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Convergence of $sum_n=1^infty sinbig((-1)^nlog(1+n^-1)+n^-2big)$
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up vote
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Study the convergence of teh following series:
$$sum_n=1^infty sinbigg((-1)^nlogbigg(1+frac1nbigg)+frac1n^2bigg)$$
Using the fact that $sin(-x)=-sin(x)$, I wrote again the series in the form:
$$sum_n=1^infty (-1)^nsinbigg(logbigg(1+frac1nbigg)+frac(-1)^nn^2bigg)$$
Then using Taylor series:
$$sum_n=1^infty (-1)^nbigg(frac1n-frac12n^2+obigg(frac1n^2bigg)+frac(-1)^nn^2bigg)$$
My aim would be that of using Leibniz, but how to handle $o$ in this case? Thank you.
real-analysis sequences-and-series convergence
asked Aug 6 at 9:19
F.inc
2958
add a comment |Â
up vote
1
down vote
favorite
Study the convergence of teh following series:
$$sum_n=1^infty sinbigg((-1)^nlogbigg(1+frac1nbigg)+frac1n^2bigg)$$
Using the fact that $sin(-x)=-sin(x)$, I wrote again the series in the form:
$$sum_n=1^infty (-1)^nsinbigg(logbigg(1+frac1nbigg)+frac(-1)^nn^2bigg)$$
Then using Taylor series:
$$sum_n=1^infty (-1)^nbigg(frac1n-frac12n^2+obigg(frac1n^2bigg)+frac(-1)^nn^2bigg)$$
My aim would be that of using Leibniz, but how to handle $o$ in this case? Thank you.
real-analysis sequences-and-series convergence
asked Aug 6 at 9:19
F.inc
2958
2
After $sin(-x)=-sin(x)$, you should have $$sum_n=1^infty (-1)^nsinbigg(logbigg(1+frac1nbigg)+frac(-1)^nn^2bigg)$$
– Robert Z
Aug 6 at 9:29
Yes, it was a total lack of attention, thank you
– F.inc
Aug 6 at 9:34
I have just corrected it, thank you.
– F.inc
Aug 6 at 9:43
1
The last expression is sufficient for your purpose; $sum (-1)^n frac1n$ converges by the alternating series test, and $sum mathcalOleft( frac1n^2 right)$ converges by the comparison test.
– Sangchul Lee
Aug 6 at 9:47
@SangchulLee In $mathcalOleft( frac1n^2right)$ did you "put" $ left(-frac(-1)^n2n^2+mathcaloleft(frac1n^2right)+frac1n^2right) $ ?
– F.inc
Aug 6 at 11:56
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Study the convergence of teh following series:
$$sum_n=1^infty sinbigg((-1)^nlogbigg(1+frac1nbigg)+frac1n^2bigg)$$
Using the fact that $sin(-x)=-sin(x)$, I wrote again the series in the form:
$$sum_n=1^infty (-1)^nsinbigg(logbigg(1+frac1nbigg)+frac(-1)^nn^2bigg)$$
Then using Taylor series:
$$sum_n=1^infty (-1)^nbigg(frac1n-frac12n^2+obigg(frac1n^2bigg)+frac(-1)^nn^2bigg)$$
My aim would be that of using Leibniz, but how to handle $o$ in this case? Thank you.
real-analysis sequences-and-series convergence
asked Aug 6 at 9:19
F.inc
2958
Study the convergence of teh following series:
$$sum_n=1^infty sinbigg((-1)^nlogbigg(1+frac1nbigg)+frac1n^2bigg)$$
Using the fact that $sin(-x)=-sin(x)$, I wrote again the series in the form:
$$sum_n=1^infty (-1)^nsinbigg(logbigg(1+frac1nbigg)+frac(-1)^nn^2bigg)$$
Then using Taylor series:
$$sum_n=1^infty (-1)^nbigg(frac1n-frac12n^2+obigg(frac1n^2bigg)+frac(-1)^nn^2bigg)$$
My aim would be that of using Leibniz, but how to handle $o$ in this case? Thank you.
real-analysis sequences-and-series convergence
asked Aug 6 at 9:19
F.inc
2958
edited Aug 6 at 9:42
asked Aug 6 at 9:19
F.inc
2958
asked Aug 6 at 9:19
F.inc
2958
asked Aug 6 at 9:19
F.inc
2958
2958
2
After $sin(-x)=-sin(x)$, you should have $$sum_n=1^infty (-1)^nsinbigg(logbigg(1+frac1nbigg)+frac(-1)^nn^2bigg)$$
– Robert Z
Aug 6 at 9:29
Yes, it was a total lack of attention, thank you
– F.inc
Aug 6 at 9:34
I have just corrected it, thank you.
– F.inc
Aug 6 at 9:43
1
The last expression is sufficient for your purpose; $sum (-1)^n frac1n$ converges by the alternating series test, and $sum mathcalOleft( frac1n^2 right)$ converges by the comparison test.
– Sangchul Lee
Aug 6 at 9:47
@SangchulLee In $mathcalOleft( frac1n^2right)$ did you "put" $ left(-frac(-1)^n2n^2+mathcaloleft(frac1n^2right)+frac1n^2right) $ ?
– F.inc
Aug 6 at 11:56
add a comment |Â
2
After $sin(-x)=-sin(x)$, you should have $$sum_n=1^infty (-1)^nsinbigg(logbigg(1+frac1nbigg)+frac(-1)^nn^2bigg)$$
– Robert Z
Aug 6 at 9:29
Yes, it was a total lack of attention, thank you
– F.inc
Aug 6 at 9:34
I have just corrected it, thank you.
– F.inc
Aug 6 at 9:43
1
The last expression is sufficient for your purpose; $sum (-1)^n frac1n$ converges by the alternating series test, and $sum mathcalOleft( frac1n^2 right)$ converges by the comparison test.
– Sangchul Lee
Aug 6 at 9:47
@SangchulLee In $mathcalOleft( frac1n^2right)$ did you "put" $ left(-frac(-1)^n2n^2+mathcaloleft(frac1n^2right)+frac1n^2right) $ ?
– F.inc
Aug 6 at 11:56
2
2
After $sin(-x)=-sin(x)$, you should have $$sum_n=1^infty (-1)^nsinbigg(logbigg(1+frac1nbigg)+frac(-1)^nn^2bigg)$$
– Robert Z
Aug 6 at 9:29
After $sin(-x)=-sin(x)$, you should have $$sum_n=1^infty (-1)^nsinbigg(logbigg(1+frac1nbigg)+frac(-1)^nn^2bigg)$$
– Robert Z
Aug 6 at 9:29
Yes, it was a total lack of attention, thank you
– F.inc
Aug 6 at 9:34
Yes, it was a total lack of attention, thank you
– F.inc
Aug 6 at 9:34
I have just corrected it, thank you.
– F.inc
Aug 6 at 9:43
I have just corrected it, thank you.
– F.inc
Aug 6 at 9:43
1
1
The last expression is sufficient for your purpose; $sum (-1)^n frac1n$ converges by the alternating series test, and $sum mathcalOleft( frac1n^2 right)$ converges by the comparison test.
– Sangchul Lee
Aug 6 at 9:47
The last expression is sufficient for your purpose; $sum (-1)^n frac1n$ converges by the alternating series test, and $sum mathcalOleft( frac1n^2 right)$ converges by the comparison test.
– Sangchul Lee
Aug 6 at 9:47
@SangchulLee In $mathcalOleft( frac1n^2right)$ did you "put" $ left(-frac(-1)^n2n^2+mathcaloleft(frac1n^2right)+frac1n^2right) $ ?
– F.inc
Aug 6 at 11:56
@SangchulLee In $mathcalOleft( frac1n^2right)$ did you "put" $ left(-frac(-1)^n2n^2+mathcaloleft(frac1n^2right)+frac1n^2right) $ ?
– F.inc
Aug 6 at 11:56
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
accepted
Note that
$$sinbigg((-1)^nlogbigg(1+frac1nbigg)+frac1n^2bigg)=(-1)^na_n+b_n$$
where
$$a_n:=sinbigg(logbigg(1+frac1nbigg)bigg)cosbigg(frac1n^2bigg),quad b_n:=cosbigg(logbigg(1+frac1nbigg)bigg)sinbigg(frac1n^2bigg)$$
Show that $xto sin(log(1+x))cos(x^2)$ is positive and increasing in a right neighbourhood of zero. Hence $a_n$ decreases to zero and $sum_n(-1)^na_n$ is convergent by Leibniz. On the other hand $sum_nb_n$ is convergent because $b_nsim 1/n^2$. Hence the given series is convergent.
answered Aug 6 at 9:48
Robert Z
84.2k955123
add a comment |Â
up vote
1
down vote
Let $a_n:= sinbigg(logbigg(1+frac1nbigg)+frac1n^2bigg)$. Then $(a_n)$ is decreasing and $a_n to 0$. Hence , by Leibniz (not Leibnitz !):
the series is convergent.
answered Aug 6 at 9:27
Fred
37.6k1237
Here $a_n:= sinbigg(logbigg(1+frac1nbigg)+frac(-1)^nn^2bigg)$!! AND such $a_n$ is NOT decreasing.
– Robert Z
Aug 6 at 9:33
add a comment |Â
up vote
1
down vote
$|sin (x +y)-sin y|leq |x|$ by MVT. Since $ sumsin (frac 1 n^2)$ is absoluetly convergent (because $|sin x | leq |x|$) it suffices to consider the series withous the $frac 1 n^2$ term. Here you can pull out the $(-1)^n$ and use the theorem that $sum (-1)^n a_n$ is convergent of $a_n$ decreases to $0$.
answered Aug 6 at 10:08
Kavi Rama Murthy
21k2830
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Note that
$$sinbigg((-1)^nlogbigg(1+frac1nbigg)+frac1n^2bigg)=(-1)^na_n+b_n$$
where
$$a_n:=sinbigg(logbigg(1+frac1nbigg)bigg)cosbigg(frac1n^2bigg),quad b_n:=cosbigg(logbigg(1+frac1nbigg)bigg)sinbigg(frac1n^2bigg)$$
Show that $xto sin(log(1+x))cos(x^2)$ is positive and increasing in a right neighbourhood of zero. Hence $a_n$ decreases to zero and $sum_n(-1)^na_n$ is convergent by Leibniz. On the other hand $sum_nb_n$ is convergent because $b_nsim 1/n^2$. Hence the given series is convergent.
answered Aug 6 at 9:48
Robert Z
84.2k955123
add a comment |Â
up vote
2
down vote
accepted
Note that
$$sinbigg((-1)^nlogbigg(1+frac1nbigg)+frac1n^2bigg)=(-1)^na_n+b_n$$
where
$$a_n:=sinbigg(logbigg(1+frac1nbigg)bigg)cosbigg(frac1n^2bigg),quad b_n:=cosbigg(logbigg(1+frac1nbigg)bigg)sinbigg(frac1n^2bigg)$$
Show that $xto sin(log(1+x))cos(x^2)$ is positive and increasing in a right neighbourhood of zero. Hence $a_n$ decreases to zero and $sum_n(-1)^na_n$ is convergent by Leibniz. On the other hand $sum_nb_n$ is convergent because $b_nsim 1/n^2$. Hence the given series is convergent.
answered Aug 6 at 9:48
Robert Z
84.2k955123
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Note that
$$sinbigg((-1)^nlogbigg(1+frac1nbigg)+frac1n^2bigg)=(-1)^na_n+b_n$$
where
$$a_n:=sinbigg(logbigg(1+frac1nbigg)bigg)cosbigg(frac1n^2bigg),quad b_n:=cosbigg(logbigg(1+frac1nbigg)bigg)sinbigg(frac1n^2bigg)$$
Show that $xto sin(log(1+x))cos(x^2)$ is positive and increasing in a right neighbourhood of zero. Hence $a_n$ decreases to zero and $sum_n(-1)^na_n$ is convergent by Leibniz. On the other hand $sum_nb_n$ is convergent because $b_nsim 1/n^2$. Hence the given series is convergent.
answered Aug 6 at 9:48
Robert Z
84.2k955123
Note that
$$sinbigg((-1)^nlogbigg(1+frac1nbigg)+frac1n^2bigg)=(-1)^na_n+b_n$$
where
$$a_n:=sinbigg(logbigg(1+frac1nbigg)bigg)cosbigg(frac1n^2bigg),quad b_n:=cosbigg(logbigg(1+frac1nbigg)bigg)sinbigg(frac1n^2bigg)$$
Show that $xto sin(log(1+x))cos(x^2)$ is positive and increasing in a right neighbourhood of zero. Hence $a_n$ decreases to zero and $sum_n(-1)^na_n$ is convergent by Leibniz. On the other hand $sum_nb_n$ is convergent because $b_nsim 1/n^2$. Hence the given series is convergent.
answered Aug 6 at 9:48
Robert Z
84.2k955123
edited Aug 6 at 10:03
answered Aug 6 at 9:48
Robert Z
84.2k955123
answered Aug 6 at 9:48
Robert Z
84.2k955123
answered Aug 6 at 9:48
Robert Z
84.2k955123
84.2k955123
add a comment |Â
add a comment |Â
up vote
1
down vote
Let $a_n:= sinbigg(logbigg(1+frac1nbigg)+frac1n^2bigg)$. Then $(a_n)$ is decreasing and $a_n to 0$. Hence , by Leibniz (not Leibnitz !):
the series is convergent.
answered Aug 6 at 9:27
Fred
37.6k1237
Here $a_n:= sinbigg(logbigg(1+frac1nbigg)+frac(-1)^nn^2bigg)$!! AND such $a_n$ is NOT decreasing.
– Robert Z
Aug 6 at 9:33
add a comment |Â
up vote
1
down vote
Let $a_n:= sinbigg(logbigg(1+frac1nbigg)+frac1n^2bigg)$. Then $(a_n)$ is decreasing and $a_n to 0$. Hence , by Leibniz (not Leibnitz !):
the series is convergent.
answered Aug 6 at 9:27
Fred
37.6k1237
Here $a_n:= sinbigg(logbigg(1+frac1nbigg)+frac(-1)^nn^2bigg)$!! AND such $a_n$ is NOT decreasing.
– Robert Z
Aug 6 at 9:33
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Let $a_n:= sinbigg(logbigg(1+frac1nbigg)+frac1n^2bigg)$. Then $(a_n)$ is decreasing and $a_n to 0$. Hence , by Leibniz (not Leibnitz !):
the series is convergent.
answered Aug 6 at 9:27
Fred
37.6k1237
Let $a_n:= sinbigg(logbigg(1+frac1nbigg)+frac1n^2bigg)$. Then $(a_n)$ is decreasing and $a_n to 0$. Hence , by Leibniz (not Leibnitz !):
the series is convergent.
answered Aug 6 at 9:27
Fred
37.6k1237
answered Aug 6 at 9:27
Fred
37.6k1237
answered Aug 6 at 9:27
Fred
37.6k1237
answered Aug 6 at 9:27
Fred
37.6k1237
37.6k1237
Here $a_n:= sinbigg(logbigg(1+frac1nbigg)+frac(-1)^nn^2bigg)$!! AND such $a_n$ is NOT decreasing.
– Robert Z
Aug 6 at 9:33
add a comment |Â
Here $a_n:= sinbigg(logbigg(1+frac1nbigg)+frac(-1)^nn^2bigg)$!! AND such $a_n$ is NOT decreasing.
– Robert Z
Aug 6 at 9:33
Here $a_n:= sinbigg(logbigg(1+frac1nbigg)+frac(-1)^nn^2bigg)$!! AND such $a_n$ is NOT decreasing.
– Robert Z
Aug 6 at 9:33
Here $a_n:= sinbigg(logbigg(1+frac1nbigg)+frac(-1)^nn^2bigg)$!! AND such $a_n$ is NOT decreasing.
– Robert Z
Aug 6 at 9:33
add a comment |Â
up vote
1
down vote
$|sin (x +y)-sin y|leq |x|$ by MVT. Since $ sumsin (frac 1 n^2)$ is absoluetly convergent (because $|sin x | leq |x|$) it suffices to consider the series withous the $frac 1 n^2$ term. Here you can pull out the $(-1)^n$ and use the theorem that $sum (-1)^n a_n$ is convergent of $a_n$ decreases to $0$.
answered Aug 6 at 10:08
Kavi Rama Murthy
21k2830
add a comment |Â
up vote
1
down vote
$|sin (x +y)-sin y|leq |x|$ by MVT. Since $ sumsin (frac 1 n^2)$ is absoluetly convergent (because $|sin x | leq |x|$) it suffices to consider the series withous the $frac 1 n^2$ term. Here you can pull out the $(-1)^n$ and use the theorem that $sum (-1)^n a_n$ is convergent of $a_n$ decreases to $0$.
answered Aug 6 at 10:08
Kavi Rama Murthy
21k2830
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$|sin (x +y)-sin y|leq |x|$ by MVT. Since $ sumsin (frac 1 n^2)$ is absoluetly convergent (because $|sin x | leq |x|$) it suffices to consider the series withous the $frac 1 n^2$ term. Here you can pull out the $(-1)^n$ and use the theorem that $sum (-1)^n a_n$ is convergent of $a_n$ decreases to $0$.
answered Aug 6 at 10:08
Kavi Rama Murthy
21k2830
$|sin (x +y)-sin y|leq |x|$ by MVT. Since $ sumsin (frac 1 n^2)$ is absoluetly convergent (because $|sin x | leq |x|$) it suffices to consider the series withous the $frac 1 n^2$ term. Here you can pull out the $(-1)^n$ and use the theorem that $sum (-1)^n a_n$ is convergent of $a_n$ decreases to $0$.
answered Aug 6 at 10:08
Kavi Rama Murthy
21k2830
answered Aug 6 at 10:08
Kavi Rama Murthy
21k2830
answered Aug 6 at 10:08
Kavi Rama Murthy
21k2830
answered Aug 6 at 10:08
Kavi Rama Murthy
21k2830
21k2830
add a comment |Â
add a comment |Â
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2
After $sin(-x)=-sin(x)$, you should have $$sum_n=1^infty (-1)^nsinbigg(logbigg(1+frac1nbigg)+frac(-1)^nn^2bigg)$$
– Robert Z
Aug 6 at 9:29
Yes, it was a total lack of attention, thank you
– F.inc
Aug 6 at 9:34
I have just corrected it, thank you.
– F.inc
Aug 6 at 9:43
1
The last expression is sufficient for your purpose; $sum (-1)^n frac1n$ converges by the alternating series test, and $sum mathcalOleft( frac1n^2 right)$ converges by the comparison test.
– Sangchul Lee
Aug 6 at 9:47
@SangchulLee In $mathcalOleft( frac1n^2right)$ did you "put" $ left(-frac(-1)^n2n^2+mathcaloleft(frac1n^2right)+frac1n^2right) $ ?
– F.inc
Aug 6 at 11:56