Showing that $sum_{n=1}^{infty}frac{a_n}{a_n+b_n}$ converges. [duplicate]
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This question already has an answer here:
how prove $sum_{n=1}^inftyfrac{a_n}{b_n+a_n} $is convergent?
4 answers
Show that if $a_n,b_ninmathbb{R}$, $(a_n+b_n)b_nneq0$ and both $displaystylesum_{n=1}^{infty}frac{a_n}{b_n}$ and $displaystylesum_{n=1}^{infty}left(frac{a_n}{b_n}right)^2$ converge, then $displaystylesum_{n=1}^{infty}frac{a_n}{a_n+b_n}$ converges.
If $a_n$ is positive, I have been able to solve. How we can solve in general?
real-analysis sequences-and-series
edited Mar 29 at 20:23
TheSimpliFire
13.1k62464
asked Mar 29 at 17:07
J.DoeJ.Doe
943
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marked as duplicate by Martin R, Lord Shark the Unknown, FredH, Jyrki Lahtonen, Leucippus Mar 30 at 6:33
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
how prove $sum_{n=1}^inftyfrac{a_n}{b_n+a_n} $is convergent?
4 answers
Show that if $a_n,b_ninmathbb{R}$, $(a_n+b_n)b_nneq0$ and both $displaystylesum_{n=1}^{infty}frac{a_n}{b_n}$ and $displaystylesum_{n=1}^{infty}left(frac{a_n}{b_n}right)^2$ converge, then $displaystylesum_{n=1}^{infty}frac{a_n}{a_n+b_n}$ converges.
If $a_n$ is positive, I have been able to solve. How we can solve in general?
real-analysis sequences-and-series
edited Mar 29 at 20:23
TheSimpliFire
13.1k62464
asked Mar 29 at 17:07
J.DoeJ.Doe
943
$endgroup$
marked as duplicate by Martin R, Lord Shark the Unknown, FredH, Jyrki Lahtonen, Leucippus Mar 30 at 6:33
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
Also: math.stackexchange.com/q/2154959/42969.
$endgroup$
– Martin R
Mar 30 at 2:31
add a comment |
$begingroup$
This question already has an answer here:
how prove $sum_{n=1}^inftyfrac{a_n}{b_n+a_n} $is convergent?
4 answers
Show that if $a_n,b_ninmathbb{R}$, $(a_n+b_n)b_nneq0$ and both $displaystylesum_{n=1}^{infty}frac{a_n}{b_n}$ and $displaystylesum_{n=1}^{infty}left(frac{a_n}{b_n}right)^2$ converge, then $displaystylesum_{n=1}^{infty}frac{a_n}{a_n+b_n}$ converges.
If $a_n$ is positive, I have been able to solve. How we can solve in general?
real-analysis sequences-and-series
edited Mar 29 at 20:23
TheSimpliFire
13.1k62464
asked Mar 29 at 17:07
J.DoeJ.Doe
943
$endgroup$
This question already has an answer here:
how prove $sum_{n=1}^inftyfrac{a_n}{b_n+a_n} $is convergent?
4 answers
Show that if $a_n,b_ninmathbb{R}$, $(a_n+b_n)b_nneq0$ and both $displaystylesum_{n=1}^{infty}frac{a_n}{b_n}$ and $displaystylesum_{n=1}^{infty}left(frac{a_n}{b_n}right)^2$ converge, then $displaystylesum_{n=1}^{infty}frac{a_n}{a_n+b_n}$ converges.
If $a_n$ is positive, I have been able to solve. How we can solve in general?
This question already has an answer here:
how prove $sum_{n=1}^inftyfrac{a_n}{b_n+a_n} $is convergent?
4 answers
real-analysis sequences-and-series
real-analysis sequences-and-series
edited Mar 29 at 20:23
TheSimpliFire
13.1k62464
asked Mar 29 at 17:07
J.DoeJ.Doe
943
edited Mar 29 at 20:23
TheSimpliFire
13.1k62464
asked Mar 29 at 17:07
J.DoeJ.Doe
943
edited Mar 29 at 20:23
TheSimpliFire
13.1k62464
edited Mar 29 at 20:23
TheSimpliFire
13.1k62464
edited Mar 29 at 20:23
TheSimpliFire
13.1k62464
13.1k62464
asked Mar 29 at 17:07
J.DoeJ.Doe
943
asked Mar 29 at 17:07
J.DoeJ.Doe
943
asked Mar 29 at 17:07
J.DoeJ.Doe
943
943
marked as duplicate by Martin R, Lord Shark the Unknown, FredH, Jyrki Lahtonen, Leucippus Mar 30 at 6:33
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Martin R, Lord Shark the Unknown, FredH, Jyrki Lahtonen, Leucippus Mar 30 at 6:33
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
Also: math.stackexchange.com/q/2154959/42969.
$endgroup$
– Martin R
Mar 30 at 2:31
add a comment |
$begingroup$
Also: math.stackexchange.com/q/2154959/42969.
$endgroup$
– Martin R
Mar 30 at 2:31
$begingroup$
Also: math.stackexchange.com/q/2154959/42969.
$endgroup$
– Martin R
Mar 30 at 2:31
$begingroup$
Also: math.stackexchange.com/q/2154959/42969.
$endgroup$
– Martin R
Mar 30 at 2:31
add a comment |
1 Answer
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Write $c_n=frac{a_n}{b_n}$. Then we have $c_nne -1$, and also $sum c_n$, $sum c_n^2$ converge. We need to show $sum frac{c_n}{1+c_n}$ converges.
It suffices to show that the sum of
$$c_n-frac{c_n}{1+c_n}=frac{c_n^2}{1+c_n}.$$
converges, since $sum c_n$ converges.
But $1+c_nto 1$. Then $sumfrac{c_n^2}{1+c_n}$ converges by comparison to $sum c_n^2 $.
answered Mar 29 at 17:19
Eclipse SunEclipse Sun
8,0151438
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Write $c_n=frac{a_n}{b_n}$. Then we have $c_nne -1$, and also $sum c_n$, $sum c_n^2$ converge. We need to show $sum frac{c_n}{1+c_n}$ converges.
It suffices to show that the sum of
$$c_n-frac{c_n}{1+c_n}=frac{c_n^2}{1+c_n}.$$
converges, since $sum c_n$ converges.
But $1+c_nto 1$. Then $sumfrac{c_n^2}{1+c_n}$ converges by comparison to $sum c_n^2 $.
answered Mar 29 at 17:19
Eclipse SunEclipse Sun
8,0151438
$endgroup$
add a comment |
$begingroup$
Write $c_n=frac{a_n}{b_n}$. Then we have $c_nne -1$, and also $sum c_n$, $sum c_n^2$ converge. We need to show $sum frac{c_n}{1+c_n}$ converges.
It suffices to show that the sum of
$$c_n-frac{c_n}{1+c_n}=frac{c_n^2}{1+c_n}.$$
converges, since $sum c_n$ converges.
But $1+c_nto 1$. Then $sumfrac{c_n^2}{1+c_n}$ converges by comparison to $sum c_n^2 $.
answered Mar 29 at 17:19
Eclipse SunEclipse Sun
8,0151438
$endgroup$
add a comment |
$begingroup$
Write $c_n=frac{a_n}{b_n}$. Then we have $c_nne -1$, and also $sum c_n$, $sum c_n^2$ converge. We need to show $sum frac{c_n}{1+c_n}$ converges.
It suffices to show that the sum of
$$c_n-frac{c_n}{1+c_n}=frac{c_n^2}{1+c_n}.$$
converges, since $sum c_n$ converges.
But $1+c_nto 1$. Then $sumfrac{c_n^2}{1+c_n}$ converges by comparison to $sum c_n^2 $.
answered Mar 29 at 17:19
Eclipse SunEclipse Sun
8,0151438
$endgroup$
Write $c_n=frac{a_n}{b_n}$. Then we have $c_nne -1$, and also $sum c_n$, $sum c_n^2$ converge. We need to show $sum frac{c_n}{1+c_n}$ converges.
It suffices to show that the sum of
$$c_n-frac{c_n}{1+c_n}=frac{c_n^2}{1+c_n}.$$
converges, since $sum c_n$ converges.
But $1+c_nto 1$. Then $sumfrac{c_n^2}{1+c_n}$ converges by comparison to $sum c_n^2 $.
answered Mar 29 at 17:19
Eclipse SunEclipse Sun
8,0151438
answered Mar 29 at 17:19
Eclipse SunEclipse Sun
8,0151438
answered Mar 29 at 17:19
Eclipse SunEclipse Sun
8,0151438
answered Mar 29 at 17:19
Eclipse SunEclipse Sun
8,0151438
8,0151438
add a comment |
add a comment |
$begingroup$
Also: math.stackexchange.com/q/2154959/42969.
$endgroup$
– Martin R
Mar 30 at 2:31