Mixing
Convergence or divergence of a series given divergent series
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If $displaystylesumlimits_{n=0}^{infty}a_n$ is divergent and $a_n > 0$ for all $n$, then does $displaystylesumlimits_{n=0}^{infty} dfrac{a_n}{1+n^2 a_n}$ converge or diverge?
The only progress I have is that if you consider the harmonic series, then we get the series with terms $dfrac{1}{n(n+1)}$, which converges.
sequences-and-series convergence
asked Jan 17 at 21:56
A. SmithA. Smith
585
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add a comment |
$begingroup$
If $displaystylesumlimits_{n=0}^{infty}a_n$ is divergent and $a_n > 0$ for all $n$, then does $displaystylesumlimits_{n=0}^{infty} dfrac{a_n}{1+n^2 a_n}$ converge or diverge?
The only progress I have is that if you consider the harmonic series, then we get the series with terms $dfrac{1}{n(n+1)}$, which converges.
sequences-and-series convergence
asked Jan 17 at 21:56
A. SmithA. Smith
585
$endgroup$
add a comment |
$begingroup$
If $displaystylesumlimits_{n=0}^{infty}a_n$ is divergent and $a_n > 0$ for all $n$, then does $displaystylesumlimits_{n=0}^{infty} dfrac{a_n}{1+n^2 a_n}$ converge or diverge?
The only progress I have is that if you consider the harmonic series, then we get the series with terms $dfrac{1}{n(n+1)}$, which converges.
sequences-and-series convergence
asked Jan 17 at 21:56
A. SmithA. Smith
585
$endgroup$
If $displaystylesumlimits_{n=0}^{infty}a_n$ is divergent and $a_n > 0$ for all $n$, then does $displaystylesumlimits_{n=0}^{infty} dfrac{a_n}{1+n^2 a_n}$ converge or diverge?
The only progress I have is that if you consider the harmonic series, then we get the series with terms $dfrac{1}{n(n+1)}$, which converges.
sequences-and-series convergence
sequences-and-series convergence
asked Jan 17 at 21:56
A. SmithA. Smith
585
asked Jan 17 at 21:56
A. SmithA. Smith
585
asked Jan 17 at 21:56
A. SmithA. Smith
585
asked Jan 17 at 21:56
A. SmithA. Smith
585
asked Jan 17 at 21:56
A. SmithA. Smith
585
585
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
HINT
You have
$$
0<sum frac{a_n}{1+n^2a_n} < sum frac{a_n}{n^2a_n} = sum frac{1}{n^2}
$$
answered Jan 17 at 21:59
gt6989bgt6989b
34.4k22456
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1
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(+1) Looks like the assumption that $sum a_n$ diverges is unnecessary.
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– Yanko
Jan 17 at 22:00
$begingroup$
@Yanko likely professor tried to trick them into thinking this may diverge as well, not sure. Definitely not needed...
$endgroup$
– gt6989b
Jan 17 at 22:00
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
HINT
You have
$$
0<sum frac{a_n}{1+n^2a_n} < sum frac{a_n}{n^2a_n} = sum frac{1}{n^2}
$$
answered Jan 17 at 21:59
gt6989bgt6989b
34.4k22456
$endgroup$
1
$begingroup$
(+1) Looks like the assumption that $sum a_n$ diverges is unnecessary.
$endgroup$
– Yanko
Jan 17 at 22:00
$begingroup$
@Yanko likely professor tried to trick them into thinking this may diverge as well, not sure. Definitely not needed...
$endgroup$
– gt6989b
Jan 17 at 22:00
add a comment |
$begingroup$
HINT
You have
$$
0<sum frac{a_n}{1+n^2a_n} < sum frac{a_n}{n^2a_n} = sum frac{1}{n^2}
$$
answered Jan 17 at 21:59
gt6989bgt6989b
34.4k22456
$endgroup$
1
$begingroup$
(+1) Looks like the assumption that $sum a_n$ diverges is unnecessary.
$endgroup$
– Yanko
Jan 17 at 22:00
$begingroup$
@Yanko likely professor tried to trick them into thinking this may diverge as well, not sure. Definitely not needed...
$endgroup$
– gt6989b
Jan 17 at 22:00
add a comment |
$begingroup$
HINT
You have
$$
0<sum frac{a_n}{1+n^2a_n} < sum frac{a_n}{n^2a_n} = sum frac{1}{n^2}
$$
answered Jan 17 at 21:59
gt6989bgt6989b
34.4k22456
$endgroup$
HINT
You have
$$
0<sum frac{a_n}{1+n^2a_n} < sum frac{a_n}{n^2a_n} = sum frac{1}{n^2}
$$
answered Jan 17 at 21:59
gt6989bgt6989b
34.4k22456
answered Jan 17 at 21:59
gt6989bgt6989b
34.4k22456
answered Jan 17 at 21:59
gt6989bgt6989b
34.4k22456
answered Jan 17 at 21:59
gt6989bgt6989b
34.4k22456
34.4k22456
1
$begingroup$
(+1) Looks like the assumption that $sum a_n$ diverges is unnecessary.
$endgroup$
– Yanko
Jan 17 at 22:00
$begingroup$
@Yanko likely professor tried to trick them into thinking this may diverge as well, not sure. Definitely not needed...
$endgroup$
– gt6989b
Jan 17 at 22:00
add a comment |
1
$begingroup$
(+1) Looks like the assumption that $sum a_n$ diverges is unnecessary.
$endgroup$
– Yanko
Jan 17 at 22:00
$begingroup$
@Yanko likely professor tried to trick them into thinking this may diverge as well, not sure. Definitely not needed...
$endgroup$
– gt6989b
Jan 17 at 22:00
1
1
$begingroup$
(+1) Looks like the assumption that $sum a_n$ diverges is unnecessary.
$endgroup$
– Yanko
Jan 17 at 22:00
$begingroup$
(+1) Looks like the assumption that $sum a_n$ diverges is unnecessary.
$endgroup$
– Yanko
Jan 17 at 22:00
$begingroup$
@Yanko likely professor tried to trick them into thinking this may diverge as well, not sure. Definitely not needed...
$endgroup$
– gt6989b
Jan 17 at 22:00
$begingroup$
@Yanko likely professor tried to trick them into thinking this may diverge as well, not sure. Definitely not needed...
$endgroup$
– gt6989b
Jan 17 at 22:00
add a comment |
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