Convergence or divergence of a series given divergent series
1
$begingroup$
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 1 hour ago
A. SmithA. Smith
435
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add a comment |
1
$begingroup$
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 1 hour ago
A. SmithA. Smith
435
$endgroup$
add a comment |
1
1
1
0
$begingroup$
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 1 hour ago
A. SmithA. Smith
435
$endgroup$
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 1 hour ago
A. SmithA. Smith
435
asked 1 hour ago
A. SmithA. Smith
435
asked 1 hour ago
A. SmithA. Smith
435
asked 1 hour ago
A. SmithA. Smith
435
asked 1 hour ago
A. SmithA. Smith
435
435
add a comment |
add a comment |
1 Answer
1
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oldest
votes
7
$begingroup$
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 1 hour ago
gt6989bgt6989b
33.6k22453
$endgroup$
1
$begingroup$
(+1) Looks like the assumption that $sum a_n$ diverges is unnecessary.
$endgroup$
– Yanko
1 hour ago
$begingroup$
@Yanko likely professor tried to trick them into thinking this may diverge as well, not sure. Definitely not needed...
$endgroup$
– gt6989b
1 hour ago
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
7
$begingroup$
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 1 hour ago
gt6989bgt6989b
33.6k22453
$endgroup$
1
$begingroup$
(+1) Looks like the assumption that $sum a_n$ diverges is unnecessary.
$endgroup$
– Yanko
1 hour ago
$begingroup$
@Yanko likely professor tried to trick them into thinking this may diverge as well, not sure. Definitely not needed...
$endgroup$
– gt6989b
1 hour ago
add a comment |
7
$begingroup$
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 1 hour ago
gt6989bgt6989b
33.6k22453
$endgroup$
1
$begingroup$
(+1) Looks like the assumption that $sum a_n$ diverges is unnecessary.
$endgroup$
– Yanko
1 hour ago
$begingroup$
@Yanko likely professor tried to trick them into thinking this may diverge as well, not sure. Definitely not needed...
$endgroup$
– gt6989b
1 hour ago
add a comment |
7
7
7
$begingroup$
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 1 hour ago
gt6989bgt6989b
33.6k22453
$endgroup$
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 1 hour ago
gt6989bgt6989b
33.6k22453
answered 1 hour ago
gt6989bgt6989b
33.6k22453
answered 1 hour ago
gt6989bgt6989b
33.6k22453
answered 1 hour ago
gt6989bgt6989b
33.6k22453
33.6k22453
1
$begingroup$
(+1) Looks like the assumption that $sum a_n$ diverges is unnecessary.
$endgroup$
– Yanko
1 hour ago
$begingroup$
@Yanko likely professor tried to trick them into thinking this may diverge as well, not sure. Definitely not needed...
$endgroup$
– gt6989b
1 hour ago
add a comment |
1
$begingroup$
(+1) Looks like the assumption that $sum a_n$ diverges is unnecessary.
$endgroup$
– Yanko
1 hour ago
$begingroup$
@Yanko likely professor tried to trick them into thinking this may diverge as well, not sure. Definitely not needed...
$endgroup$
– gt6989b
1 hour ago
1
1
$begingroup$
(+1) Looks like the assumption that $sum a_n$ diverges is unnecessary.
$endgroup$
– Yanko
1 hour ago
$begingroup$
(+1) Looks like the assumption that $sum a_n$ diverges is unnecessary.
$endgroup$
– Yanko
1 hour ago
$begingroup$
@Yanko likely professor tried to trick them into thinking this may diverge as well, not sure. Definitely not needed...
$endgroup$
– gt6989b
1 hour ago
$begingroup$
@Yanko likely professor tried to trick them into thinking this may diverge as well, not sure. Definitely not needed...
$endgroup$
– gt6989b
1 hour ago
add a comment |
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