Determining the uniform convergence of given series [closed]
Show that the following series is uniformly convergent-
e^x + e^(2x) + e^(3x) +.... for -1/4 « x « 1/4
But the series is increasing and not bounded above..so how it can be convergent?
calculus
closed as off-topic by RRL, Shubham Johri, amWhy, jgon, Davide Giraudo Jan 4 at 23:36
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Show that the following series is uniformly convergent-
e^x + e^(2x) + e^(3x) +.... for -1/4 « x « 1/4
But the series is increasing and not bounded above..so how it can be convergent?
calculus
closed as off-topic by RRL, Shubham Johri, amWhy, jgon, Davide Giraudo Jan 4 at 23:36
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Shubham Johri, amWhy, jgon, Davide Giraudo
If this question can be reworded to fit the rules in the help center, please edit the question.
5
It's simple: it cannot.
– José Carlos Santos
Jan 4 at 17:30
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Show that the following series is uniformly convergent-
e^x + e^(2x) + e^(3x) +.... for -1/4 « x « 1/4
But the series is increasing and not bounded above..so how it can be convergent?
calculus
Show that the following series is uniformly convergent-
e^x + e^(2x) + e^(3x) +.... for -1/4 « x « 1/4
But the series is increasing and not bounded above..so how it can be convergent?
calculus
calculus
asked Jan 4 at 17:29
KashmiraKashmira
443
443
closed as off-topic by RRL, Shubham Johri, amWhy, jgon, Davide Giraudo Jan 4 at 23:36
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Shubham Johri, amWhy, jgon, Davide Giraudo
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by RRL, Shubham Johri, amWhy, jgon, Davide Giraudo Jan 4 at 23:36
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Shubham Johri, amWhy, jgon, Davide Giraudo
If this question can be reworded to fit the rules in the help center, please edit the question.
5
It's simple: it cannot.
– José Carlos Santos
Jan 4 at 17:30
add a comment |
5
It's simple: it cannot.
– José Carlos Santos
Jan 4 at 17:30
5
5
It's simple: it cannot.
– José Carlos Santos
Jan 4 at 17:30
It's simple: it cannot.
– José Carlos Santos
Jan 4 at 17:30
add a comment |
1 Answer
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Let $$u_n(x)=e^{nx}.$$
$$u_n(0)=1implies sum u_n(0) text{ diverges}.$$
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Let $$u_n(x)=e^{nx}.$$
$$u_n(0)=1implies sum u_n(0) text{ diverges}.$$
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Let $$u_n(x)=e^{nx}.$$
$$u_n(0)=1implies sum u_n(0) text{ diverges}.$$
add a comment |
Let $$u_n(x)=e^{nx}.$$
$$u_n(0)=1implies sum u_n(0) text{ diverges}.$$
Let $$u_n(x)=e^{nx}.$$
$$u_n(0)=1implies sum u_n(0) text{ diverges}.$$
answered Jan 4 at 17:44
hamam_Abdallahhamam_Abdallah
38.1k21634
38.1k21634
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add a comment |
5
It's simple: it cannot.
– José Carlos Santos
Jan 4 at 17:30