Convergence of the series $sum^infty_n=2 left(lnleft(fracnn-1right) - frac1nright) $ The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Series $sum fracsin(n)n cdot left(1+cdots +frac1nright)$ convergence questionAm I right in my conclusions about these series?How to deal with series of the form $sum^infty_n=1f(n)^(-1)^n$does the series converge? $sum_n=1^inftyleft(frac 35^n+frac 2nright) $Convergence for $sum _n=1^infty :fracsqrtn+1-sqrtnn^a$Does the series convergeShow that the series $sumlimits_k=1^inftyfracx^kk$ does not converge uniformly on $(-1,1)$.Calculate the following convergent series: $sum _n=1^infty :frac1nleft(n+3right)$Convergence or divergence of a series given divergent seriesHow can I show that the series $sum _k =1^infty frac1k (1 + (1/k))^k$ is divergent.
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Convergence of the series $sum^infty_n=2 left(lnleft(fracnn-1right) - frac1nright) $
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Series $sum fracsin(n)n cdot left(1+cdots +frac1nright)$ convergence questionAm I right in my conclusions about these series?How to deal with series of the form $sum^infty_n=1f(n)^(-1)^n$does the series converge? $sum_n=1^inftyleft(frac 35^n+frac 2nright) $Convergence for $sum _n=1^infty :fracsqrtn+1-sqrtnn^a$Does the series convergeShow that the series $sumlimits_k=1^inftyfracx^kk$ does not converge uniformly on $(-1,1)$.Calculate the following convergent series: $sum _n=1^infty :frac1nleft(n+3right)$Convergence or divergence of a series given divergent seriesHow can I show that the series $sum _k =1^infty frac1k (1 + (1/k))^k$ is divergent.
$begingroup$
Does the following series converge or diverge?
beginequation
sum^infty_n=2 left(lnleft(fracnn-1right) - frac1nright)
endequation
I have noticed that each of partial sum telescopes leaving me with:
beginequation
S_n = ln(n) - sum^n_k=2frac1k
endequation
I know that harmonic series are divergent, I am not sure how to use that fact in this case though. How should one follow from here?
real-analysis sequences-and-series divergent-series
edited Mar 24 at 17:19
rtybase
11.6k31534
asked Mar 24 at 16:15
TomaszTomasz
1646
$endgroup$
add a comment |
$begingroup$
Does the following series converge or diverge?
beginequation
sum^infty_n=2 left(lnleft(fracnn-1right) - frac1nright)
endequation
I have noticed that each of partial sum telescopes leaving me with:
beginequation
S_n = ln(n) - sum^n_k=2frac1k
endequation
I know that harmonic series are divergent, I am not sure how to use that fact in this case though. How should one follow from here?
real-analysis sequences-and-series divergent-series
edited Mar 24 at 17:19
rtybase
11.6k31534
asked Mar 24 at 16:15
TomaszTomasz
1646
$endgroup$
2
$begingroup$
cf. Euler-Mascheroni constant
$endgroup$
– J. W. Tanner
Mar 24 at 16:19
2
$begingroup$
Have you heard about Euler-Masheroni constant? en.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant
$endgroup$
– Tito Eliatron
Mar 24 at 16:19
add a comment |
$begingroup$
Does the following series converge or diverge?
beginequation
sum^infty_n=2 left(lnleft(fracnn-1right) - frac1nright)
endequation
I have noticed that each of partial sum telescopes leaving me with:
beginequation
S_n = ln(n) - sum^n_k=2frac1k
endequation
I know that harmonic series are divergent, I am not sure how to use that fact in this case though. How should one follow from here?
real-analysis sequences-and-series divergent-series
edited Mar 24 at 17:19
rtybase
11.6k31534
asked Mar 24 at 16:15
TomaszTomasz
1646
$endgroup$
Does the following series converge or diverge?
beginequation
sum^infty_n=2 left(lnleft(fracnn-1right) - frac1nright)
endequation
I have noticed that each of partial sum telescopes leaving me with:
beginequation
S_n = ln(n) - sum^n_k=2frac1k
endequation
I know that harmonic series are divergent, I am not sure how to use that fact in this case though. How should one follow from here?
real-analysis sequences-and-series divergent-series
real-analysis sequences-and-series divergent-series
edited Mar 24 at 17:19
rtybase
11.6k31534
asked Mar 24 at 16:15
TomaszTomasz
1646
edited Mar 24 at 17:19
rtybase
11.6k31534
asked Mar 24 at 16:15
TomaszTomasz
1646
edited Mar 24 at 17:19
rtybase
11.6k31534
edited Mar 24 at 17:19
rtybase
11.6k31534
edited Mar 24 at 17:19
rtybase
11.6k31534
11.6k31534
asked Mar 24 at 16:15
TomaszTomasz
1646
asked Mar 24 at 16:15
TomaszTomasz
1646
asked Mar 24 at 16:15
TomaszTomasz
1646
1646
2
$begingroup$
cf. Euler-Mascheroni constant
$endgroup$
– J. W. Tanner
Mar 24 at 16:19
2
$begingroup$
Have you heard about Euler-Masheroni constant? en.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant
$endgroup$
– Tito Eliatron
Mar 24 at 16:19
add a comment |
2
$begingroup$
cf. Euler-Mascheroni constant
$endgroup$
– J. W. Tanner
Mar 24 at 16:19
2
$begingroup$
Have you heard about Euler-Masheroni constant? en.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant
$endgroup$
– Tito Eliatron
Mar 24 at 16:19
2
2
$begingroup$
cf. Euler-Mascheroni constant
$endgroup$
– J. W. Tanner
Mar 24 at 16:19
$begingroup$
cf. Euler-Mascheroni constant
$endgroup$
– J. W. Tanner
Mar 24 at 16:19
2
2
$begingroup$
Have you heard about Euler-Masheroni constant? en.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant
$endgroup$
– Tito Eliatron
Mar 24 at 16:19
$begingroup$
Have you heard about Euler-Masheroni constant? en.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant
$endgroup$
– Tito Eliatron
Mar 24 at 16:19
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
$$logleft( fracnn-1right) -frac1n=-logleft(fracn-1nright)-frac1n= -logleft(1-frac1nright)-frac1n$$
$$ -logleft( 1-frac1nright)=frac1n+frac12n^2+oleft( frac1n^2right)$$
$$logleft(fracnn-1right)-frac1n =frac12n^2+oleft( frac1n^2right)$$ The series is therefore convergent.
$$underlinetextbfAbout the limit of this sum:$$
Let $gamma$ be the limit. Using partial summation Lemma:
$$sum_nle xfrac1n=fraclfloorxrfloorx+int_1^xfraclfloortrfloort^2dt= fraclfloorxrfloorx+int_1^xfract-tt^2dt $$
$$= fraclfloorxrfloorx+logx-int_1^xfractt^2dt $$ so:
$$sum_nle xfrac1n-logx= fracx+O(1)x-int_1^xfractt^2dt$$
$$=1-int_1^inftyfractt^2dt+ underbraceint_x^inftyfractt^2dt_=O(frac1x)+O(frac1x)$$
$$ sum_nle xfrac1n-logx=gamma+O(frac1x)$$ where: $$gamma= 1-int_1^inftyfractt^2dtapprox 0.57721$$
answered Mar 24 at 16:22
HAMIDINE SOUMAREHAMIDINE SOUMARE
2,076212
$endgroup$
add a comment |
$begingroup$
Notice that $$lnleft(fracnn-1 right) = - lnleft(1-frac1nright) = -left(-frac 1 n - frac 1 2n^2 - textsmaller termsright).$$ Thus $$left( lnleft(fracnn-1 right) - frac 1 nright) sim frac 1 2n^2.$$ Thus the series converges by comparison with $sum frac 1 n^2$, and as pointed out in the comments, the series sums to $gamma$, the Euler-Mascheroni constant.
answered Mar 24 at 16:24
User8128User8128
10.9k1622
$endgroup$
2
$begingroup$
It sums to $1-gamma$ as the $frac11$ term is not included and the terms are reversed.
$endgroup$
– Peter Foreman
Mar 24 at 16:39
add a comment |
$begingroup$
Using this inequality
$$xgeqln(1+x)geq fracx1+x, forall x>-1 tag1$$
and
$$logleft(fracnn-1right)-frac1n=
logleft(1+frac1n-1right)-frac1n tag2$$
we have
$$0=frac1n-frac1n=
fracfrac1n-11+frac1n-1-frac1noverset(1)leq
logleft(1+frac1n-1right)-frac1noverset(1)leq
frac1n-1-frac1n$$
and as a result, from $(2)$
$$0leq sumlimits_ngeq2left(logleft(fracnn-1right)-frac1nright)leq
sumlimits_ngeq2left(frac1n-1-frac1nright)=\
limlimits_krightarrowinftysumlimits_n=2^kleft(frac1n-1-frac1nright)=
limlimits_krightarrowinftyleft(1-frac1kright)=1$$
and the original series converges.
answered Mar 24 at 17:15
rtybasertybase
11.6k31534
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$$logleft( fracnn-1right) -frac1n=-logleft(fracn-1nright)-frac1n= -logleft(1-frac1nright)-frac1n$$
$$ -logleft( 1-frac1nright)=frac1n+frac12n^2+oleft( frac1n^2right)$$
$$logleft(fracnn-1right)-frac1n =frac12n^2+oleft( frac1n^2right)$$ The series is therefore convergent.
$$underlinetextbfAbout the limit of this sum:$$
Let $gamma$ be the limit. Using partial summation Lemma:
$$sum_nle xfrac1n=fraclfloorxrfloorx+int_1^xfraclfloortrfloort^2dt= fraclfloorxrfloorx+int_1^xfract-tt^2dt $$
$$= fraclfloorxrfloorx+logx-int_1^xfractt^2dt $$ so:
$$sum_nle xfrac1n-logx= fracx+O(1)x-int_1^xfractt^2dt$$
$$=1-int_1^inftyfractt^2dt+ underbraceint_x^inftyfractt^2dt_=O(frac1x)+O(frac1x)$$
$$ sum_nle xfrac1n-logx=gamma+O(frac1x)$$ where: $$gamma= 1-int_1^inftyfractt^2dtapprox 0.57721$$
answered Mar 24 at 16:22
HAMIDINE SOUMAREHAMIDINE SOUMARE
2,076212
$endgroup$
add a comment |
$begingroup$
$$logleft( fracnn-1right) -frac1n=-logleft(fracn-1nright)-frac1n= -logleft(1-frac1nright)-frac1n$$
$$ -logleft( 1-frac1nright)=frac1n+frac12n^2+oleft( frac1n^2right)$$
$$logleft(fracnn-1right)-frac1n =frac12n^2+oleft( frac1n^2right)$$ The series is therefore convergent.
$$underlinetextbfAbout the limit of this sum:$$
Let $gamma$ be the limit. Using partial summation Lemma:
$$sum_nle xfrac1n=fraclfloorxrfloorx+int_1^xfraclfloortrfloort^2dt= fraclfloorxrfloorx+int_1^xfract-tt^2dt $$
$$= fraclfloorxrfloorx+logx-int_1^xfractt^2dt $$ so:
$$sum_nle xfrac1n-logx= fracx+O(1)x-int_1^xfractt^2dt$$
$$=1-int_1^inftyfractt^2dt+ underbraceint_x^inftyfractt^2dt_=O(frac1x)+O(frac1x)$$
$$ sum_nle xfrac1n-logx=gamma+O(frac1x)$$ where: $$gamma= 1-int_1^inftyfractt^2dtapprox 0.57721$$
answered Mar 24 at 16:22
HAMIDINE SOUMAREHAMIDINE SOUMARE
2,076212
$endgroup$
add a comment |
$begingroup$
$$logleft( fracnn-1right) -frac1n=-logleft(fracn-1nright)-frac1n= -logleft(1-frac1nright)-frac1n$$
$$ -logleft( 1-frac1nright)=frac1n+frac12n^2+oleft( frac1n^2right)$$
$$logleft(fracnn-1right)-frac1n =frac12n^2+oleft( frac1n^2right)$$ The series is therefore convergent.
$$underlinetextbfAbout the limit of this sum:$$
Let $gamma$ be the limit. Using partial summation Lemma:
$$sum_nle xfrac1n=fraclfloorxrfloorx+int_1^xfraclfloortrfloort^2dt= fraclfloorxrfloorx+int_1^xfract-tt^2dt $$
$$= fraclfloorxrfloorx+logx-int_1^xfractt^2dt $$ so:
$$sum_nle xfrac1n-logx= fracx+O(1)x-int_1^xfractt^2dt$$
$$=1-int_1^inftyfractt^2dt+ underbraceint_x^inftyfractt^2dt_=O(frac1x)+O(frac1x)$$
$$ sum_nle xfrac1n-logx=gamma+O(frac1x)$$ where: $$gamma= 1-int_1^inftyfractt^2dtapprox 0.57721$$
answered Mar 24 at 16:22
HAMIDINE SOUMAREHAMIDINE SOUMARE
2,076212
$endgroup$
$$logleft( fracnn-1right) -frac1n=-logleft(fracn-1nright)-frac1n= -logleft(1-frac1nright)-frac1n$$
$$ -logleft( 1-frac1nright)=frac1n+frac12n^2+oleft( frac1n^2right)$$
$$logleft(fracnn-1right)-frac1n =frac12n^2+oleft( frac1n^2right)$$ The series is therefore convergent.
$$underlinetextbfAbout the limit of this sum:$$
Let $gamma$ be the limit. Using partial summation Lemma:
$$sum_nle xfrac1n=fraclfloorxrfloorx+int_1^xfraclfloortrfloort^2dt= fraclfloorxrfloorx+int_1^xfract-tt^2dt $$
$$= fraclfloorxrfloorx+logx-int_1^xfractt^2dt $$ so:
$$sum_nle xfrac1n-logx= fracx+O(1)x-int_1^xfractt^2dt$$
$$=1-int_1^inftyfractt^2dt+ underbraceint_x^inftyfractt^2dt_=O(frac1x)+O(frac1x)$$
$$ sum_nle xfrac1n-logx=gamma+O(frac1x)$$ where: $$gamma= 1-int_1^inftyfractt^2dtapprox 0.57721$$
answered Mar 24 at 16:22
HAMIDINE SOUMAREHAMIDINE SOUMARE
2,076212
edited Mar 24 at 17:49
answered Mar 24 at 16:22
HAMIDINE SOUMAREHAMIDINE SOUMARE
2,076212
answered Mar 24 at 16:22
HAMIDINE SOUMAREHAMIDINE SOUMARE
2,076212
answered Mar 24 at 16:22
HAMIDINE SOUMAREHAMIDINE SOUMARE
2,076212
2,076212
add a comment |
add a comment |
$begingroup$
Notice that $$lnleft(fracnn-1 right) = - lnleft(1-frac1nright) = -left(-frac 1 n - frac 1 2n^2 - textsmaller termsright).$$ Thus $$left( lnleft(fracnn-1 right) - frac 1 nright) sim frac 1 2n^2.$$ Thus the series converges by comparison with $sum frac 1 n^2$, and as pointed out in the comments, the series sums to $gamma$, the Euler-Mascheroni constant.
answered Mar 24 at 16:24
User8128User8128
10.9k1622
$endgroup$
2
$begingroup$
It sums to $1-gamma$ as the $frac11$ term is not included and the terms are reversed.
$endgroup$
– Peter Foreman
Mar 24 at 16:39
add a comment |
$begingroup$
Notice that $$lnleft(fracnn-1 right) = - lnleft(1-frac1nright) = -left(-frac 1 n - frac 1 2n^2 - textsmaller termsright).$$ Thus $$left( lnleft(fracnn-1 right) - frac 1 nright) sim frac 1 2n^2.$$ Thus the series converges by comparison with $sum frac 1 n^2$, and as pointed out in the comments, the series sums to $gamma$, the Euler-Mascheroni constant.
answered Mar 24 at 16:24
User8128User8128
10.9k1622
$endgroup$
2
$begingroup$
It sums to $1-gamma$ as the $frac11$ term is not included and the terms are reversed.
$endgroup$
– Peter Foreman
Mar 24 at 16:39
add a comment |
$begingroup$
Notice that $$lnleft(fracnn-1 right) = - lnleft(1-frac1nright) = -left(-frac 1 n - frac 1 2n^2 - textsmaller termsright).$$ Thus $$left( lnleft(fracnn-1 right) - frac 1 nright) sim frac 1 2n^2.$$ Thus the series converges by comparison with $sum frac 1 n^2$, and as pointed out in the comments, the series sums to $gamma$, the Euler-Mascheroni constant.
answered Mar 24 at 16:24
User8128User8128
10.9k1622
$endgroup$
Notice that $$lnleft(fracnn-1 right) = - lnleft(1-frac1nright) = -left(-frac 1 n - frac 1 2n^2 - textsmaller termsright).$$ Thus $$left( lnleft(fracnn-1 right) - frac 1 nright) sim frac 1 2n^2.$$ Thus the series converges by comparison with $sum frac 1 n^2$, and as pointed out in the comments, the series sums to $gamma$, the Euler-Mascheroni constant.
answered Mar 24 at 16:24
User8128User8128
10.9k1622
answered Mar 24 at 16:24
User8128User8128
10.9k1622
answered Mar 24 at 16:24
User8128User8128
10.9k1622
answered Mar 24 at 16:24
User8128User8128
10.9k1622
10.9k1622
2
$begingroup$
It sums to $1-gamma$ as the $frac11$ term is not included and the terms are reversed.
$endgroup$
– Peter Foreman
Mar 24 at 16:39
add a comment |
2
$begingroup$
It sums to $1-gamma$ as the $frac11$ term is not included and the terms are reversed.
$endgroup$
– Peter Foreman
Mar 24 at 16:39
2
2
$begingroup$
It sums to $1-gamma$ as the $frac11$ term is not included and the terms are reversed.
$endgroup$
– Peter Foreman
Mar 24 at 16:39
$begingroup$
It sums to $1-gamma$ as the $frac11$ term is not included and the terms are reversed.
$endgroup$
– Peter Foreman
Mar 24 at 16:39
add a comment |
$begingroup$
Using this inequality
$$xgeqln(1+x)geq fracx1+x, forall x>-1 tag1$$
and
$$logleft(fracnn-1right)-frac1n=
logleft(1+frac1n-1right)-frac1n tag2$$
we have
$$0=frac1n-frac1n=
fracfrac1n-11+frac1n-1-frac1noverset(1)leq
logleft(1+frac1n-1right)-frac1noverset(1)leq
frac1n-1-frac1n$$
and as a result, from $(2)$
$$0leq sumlimits_ngeq2left(logleft(fracnn-1right)-frac1nright)leq
sumlimits_ngeq2left(frac1n-1-frac1nright)=\
limlimits_krightarrowinftysumlimits_n=2^kleft(frac1n-1-frac1nright)=
limlimits_krightarrowinftyleft(1-frac1kright)=1$$
and the original series converges.
answered Mar 24 at 17:15
rtybasertybase
11.6k31534
$endgroup$
add a comment |
$begingroup$
Using this inequality
$$xgeqln(1+x)geq fracx1+x, forall x>-1 tag1$$
and
$$logleft(fracnn-1right)-frac1n=
logleft(1+frac1n-1right)-frac1n tag2$$
we have
$$0=frac1n-frac1n=
fracfrac1n-11+frac1n-1-frac1noverset(1)leq
logleft(1+frac1n-1right)-frac1noverset(1)leq
frac1n-1-frac1n$$
and as a result, from $(2)$
$$0leq sumlimits_ngeq2left(logleft(fracnn-1right)-frac1nright)leq
sumlimits_ngeq2left(frac1n-1-frac1nright)=\
limlimits_krightarrowinftysumlimits_n=2^kleft(frac1n-1-frac1nright)=
limlimits_krightarrowinftyleft(1-frac1kright)=1$$
and the original series converges.
answered Mar 24 at 17:15
rtybasertybase
11.6k31534
$endgroup$
add a comment |
$begingroup$
Using this inequality
$$xgeqln(1+x)geq fracx1+x, forall x>-1 tag1$$
and
$$logleft(fracnn-1right)-frac1n=
logleft(1+frac1n-1right)-frac1n tag2$$
we have
$$0=frac1n-frac1n=
fracfrac1n-11+frac1n-1-frac1noverset(1)leq
logleft(1+frac1n-1right)-frac1noverset(1)leq
frac1n-1-frac1n$$
and as a result, from $(2)$
$$0leq sumlimits_ngeq2left(logleft(fracnn-1right)-frac1nright)leq
sumlimits_ngeq2left(frac1n-1-frac1nright)=\
limlimits_krightarrowinftysumlimits_n=2^kleft(frac1n-1-frac1nright)=
limlimits_krightarrowinftyleft(1-frac1kright)=1$$
and the original series converges.
answered Mar 24 at 17:15
rtybasertybase
11.6k31534
$endgroup$
Using this inequality
$$xgeqln(1+x)geq fracx1+x, forall x>-1 tag1$$
and
$$logleft(fracnn-1right)-frac1n=
logleft(1+frac1n-1right)-frac1n tag2$$
we have
$$0=frac1n-frac1n=
fracfrac1n-11+frac1n-1-frac1noverset(1)leq
logleft(1+frac1n-1right)-frac1noverset(1)leq
frac1n-1-frac1n$$
and as a result, from $(2)$
$$0leq sumlimits_ngeq2left(logleft(fracnn-1right)-frac1nright)leq
sumlimits_ngeq2left(frac1n-1-frac1nright)=\
limlimits_krightarrowinftysumlimits_n=2^kleft(frac1n-1-frac1nright)=
limlimits_krightarrowinftyleft(1-frac1kright)=1$$
and the original series converges.
answered Mar 24 at 17:15
rtybasertybase
11.6k31534
answered Mar 24 at 17:15
rtybasertybase
11.6k31534
answered Mar 24 at 17:15
rtybasertybase
11.6k31534
answered Mar 24 at 17:15
rtybasertybase
11.6k31534
11.6k31534
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2
$begingroup$
cf. Euler-Mascheroni constant
$endgroup$
– J. W. Tanner
Mar 24 at 16:19
2
$begingroup$
Have you heard about Euler-Masheroni constant? en.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant
$endgroup$
– Tito Eliatron
Mar 24 at 16:19