Show that the series $sum fracsin(n)1+n^2x^2$ converges uniformly. The Next CEO of Stack Overflowproving that some series converges uniformlyDoes $sum_nge1frac1nsinleft(frac xnright)$ converges uniformly?Uniform convergence of $sumlimits_n=1^infty sin left(fracxn^2right)$Show if the series $f(x)=sumlimits_k=1^infty frac1k sin(fracxk)$ converges uniformly or not.Show the series converges uniformlyHow can I show this series converges uniformly?Show that $Nsum_n = 2^infty fract^nnN^n$ converges uniformly to $0$Show that the series converges absolutely $ sum fracsin ntheta2^n$Is the sum of the series $sum fracsin nx^21 + n^3$ continuously differentiable?Show that $sum_n=1^infty fracxn(1+nx^2)$ converges uniformly on $R$.
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Show that the series $sum fracsin(n)1+n^2x^2$ converges uniformly.
The Next CEO of Stack Overflowproving that some series converges uniformlyDoes $sum_nge1frac1nsinleft(frac xnright)$ converges uniformly?Uniform convergence of $sumlimits_n=1^infty sin left(fracxn^2right)$Show if the series $f(x)=sumlimits_k=1^infty frac1k sin(fracxk)$ converges uniformly or not.Show the series converges uniformlyHow can I show this series converges uniformly?Show that $Nsum_n = 2^infty fract^nnN^n$ converges uniformly to $0$Show that the series converges absolutely $ sum fracsin ntheta2^n$Is the sum of the series $sum fracsin nx^21 + n^3$ continuously differentiable?Show that $sum_n=1^infty fracxn(1+nx^2)$ converges uniformly on $R$.
$begingroup$
Show that the series $sum fracsin(n)1+n^2x^2$ converges uniformly on the set where $x$ is real and $|x|geq varepsilon$.
I wanted to do the Weierstrass M-test. Can I just say that $left | fracsin(n)1+n^2x^2 right |leq left | frac11+n^2 right |$, and $sum frac11+n^2$ converges by the comparison test or am I supposed to say something about $x$?
real-analysis convergence uniform-convergence
edited Mar 28 at 9:00
Bernard
124k741118
asked Mar 28 at 6:59
numericalorangenumericalorange
1,855312
$endgroup$
add a comment |
$begingroup$
Show that the series $sum fracsin(n)1+n^2x^2$ converges uniformly on the set where $x$ is real and $|x|geq varepsilon$.
I wanted to do the Weierstrass M-test. Can I just say that $left | fracsin(n)1+n^2x^2 right |leq left | frac11+n^2 right |$, and $sum frac11+n^2$ converges by the comparison test or am I supposed to say something about $x$?
real-analysis convergence uniform-convergence
edited Mar 28 at 9:00
Bernard
124k741118
asked Mar 28 at 6:59
numericalorangenumericalorange
1,855312
$endgroup$
add a comment |
$begingroup$
Show that the series $sum fracsin(n)1+n^2x^2$ converges uniformly on the set where $x$ is real and $|x|geq varepsilon$.
I wanted to do the Weierstrass M-test. Can I just say that $left | fracsin(n)1+n^2x^2 right |leq left | frac11+n^2 right |$, and $sum frac11+n^2$ converges by the comparison test or am I supposed to say something about $x$?
real-analysis convergence uniform-convergence
edited Mar 28 at 9:00
Bernard
124k741118
asked Mar 28 at 6:59
numericalorangenumericalorange
1,855312
$endgroup$
Show that the series $sum fracsin(n)1+n^2x^2$ converges uniformly on the set where $x$ is real and $|x|geq varepsilon$.
I wanted to do the Weierstrass M-test. Can I just say that $left | fracsin(n)1+n^2x^2 right |leq left | frac11+n^2 right |$, and $sum frac11+n^2$ converges by the comparison test or am I supposed to say something about $x$?
real-analysis convergence uniform-convergence
real-analysis convergence uniform-convergence
edited Mar 28 at 9:00
Bernard
124k741118
asked Mar 28 at 6:59
numericalorangenumericalorange
1,855312
edited Mar 28 at 9:00
Bernard
124k741118
asked Mar 28 at 6:59
numericalorangenumericalorange
1,855312
edited Mar 28 at 9:00
Bernard
124k741118
edited Mar 28 at 9:00
Bernard
124k741118
edited Mar 28 at 9:00
Bernard
124k741118
124k741118
asked Mar 28 at 6:59
numericalorangenumericalorange
1,855312
asked Mar 28 at 6:59
numericalorangenumericalorange
1,855312
asked Mar 28 at 6:59
numericalorangenumericalorange
1,855312
1,855312
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Since it doesn't converge uniformly if we take an interval containing zero - the series at $x=0$ is $sum_n sin n$, which doesn't converge at all since its terms don't go to zero - we absolutely need something about $x$ in there.
Indeed, your inequality $left|fracsin n1+n^2x^2right| le frac11+n^2$ is only true for $|x|ge 1$. For the domain $|x|geepsilon$ we're working with here, we have to settle for the weaker inequality
$$left|fracsin n1+n^2x^2right| le left|frac11+n^2x^2right| le frac11+n^2epsilon^2$$
That series converges by limit comparison, and then we can apply the M-test.
answered Mar 28 at 7:14
jmerryjmerry
16.9k11633
$endgroup$
$begingroup$
Awesome! Very easy to understand, thanks.
$endgroup$
– numericalorange
Mar 28 at 13:08
add a comment |
$begingroup$
Let $varepsilon>0.$ One may use the inequality
$$
left | fracsin(n)1+n^2x^2 right |leq left | frac11+n^2x^2 right |leqfrac11+varepsilon^2n^2 , qquad |x|gevarepsilon,
$$ then concluding with the Weierstrass M-test.
answered Mar 28 at 7:03
Olivier OloaOlivier Oloa
109k17178294
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Since it doesn't converge uniformly if we take an interval containing zero - the series at $x=0$ is $sum_n sin n$, which doesn't converge at all since its terms don't go to zero - we absolutely need something about $x$ in there.
Indeed, your inequality $left|fracsin n1+n^2x^2right| le frac11+n^2$ is only true for $|x|ge 1$. For the domain $|x|geepsilon$ we're working with here, we have to settle for the weaker inequality
$$left|fracsin n1+n^2x^2right| le left|frac11+n^2x^2right| le frac11+n^2epsilon^2$$
That series converges by limit comparison, and then we can apply the M-test.
answered Mar 28 at 7:14
jmerryjmerry
16.9k11633
$endgroup$
$begingroup$
Awesome! Very easy to understand, thanks.
$endgroup$
– numericalorange
Mar 28 at 13:08
add a comment |
$begingroup$
Since it doesn't converge uniformly if we take an interval containing zero - the series at $x=0$ is $sum_n sin n$, which doesn't converge at all since its terms don't go to zero - we absolutely need something about $x$ in there.
Indeed, your inequality $left|fracsin n1+n^2x^2right| le frac11+n^2$ is only true for $|x|ge 1$. For the domain $|x|geepsilon$ we're working with here, we have to settle for the weaker inequality
$$left|fracsin n1+n^2x^2right| le left|frac11+n^2x^2right| le frac11+n^2epsilon^2$$
That series converges by limit comparison, and then we can apply the M-test.
answered Mar 28 at 7:14
jmerryjmerry
16.9k11633
$endgroup$
$begingroup$
Awesome! Very easy to understand, thanks.
$endgroup$
– numericalorange
Mar 28 at 13:08
add a comment |
$begingroup$
Since it doesn't converge uniformly if we take an interval containing zero - the series at $x=0$ is $sum_n sin n$, which doesn't converge at all since its terms don't go to zero - we absolutely need something about $x$ in there.
Indeed, your inequality $left|fracsin n1+n^2x^2right| le frac11+n^2$ is only true for $|x|ge 1$. For the domain $|x|geepsilon$ we're working with here, we have to settle for the weaker inequality
$$left|fracsin n1+n^2x^2right| le left|frac11+n^2x^2right| le frac11+n^2epsilon^2$$
That series converges by limit comparison, and then we can apply the M-test.
answered Mar 28 at 7:14
jmerryjmerry
16.9k11633
$endgroup$
Since it doesn't converge uniformly if we take an interval containing zero - the series at $x=0$ is $sum_n sin n$, which doesn't converge at all since its terms don't go to zero - we absolutely need something about $x$ in there.
Indeed, your inequality $left|fracsin n1+n^2x^2right| le frac11+n^2$ is only true for $|x|ge 1$. For the domain $|x|geepsilon$ we're working with here, we have to settle for the weaker inequality
$$left|fracsin n1+n^2x^2right| le left|frac11+n^2x^2right| le frac11+n^2epsilon^2$$
That series converges by limit comparison, and then we can apply the M-test.
answered Mar 28 at 7:14
jmerryjmerry
16.9k11633
answered Mar 28 at 7:14
jmerryjmerry
16.9k11633
answered Mar 28 at 7:14
jmerryjmerry
16.9k11633
answered Mar 28 at 7:14
jmerryjmerry
16.9k11633
16.9k11633
$begingroup$
Awesome! Very easy to understand, thanks.
$endgroup$
– numericalorange
Mar 28 at 13:08
add a comment |
$begingroup$
Awesome! Very easy to understand, thanks.
$endgroup$
– numericalorange
Mar 28 at 13:08
$begingroup$
Awesome! Very easy to understand, thanks.
$endgroup$
– numericalorange
Mar 28 at 13:08
$begingroup$
Awesome! Very easy to understand, thanks.
$endgroup$
– numericalorange
Mar 28 at 13:08
add a comment |
$begingroup$
Let $varepsilon>0.$ One may use the inequality
$$
left | fracsin(n)1+n^2x^2 right |leq left | frac11+n^2x^2 right |leqfrac11+varepsilon^2n^2 , qquad |x|gevarepsilon,
$$ then concluding with the Weierstrass M-test.
answered Mar 28 at 7:03
Olivier OloaOlivier Oloa
109k17178294
$endgroup$
add a comment |
$begingroup$
Let $varepsilon>0.$ One may use the inequality
$$
left | fracsin(n)1+n^2x^2 right |leq left | frac11+n^2x^2 right |leqfrac11+varepsilon^2n^2 , qquad |x|gevarepsilon,
$$ then concluding with the Weierstrass M-test.
answered Mar 28 at 7:03
Olivier OloaOlivier Oloa
109k17178294
$endgroup$
add a comment |
$begingroup$
Let $varepsilon>0.$ One may use the inequality
$$
left | fracsin(n)1+n^2x^2 right |leq left | frac11+n^2x^2 right |leqfrac11+varepsilon^2n^2 , qquad |x|gevarepsilon,
$$ then concluding with the Weierstrass M-test.
answered Mar 28 at 7:03
Olivier OloaOlivier Oloa
109k17178294
$endgroup$
Let $varepsilon>0.$ One may use the inequality
$$
left | fracsin(n)1+n^2x^2 right |leq left | frac11+n^2x^2 right |leqfrac11+varepsilon^2n^2 , qquad |x|gevarepsilon,
$$ then concluding with the Weierstrass M-test.
answered Mar 28 at 7:03
Olivier OloaOlivier Oloa
109k17178294
answered Mar 28 at 7:03
Olivier OloaOlivier Oloa
109k17178294
answered Mar 28 at 7:03
Olivier OloaOlivier Oloa
109k17178294
answered Mar 28 at 7:03
Olivier OloaOlivier Oloa
109k17178294
109k17178294
add a comment |
add a comment |
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