Example of Series such that $sum a_n$ converges but $sum a_n^4$ diverges
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I am interested in finding
Example of Series such that $sum a_n$ converges but $sum a_n^4$ diverges
I am able to find converse like if $sum a_n^4$ converges but $sum a_n$ diverges using harmonic series
But I not able to find suitable example for above
Any help will be appreciated
real-analysis sequences-and-series examples-counterexamples
asked Dec 2 '18 at 12:51
MathLoverMathLover
49710
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add a comment |
$begingroup$
I am interested in finding
Example of Series such that $sum a_n$ converges but $sum a_n^4$ diverges
I am able to find converse like if $sum a_n^4$ converges but $sum a_n$ diverges using harmonic series
But I not able to find suitable example for above
Any help will be appreciated
real-analysis sequences-and-series examples-counterexamples
asked Dec 2 '18 at 12:51
MathLoverMathLover
49710
$endgroup$
add a comment |
$begingroup$
I am interested in finding
Example of Series such that $sum a_n$ converges but $sum a_n^4$ diverges
I am able to find converse like if $sum a_n^4$ converges but $sum a_n$ diverges using harmonic series
But I not able to find suitable example for above
Any help will be appreciated
real-analysis sequences-and-series examples-counterexamples
asked Dec 2 '18 at 12:51
MathLoverMathLover
49710
$endgroup$
I am interested in finding
Example of Series such that $sum a_n$ converges but $sum a_n^4$ diverges
I am able to find converse like if $sum a_n^4$ converges but $sum a_n$ diverges using harmonic series
But I not able to find suitable example for above
Any help will be appreciated
real-analysis sequences-and-series examples-counterexamples
real-analysis sequences-and-series examples-counterexamples
asked Dec 2 '18 at 12:51
MathLoverMathLover
49710
asked Dec 2 '18 at 12:51
MathLoverMathLover
49710
asked Dec 2 '18 at 12:51
MathLoverMathLover
49710
asked Dec 2 '18 at 12:51
MathLoverMathLover
49710
asked Dec 2 '18 at 12:51
MathLoverMathLover
49710
49710
add a comment |
add a comment |
2 Answers
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Hint: Try something alternating, e.g.
$$ a_n = frac{(-1)^n}{sqrt[4]{n}}. $$
answered Dec 2 '18 at 12:56
MisterRiemannMisterRiemann
5,8451624
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thanks again I missed this
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– MathLover
Dec 2 '18 at 12:58
2
$begingroup$
Wow, I gave precisely the same example. I promise I didn't just copy your answer, I wrote my answer up independently...
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– Eff
Dec 2 '18 at 12:58
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Yes, I just have realised it, I've deleted that comment.
$endgroup$
– Anik Bhowmick
Dec 2 '18 at 12:59
add a comment |
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When $|a_n|<1$ you have that $|a_n|^4 < |a_n|$. Therefore, the convergence of $sum |a_n|$ guarantees the convergence of $sum |a_n|^4$.
Hence, to answer your question you must use negative terms. One approach is to use alternating series.
For example, consider the series of the sequence
$$a_n = (-1)^n frac{1}{sqrt[4]{n}} $$
such that $a_n^4 = 1/n$ gives rise to the Harmonic series.
The convergence of $sum a_n$ can be established (without finding the exact sum) using the alternating series test, whereas $sum a_n^4$ is the Harmonic series which diverges.
answered Dec 2 '18 at 12:57
EffEff
11.6k21638
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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active
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active
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votes
$begingroup$
Hint: Try something alternating, e.g.
$$ a_n = frac{(-1)^n}{sqrt[4]{n}}. $$
answered Dec 2 '18 at 12:56
MisterRiemannMisterRiemann
5,8451624
$endgroup$
$begingroup$
thanks again I missed this
$endgroup$
– MathLover
Dec 2 '18 at 12:58
2
$begingroup$
Wow, I gave precisely the same example. I promise I didn't just copy your answer, I wrote my answer up independently...
$endgroup$
– Eff
Dec 2 '18 at 12:58
$begingroup$
Yes, I just have realised it, I've deleted that comment.
$endgroup$
– Anik Bhowmick
Dec 2 '18 at 12:59
add a comment |
$begingroup$
Hint: Try something alternating, e.g.
$$ a_n = frac{(-1)^n}{sqrt[4]{n}}. $$
answered Dec 2 '18 at 12:56
MisterRiemannMisterRiemann
5,8451624
$endgroup$
$begingroup$
thanks again I missed this
$endgroup$
– MathLover
Dec 2 '18 at 12:58
2
$begingroup$
Wow, I gave precisely the same example. I promise I didn't just copy your answer, I wrote my answer up independently...
$endgroup$
– Eff
Dec 2 '18 at 12:58
$begingroup$
Yes, I just have realised it, I've deleted that comment.
$endgroup$
– Anik Bhowmick
Dec 2 '18 at 12:59
add a comment |
$begingroup$
Hint: Try something alternating, e.g.
$$ a_n = frac{(-1)^n}{sqrt[4]{n}}. $$
answered Dec 2 '18 at 12:56
MisterRiemannMisterRiemann
5,8451624
$endgroup$
Hint: Try something alternating, e.g.
$$ a_n = frac{(-1)^n}{sqrt[4]{n}}. $$
answered Dec 2 '18 at 12:56
MisterRiemannMisterRiemann
5,8451624
answered Dec 2 '18 at 12:56
MisterRiemannMisterRiemann
5,8451624
answered Dec 2 '18 at 12:56
MisterRiemannMisterRiemann
5,8451624
answered Dec 2 '18 at 12:56
MisterRiemannMisterRiemann
5,8451624
5,8451624
$begingroup$
thanks again I missed this
$endgroup$
– MathLover
Dec 2 '18 at 12:58
2
$begingroup$
Wow, I gave precisely the same example. I promise I didn't just copy your answer, I wrote my answer up independently...
$endgroup$
– Eff
Dec 2 '18 at 12:58
$begingroup$
Yes, I just have realised it, I've deleted that comment.
$endgroup$
– Anik Bhowmick
Dec 2 '18 at 12:59
add a comment |
$begingroup$
thanks again I missed this
$endgroup$
– MathLover
Dec 2 '18 at 12:58
2
$begingroup$
Wow, I gave precisely the same example. I promise I didn't just copy your answer, I wrote my answer up independently...
$endgroup$
– Eff
Dec 2 '18 at 12:58
$begingroup$
Yes, I just have realised it, I've deleted that comment.
$endgroup$
– Anik Bhowmick
Dec 2 '18 at 12:59
$begingroup$
thanks again I missed this
$endgroup$
– MathLover
Dec 2 '18 at 12:58
$begingroup$
thanks again I missed this
$endgroup$
– MathLover
Dec 2 '18 at 12:58
2
2
$begingroup$
Wow, I gave precisely the same example. I promise I didn't just copy your answer, I wrote my answer up independently...
$endgroup$
– Eff
Dec 2 '18 at 12:58
$begingroup$
Wow, I gave precisely the same example. I promise I didn't just copy your answer, I wrote my answer up independently...
$endgroup$
– Eff
Dec 2 '18 at 12:58
$begingroup$
Yes, I just have realised it, I've deleted that comment.
$endgroup$
– Anik Bhowmick
Dec 2 '18 at 12:59
$begingroup$
Yes, I just have realised it, I've deleted that comment.
$endgroup$
– Anik Bhowmick
Dec 2 '18 at 12:59
add a comment |
$begingroup$
When $|a_n|<1$ you have that $|a_n|^4 < |a_n|$. Therefore, the convergence of $sum |a_n|$ guarantees the convergence of $sum |a_n|^4$.
Hence, to answer your question you must use negative terms. One approach is to use alternating series.
For example, consider the series of the sequence
$$a_n = (-1)^n frac{1}{sqrt[4]{n}} $$
such that $a_n^4 = 1/n$ gives rise to the Harmonic series.
The convergence of $sum a_n$ can be established (without finding the exact sum) using the alternating series test, whereas $sum a_n^4$ is the Harmonic series which diverges.
answered Dec 2 '18 at 12:57
EffEff
11.6k21638
$endgroup$
add a comment |
$begingroup$
When $|a_n|<1$ you have that $|a_n|^4 < |a_n|$. Therefore, the convergence of $sum |a_n|$ guarantees the convergence of $sum |a_n|^4$.
Hence, to answer your question you must use negative terms. One approach is to use alternating series.
For example, consider the series of the sequence
$$a_n = (-1)^n frac{1}{sqrt[4]{n}} $$
such that $a_n^4 = 1/n$ gives rise to the Harmonic series.
The convergence of $sum a_n$ can be established (without finding the exact sum) using the alternating series test, whereas $sum a_n^4$ is the Harmonic series which diverges.
answered Dec 2 '18 at 12:57
EffEff
11.6k21638
$endgroup$
add a comment |
$begingroup$
When $|a_n|<1$ you have that $|a_n|^4 < |a_n|$. Therefore, the convergence of $sum |a_n|$ guarantees the convergence of $sum |a_n|^4$.
Hence, to answer your question you must use negative terms. One approach is to use alternating series.
For example, consider the series of the sequence
$$a_n = (-1)^n frac{1}{sqrt[4]{n}} $$
such that $a_n^4 = 1/n$ gives rise to the Harmonic series.
The convergence of $sum a_n$ can be established (without finding the exact sum) using the alternating series test, whereas $sum a_n^4$ is the Harmonic series which diverges.
answered Dec 2 '18 at 12:57
EffEff
11.6k21638
$endgroup$
When $|a_n|<1$ you have that $|a_n|^4 < |a_n|$. Therefore, the convergence of $sum |a_n|$ guarantees the convergence of $sum |a_n|^4$.
Hence, to answer your question you must use negative terms. One approach is to use alternating series.
For example, consider the series of the sequence
$$a_n = (-1)^n frac{1}{sqrt[4]{n}} $$
such that $a_n^4 = 1/n$ gives rise to the Harmonic series.
The convergence of $sum a_n$ can be established (without finding the exact sum) using the alternating series test, whereas $sum a_n^4$ is the Harmonic series which diverges.
answered Dec 2 '18 at 12:57
EffEff
11.6k21638
edited Dec 4 '18 at 11:15
answered Dec 2 '18 at 12:57
EffEff
11.6k21638
answered Dec 2 '18 at 12:57
EffEff
11.6k21638
answered Dec 2 '18 at 12:57
EffEff
11.6k21638
11.6k21638
add a comment |
add a comment |
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