Pointwise convergence of $frac{n}{xn+1}$
$begingroup$
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
edited Dec 5 '18 at 11:25
Chinnapparaj R
5,5182928
asked Dec 5 '18 at 10:58
user613048user613048
322
$endgroup$
add a comment |
$begingroup$
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
edited Dec 5 '18 at 11:25
Chinnapparaj R
5,5182928
asked Dec 5 '18 at 10:58
user613048user613048
322
$endgroup$
$begingroup$
You have a sequence, not a series.
$endgroup$
– egreg
Dec 5 '18 at 11:04
add a comment |
$begingroup$
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
edited Dec 5 '18 at 11:25
Chinnapparaj R
5,5182928
asked Dec 5 '18 at 10:58
user613048user613048
322
$endgroup$
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
edited Dec 5 '18 at 11:25
Chinnapparaj R
5,5182928
asked Dec 5 '18 at 10:58
user613048user613048
322
edited Dec 5 '18 at 11:25
Chinnapparaj R
5,5182928
asked Dec 5 '18 at 10:58
user613048user613048
322
edited Dec 5 '18 at 11:25
Chinnapparaj R
5,5182928
edited Dec 5 '18 at 11:25
Chinnapparaj R
5,5182928
edited Dec 5 '18 at 11:25
Chinnapparaj R
5,5182928
5,5182928
asked Dec 5 '18 at 10:58
user613048user613048
322
asked Dec 5 '18 at 10:58
user613048user613048
322
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
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oldest
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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 $(0,1)$.
answered Dec 5 '18 at 11:01
Cm7F7BbCm7F7Bb
12.5k32243
$endgroup$
$begingroup$
Thank you! that makes a lotttt of sense
$endgroup$
– user613048
Dec 5 '18 at 11:03
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$f_n$'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 $(0,1)$.
answered Dec 5 '18 at 11:01
Cm7F7BbCm7F7Bb
12.5k32243
$endgroup$
$begingroup$
Thank you! that makes a lotttt of sense
$endgroup$
– user613048
Dec 5 '18 at 11:03
add a comment |
$begingroup$
$f_n$'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 $(0,1)$.
answered Dec 5 '18 at 11:01
Cm7F7BbCm7F7Bb
12.5k32243
$endgroup$
$begingroup$
Thank you! that makes a lotttt of sense
$endgroup$
– user613048
Dec 5 '18 at 11:03
add a comment |
$begingroup$
$f_n$'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 $(0,1)$.
answered Dec 5 '18 at 11:01
Cm7F7BbCm7F7Bb
12.5k32243
$endgroup$
$f_n$'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 $(0,1)$.
answered Dec 5 '18 at 11:01
Cm7F7BbCm7F7Bb
12.5k32243
edited Dec 5 '18 at 11:04
answered Dec 5 '18 at 11:01
Cm7F7BbCm7F7Bb
12.5k32243
answered Dec 5 '18 at 11:01
Cm7F7BbCm7F7Bb
12.5k32243
answered Dec 5 '18 at 11:01
Cm7F7BbCm7F7Bb
12.5k32243
12.5k32243
$begingroup$
Thank you! that makes a lotttt of sense
$endgroup$
– user613048
Dec 5 '18 at 11:03
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
You have a sequence, not a series.
$endgroup$
– egreg
Dec 5 '18 at 11:04