Prove $sum_{n=1}^{infty} x/(1+x)^n$ is uniformly convergent for $ x in [1,2]$
I was wondering if this proof and my later assertions were correct.
$f_n(x) = x/(1+x)^n$
Let $M_n = x/n(1+x)$. Here, $|f_n(x)| le M_n$ since $n(1+x) le (1+x)^n$ for $x in [1, 2]$ and for $n>0$.
$sum_{n=1}^{infty} M_n$ converges by the ratio test.
Thus, $sum_{n=1}^{infty} x/(1+x)^n$ is uniformly convergent for $ x in [1,2]$ by the Weierstrass $M$ test.
Also, since it is uniformly convergent, it is convergent.
Furthermore, $int_1^2 (sum_{1}^{infty} f_n(x)) dx = sum_{1}^{infty} int_1^2 f_n(x)dx$ since the series converges uniformly.
Thanks ahead of time.
real-analysis proof-verification uniform-convergence
asked Dec 1 at 3:56
user591271
826
add a comment |
I was wondering if this proof and my later assertions were correct.
$f_n(x) = x/(1+x)^n$
Let $M_n = x/n(1+x)$. Here, $|f_n(x)| le M_n$ since $n(1+x) le (1+x)^n$ for $x in [1, 2]$ and for $n>0$.
$sum_{n=1}^{infty} M_n$ converges by the ratio test.
Thus, $sum_{n=1}^{infty} x/(1+x)^n$ is uniformly convergent for $ x in [1,2]$ by the Weierstrass $M$ test.
Also, since it is uniformly convergent, it is convergent.
Furthermore, $int_1^2 (sum_{1}^{infty} f_n(x)) dx = sum_{1}^{infty} int_1^2 f_n(x)dx$ since the series converges uniformly.
Thanks ahead of time.
real-analysis proof-verification uniform-convergence
asked Dec 1 at 3:56
user591271
826
add a comment |
I was wondering if this proof and my later assertions were correct.
$f_n(x) = x/(1+x)^n$
Let $M_n = x/n(1+x)$. Here, $|f_n(x)| le M_n$ since $n(1+x) le (1+x)^n$ for $x in [1, 2]$ and for $n>0$.
$sum_{n=1}^{infty} M_n$ converges by the ratio test.
Thus, $sum_{n=1}^{infty} x/(1+x)^n$ is uniformly convergent for $ x in [1,2]$ by the Weierstrass $M$ test.
Also, since it is uniformly convergent, it is convergent.
Furthermore, $int_1^2 (sum_{1}^{infty} f_n(x)) dx = sum_{1}^{infty} int_1^2 f_n(x)dx$ since the series converges uniformly.
Thanks ahead of time.
real-analysis proof-verification uniform-convergence
asked Dec 1 at 3:56
user591271
826
I was wondering if this proof and my later assertions were correct.
$f_n(x) = x/(1+x)^n$
Let $M_n = x/n(1+x)$. Here, $|f_n(x)| le M_n$ since $n(1+x) le (1+x)^n$ for $x in [1, 2]$ and for $n>0$.
$sum_{n=1}^{infty} M_n$ converges by the ratio test.
Thus, $sum_{n=1}^{infty} x/(1+x)^n$ is uniformly convergent for $ x in [1,2]$ by the Weierstrass $M$ test.
Also, since it is uniformly convergent, it is convergent.
Furthermore, $int_1^2 (sum_{1}^{infty} f_n(x)) dx = sum_{1}^{infty} int_1^2 f_n(x)dx$ since the series converges uniformly.
Thanks ahead of time.
real-analysis proof-verification uniform-convergence
real-analysis proof-verification uniform-convergence
asked Dec 1 at 3:56
user591271
826
asked Dec 1 at 3:56
user591271
826
asked Dec 1 at 3:56
user591271
826
asked Dec 1 at 3:56
user591271
826
asked Dec 1 at 3:56
user591271
826
826
add a comment |
add a comment |
2 Answers
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Your application of $M_n$ test is wrong
Insted show $f_n(x)=x/(1+x)^n$ is decreasing function by derivative test
$f_n'(x)=frac{1+(1-n)x}{(1+x)^{n+1}}$
$f_n'(x)<0$ for $n>1$ as $xin [1,2]$
SO $f_n(x)=x/(1+x)^nleq f_n(1)$
i.e $M_n =1/2^n$
$sum_1^{infty}1/2^n =1$
SO given series is uniformly convergent by Weierstrass $M_n$ Test
answered Dec 1 at 4:21
Shubham
1,5851519
Ahh ok, I understand now, thank you. So then if I prove uniform convergence like this, are the other two assertions I made valid?
– user591271
Dec 1 at 17:28
1
Yes . If you have uniformly convergent series then you can interchange summation and integration over compact set
– Shubham
Dec 1 at 17:57
add a comment |
No, that is not a correct application of the M-test. First, and most importantly, $M_n$ must be a constant for the whole interval, not dependent on $x$. Without that, we wouldn't be talking about uniform convergence. Claiming "$M_n=frac{x}{n(1+x)}$" can't possibly work.
Second, that choice doesn't work because the sum doesn't converge; $sum_{n=1}^{infty} frac{x}{n(1+x)}=frac{x}{1+x}sum_{n=1}^{infty}frac1n$ is a multiple of the harmonic series, which diverges.
answered Dec 1 at 4:13
jmerry
1,48817
add a comment |
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2 Answers
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2 Answers
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active
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votes
Your application of $M_n$ test is wrong
Insted show $f_n(x)=x/(1+x)^n$ is decreasing function by derivative test
$f_n'(x)=frac{1+(1-n)x}{(1+x)^{n+1}}$
$f_n'(x)<0$ for $n>1$ as $xin [1,2]$
SO $f_n(x)=x/(1+x)^nleq f_n(1)$
i.e $M_n =1/2^n$
$sum_1^{infty}1/2^n =1$
SO given series is uniformly convergent by Weierstrass $M_n$ Test
answered Dec 1 at 4:21
Shubham
1,5851519
Ahh ok, I understand now, thank you. So then if I prove uniform convergence like this, are the other two assertions I made valid?
– user591271
Dec 1 at 17:28
1
Yes . If you have uniformly convergent series then you can interchange summation and integration over compact set
– Shubham
Dec 1 at 17:57
add a comment |
Your application of $M_n$ test is wrong
Insted show $f_n(x)=x/(1+x)^n$ is decreasing function by derivative test
$f_n'(x)=frac{1+(1-n)x}{(1+x)^{n+1}}$
$f_n'(x)<0$ for $n>1$ as $xin [1,2]$
SO $f_n(x)=x/(1+x)^nleq f_n(1)$
i.e $M_n =1/2^n$
$sum_1^{infty}1/2^n =1$
SO given series is uniformly convergent by Weierstrass $M_n$ Test
answered Dec 1 at 4:21
Shubham
1,5851519
Ahh ok, I understand now, thank you. So then if I prove uniform convergence like this, are the other two assertions I made valid?
– user591271
Dec 1 at 17:28
1
Yes . If you have uniformly convergent series then you can interchange summation and integration over compact set
– Shubham
Dec 1 at 17:57
add a comment |
Your application of $M_n$ test is wrong
Insted show $f_n(x)=x/(1+x)^n$ is decreasing function by derivative test
$f_n'(x)=frac{1+(1-n)x}{(1+x)^{n+1}}$
$f_n'(x)<0$ for $n>1$ as $xin [1,2]$
SO $f_n(x)=x/(1+x)^nleq f_n(1)$
i.e $M_n =1/2^n$
$sum_1^{infty}1/2^n =1$
SO given series is uniformly convergent by Weierstrass $M_n$ Test
answered Dec 1 at 4:21
Shubham
1,5851519
Your application of $M_n$ test is wrong
Insted show $f_n(x)=x/(1+x)^n$ is decreasing function by derivative test
$f_n'(x)=frac{1+(1-n)x}{(1+x)^{n+1}}$
$f_n'(x)<0$ for $n>1$ as $xin [1,2]$
SO $f_n(x)=x/(1+x)^nleq f_n(1)$
i.e $M_n =1/2^n$
$sum_1^{infty}1/2^n =1$
SO given series is uniformly convergent by Weierstrass $M_n$ Test
answered Dec 1 at 4:21
Shubham
1,5851519
answered Dec 1 at 4:21
Shubham
1,5851519
answered Dec 1 at 4:21
Shubham
1,5851519
answered Dec 1 at 4:21
Shubham
1,5851519
1,5851519
Ahh ok, I understand now, thank you. So then if I prove uniform convergence like this, are the other two assertions I made valid?
– user591271
Dec 1 at 17:28
1
Yes . If you have uniformly convergent series then you can interchange summation and integration over compact set
– Shubham
Dec 1 at 17:57
add a comment |
Ahh ok, I understand now, thank you. So then if I prove uniform convergence like this, are the other two assertions I made valid?
– user591271
Dec 1 at 17:28
1
Yes . If you have uniformly convergent series then you can interchange summation and integration over compact set
– Shubham
Dec 1 at 17:57
Ahh ok, I understand now, thank you. So then if I prove uniform convergence like this, are the other two assertions I made valid?
– user591271
Dec 1 at 17:28
Ahh ok, I understand now, thank you. So then if I prove uniform convergence like this, are the other two assertions I made valid?
– user591271
Dec 1 at 17:28
1
1
Yes . If you have uniformly convergent series then you can interchange summation and integration over compact set
– Shubham
Dec 1 at 17:57
Yes . If you have uniformly convergent series then you can interchange summation and integration over compact set
– Shubham
Dec 1 at 17:57
add a comment |
No, that is not a correct application of the M-test. First, and most importantly, $M_n$ must be a constant for the whole interval, not dependent on $x$. Without that, we wouldn't be talking about uniform convergence. Claiming "$M_n=frac{x}{n(1+x)}$" can't possibly work.
Second, that choice doesn't work because the sum doesn't converge; $sum_{n=1}^{infty} frac{x}{n(1+x)}=frac{x}{1+x}sum_{n=1}^{infty}frac1n$ is a multiple of the harmonic series, which diverges.
answered Dec 1 at 4:13
jmerry
1,48817
add a comment |
No, that is not a correct application of the M-test. First, and most importantly, $M_n$ must be a constant for the whole interval, not dependent on $x$. Without that, we wouldn't be talking about uniform convergence. Claiming "$M_n=frac{x}{n(1+x)}$" can't possibly work.
Second, that choice doesn't work because the sum doesn't converge; $sum_{n=1}^{infty} frac{x}{n(1+x)}=frac{x}{1+x}sum_{n=1}^{infty}frac1n$ is a multiple of the harmonic series, which diverges.
answered Dec 1 at 4:13
jmerry
1,48817
add a comment |
No, that is not a correct application of the M-test. First, and most importantly, $M_n$ must be a constant for the whole interval, not dependent on $x$. Without that, we wouldn't be talking about uniform convergence. Claiming "$M_n=frac{x}{n(1+x)}$" can't possibly work.
Second, that choice doesn't work because the sum doesn't converge; $sum_{n=1}^{infty} frac{x}{n(1+x)}=frac{x}{1+x}sum_{n=1}^{infty}frac1n$ is a multiple of the harmonic series, which diverges.
answered Dec 1 at 4:13
jmerry
1,48817
No, that is not a correct application of the M-test. First, and most importantly, $M_n$ must be a constant for the whole interval, not dependent on $x$. Without that, we wouldn't be talking about uniform convergence. Claiming "$M_n=frac{x}{n(1+x)}$" can't possibly work.
Second, that choice doesn't work because the sum doesn't converge; $sum_{n=1}^{infty} frac{x}{n(1+x)}=frac{x}{1+x}sum_{n=1}^{infty}frac1n$ is a multiple of the harmonic series, which diverges.
answered Dec 1 at 4:13
jmerry
1,48817
answered Dec 1 at 4:13
jmerry
1,48817
answered Dec 1 at 4:13
jmerry
1,48817
answered Dec 1 at 4:13
jmerry
1,48817
1,48817
add a comment |
add a comment |
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