}$ converges pointwise on $(0,1)$




That is, the $f_n$'s go to zero



I tried finding the limit as n approaches infinity of $frac{n}{nx+1}$, but that is equal to $frac{1}{x}$, and given that $x ∈ (0,1)$, this seems contrary to the idea that the sequence converges pointwise. What am I overlooking?






real-analysis sequences-and-series uniform-convergence







share|cite|improve this question











$endgroup$





For each $n ∈ Bbb N$, let $f_n : (0, 1) to Bbb R$ be defined as`$$f_n(x) = frac{n}{xn+1}$$
Prove $(f_n )^{∞}_{n=1}$ converges pointwise on $(0,1)$




That is, the $f_n$'s go to zero



I tried finding the limit as n approaches infinity of $frac{n}{nx+1}$, but that is equal to $frac{1}{x}$, and given that $x ∈ (0,1)$, this seems contrary to the idea that the sequence converges pointwise. What am I overlooking?






real-analysis sequences-and-series uniform-convergence




real-analysis sequences-and-series uniform-convergence






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 5 '18 at 11:25









Chinnapparaj R

5,4331928




5,4331928










asked Dec 5 '18 at 10:58









user613048user613048

322




322












  • $begingroup$
    You have a sequence, not a series.
    $endgroup$
    – egreg
    Dec 5 '18 at 11:04


















  • $begingroup$
    You have a sequence, not a series.
    $endgroup$
    – egreg
    Dec 5 '18 at 11:04
















$begingroup$
You have a sequence, not a series.
$endgroup$
– egreg
Dec 5 '18 at 11:04




$begingroup$
You have a sequence, not a series.
$endgroup$
– egreg
Dec 5 '18 at 11:04










1 Answer
1






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oldest

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1












$begingroup$

$f_n$'s do not go to zero. What you do is correct. We fix $xin(0,1)$ and then we investigate the limit as $ntoinfty$. We have that, for each fixed $xin(0,1)$, $f_n(x)to1/x$ as $ntoinfty$. Hence, $f_n$ converges pointwise to $f$ as $ntoinfty$, where $f$ is defined by $f(x)=1/x$ on $(0,1)$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you! that makes a lotttt of sense
    $endgroup$
    – user613048
    Dec 5 '18 at 11:03











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1 Answer
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1 Answer
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$begingroup$

$f_n$'s do not go to zero. What you do is correct. We fix $xin(0,1)$ and then we investigate the limit as $ntoinfty$. We have that, for each fixed $xin(0,1)$, $f_n(x)to1/x$ as $ntoinfty$. Hence, $f_n$ converges pointwise to $f$ as $ntoinfty$, where $f$ is defined by $f(x)=1/x$ on $(0,1)$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you! that makes a lotttt of sense
    $endgroup$
    – user613048
    Dec 5 '18 at 11:03
















1












$begingroup$

$f_n$'s do not go to zero. What you do is correct. We fix $xin(0,1)$ and then we investigate the limit as $ntoinfty$. We have that, for each fixed $xin(0,1)$, $f_n(x)to1/x$ as $ntoinfty$. Hence, $f_n$ converges pointwise to $f$ as $ntoinfty$, where $f$ is defined by $f(x)=1/x$ on $(0,1)$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you! that makes a lotttt of sense
    $endgroup$
    – user613048
    Dec 5 '18 at 11:03














1












1








1





$begingroup$

$f_n$'s do not go to zero. What you do is correct. We fix $xin(0,1)$ and then we investigate the limit as $ntoinfty$. We have that, for each fixed $xin(0,1)$, $f_n(x)to1/x$ as $ntoinfty$. Hence, $f_n$ converges pointwise to $f$ as $ntoinfty$, where $f$ is defined by $f(x)=1/x$ on $(0,1)$.






share|cite|improve this answer











$endgroup$



$f_n$'s do not go to zero. What you do is correct. We fix $xin(0,1)$ and then we investigate the limit as $ntoinfty$. We have that, for each fixed $xin(0,1)$, $f_n(x)to1/x$ as $ntoinfty$. Hence, $f_n$ converges pointwise to $f$ as $ntoinfty$, where $f$ is defined by $f(x)=1/x$ on $(0,1)$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 5 '18 at 11:04

























answered Dec 5 '18 at 11:01









Cm7F7BbCm7F7Bb

12.5k32242




12.5k32242












  • $begingroup$
    Thank you! that makes a lotttt of sense
    $endgroup$
    – user613048
    Dec 5 '18 at 11:03


















  • $begingroup$
    Thank you! that makes a lotttt of sense
    $endgroup$
    – user613048
    Dec 5 '18 at 11:03
















$begingroup$
Thank you! that makes a lotttt of sense
$endgroup$
– user613048
Dec 5 '18 at 11:03




$begingroup$
Thank you! that makes a lotttt of sense
$endgroup$
– user613048
Dec 5 '18 at 11:03


















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