Examine convergence $sum_n=1^infty left( frac1sqrtn - sqrtlnfracn+1n right) $

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I need some help with following series:



$$sum_n=1^infty left( frac1sqrtn - sqrtlnfracn+1n right)$$



I have no idea how I should modify or compare it with something else.




sequences-and-series convergence




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  • At first sight, the second term is asymptotic to $1/sqrt n$ and there is cancellation. The next term in the Taylor's development will be of order $n^-3/2$.
    – Yves Daoust
    Jul 18 at 9:45














up vote
1
down vote

favorite
1












I need some help with following series:



$$sum_n=1^infty left( frac1sqrtn - sqrtlnfracn+1n right)$$



I have no idea how I should modify or compare it with something else.




sequences-and-series convergence




share|cite|improve this question



















  • At first sight, the second term is asymptotic to $1/sqrt n$ and there is cancellation. The next term in the Taylor's development will be of order $n^-3/2$.
    – Yves Daoust
    Jul 18 at 9:45












up vote
1
down vote

favorite
1









up vote
1
down vote

favorite
1






1





I need some help with following series:



$$sum_n=1^infty left( frac1sqrtn - sqrtlnfracn+1n right)$$



I have no idea how I should modify or compare it with something else.




sequences-and-series convergence




share|cite|improve this question











I need some help with following series:



$$sum_n=1^infty left( frac1sqrtn - sqrtlnfracn+1n right)$$



I have no idea how I should modify or compare it with something else.





sequences-and-series convergence





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 18 at 6:19









Dan We

107




107











  • At first sight, the second term is asymptotic to $1/sqrt n$ and there is cancellation. The next term in the Taylor's development will be of order $n^-3/2$.
    – Yves Daoust
    Jul 18 at 9:45
















  • At first sight, the second term is asymptotic to $1/sqrt n$ and there is cancellation. The next term in the Taylor's development will be of order $n^-3/2$.
    – Yves Daoust
    Jul 18 at 9:45















At first sight, the second term is asymptotic to $1/sqrt n$ and there is cancellation. The next term in the Taylor's development will be of order $n^-3/2$.
– Yves Daoust
Jul 18 at 9:45




At first sight, the second term is asymptotic to $1/sqrt n$ and there is cancellation. The next term in the Taylor's development will be of order $n^-3/2$.
– Yves Daoust
Jul 18 at 9:45










2 Answers
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$$lnfracn+1n=frac1nleft(1-frac12n+O(n^-2)right)$$
and so
$$sqrtlnfracn+1n=frac1sqrtnleft(1-frac14n+O(n^-2)right).$$
Then
$$frac1sqrt n-sqrtlnfracn+1n=frac14n^3/2+O(n^-5/2).$$






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  • yes of course, I see now thanks
    – gimusi
    Jul 18 at 6:26

















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2
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Also, from
$$fracx1+xleq ln(1+x)leq x, forall x>-1$$
we have
$$sqrtfrac1n+1leq sqrtlnleft(1+frac1nright)leq frac1sqrtn$$
or
$$0leq frac1sqrtn-sqrtlnleft(1+frac1nright) leq frac1sqrtn-sqrtfrac1n+1$$
where
$$0leq frac1sqrtn-sqrtfrac1n+1=fracsqrtn+1-sqrtnsqrtn+1sqrtn< fracsqrtn+1-sqrtnn=frac1n(sqrtn+1+sqrtn)< frac12sqrtn^3$$
As a result
$$0leqsumlimits_n=1^infty left( frac1sqrtn - sqrtlnfracn+1n right)<
frac12sumlimits_n=1^inftyfrac1n^frac32$$
which converges, since $frac32>1$.






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  • 1




    Nice bracketing of the logarithm.
    – Yves Daoust
    Jul 18 at 9:36










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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote



accepted










$$lnfracn+1n=frac1nleft(1-frac12n+O(n^-2)right)$$
and so
$$sqrtlnfracn+1n=frac1sqrtnleft(1-frac14n+O(n^-2)right).$$
Then
$$frac1sqrt n-sqrtlnfracn+1n=frac14n^3/2+O(n^-5/2).$$






share|cite|improve this answer





















  • yes of course, I see now thanks
    – gimusi
    Jul 18 at 6:26














up vote
2
down vote



accepted










$$lnfracn+1n=frac1nleft(1-frac12n+O(n^-2)right)$$
and so
$$sqrtlnfracn+1n=frac1sqrtnleft(1-frac14n+O(n^-2)right).$$
Then
$$frac1sqrt n-sqrtlnfracn+1n=frac14n^3/2+O(n^-5/2).$$






share|cite|improve this answer





















  • yes of course, I see now thanks
    – gimusi
    Jul 18 at 6:26












up vote
2
down vote



accepted







up vote
2
down vote



accepted






$$lnfracn+1n=frac1nleft(1-frac12n+O(n^-2)right)$$
and so
$$sqrtlnfracn+1n=frac1sqrtnleft(1-frac14n+O(n^-2)right).$$
Then
$$frac1sqrt n-sqrtlnfracn+1n=frac14n^3/2+O(n^-5/2).$$






share|cite|improve this answer













$$lnfracn+1n=frac1nleft(1-frac12n+O(n^-2)right)$$
and so
$$sqrtlnfracn+1n=frac1sqrtnleft(1-frac14n+O(n^-2)right).$$
Then
$$frac1sqrt n-sqrtlnfracn+1n=frac14n^3/2+O(n^-5/2).$$







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 18 at 6:23









Lord Shark the Unknown

85.5k951112




85.5k951112











  • yes of course, I see now thanks
    – gimusi
    Jul 18 at 6:26
















  • yes of course, I see now thanks
    – gimusi
    Jul 18 at 6:26















yes of course, I see now thanks
– gimusi
Jul 18 at 6:26




yes of course, I see now thanks
– gimusi
Jul 18 at 6:26










up vote
2
down vote













Also, from
$$fracx1+xleq ln(1+x)leq x, forall x>-1$$
we have
$$sqrtfrac1n+1leq sqrtlnleft(1+frac1nright)leq frac1sqrtn$$
or
$$0leq frac1sqrtn-sqrtlnleft(1+frac1nright) leq frac1sqrtn-sqrtfrac1n+1$$
where
$$0leq frac1sqrtn-sqrtfrac1n+1=fracsqrtn+1-sqrtnsqrtn+1sqrtn< fracsqrtn+1-sqrtnn=frac1n(sqrtn+1+sqrtn)< frac12sqrtn^3$$
As a result
$$0leqsumlimits_n=1^infty left( frac1sqrtn - sqrtlnfracn+1n right)<
frac12sumlimits_n=1^inftyfrac1n^frac32$$
which converges, since $frac32>1$.






share|cite|improve this answer

















  • 1




    Nice bracketing of the logarithm.
    – Yves Daoust
    Jul 18 at 9:36














up vote
2
down vote













Also, from
$$fracx1+xleq ln(1+x)leq x, forall x>-1$$
we have
$$sqrtfrac1n+1leq sqrtlnleft(1+frac1nright)leq frac1sqrtn$$
or
$$0leq frac1sqrtn-sqrtlnleft(1+frac1nright) leq frac1sqrtn-sqrtfrac1n+1$$
where
$$0leq frac1sqrtn-sqrtfrac1n+1=fracsqrtn+1-sqrtnsqrtn+1sqrtn< fracsqrtn+1-sqrtnn=frac1n(sqrtn+1+sqrtn)< frac12sqrtn^3$$
As a result
$$0leqsumlimits_n=1^infty left( frac1sqrtn - sqrtlnfracn+1n right)<
frac12sumlimits_n=1^inftyfrac1n^frac32$$
which converges, since $frac32>1$.






share|cite|improve this answer

















  • 1




    Nice bracketing of the logarithm.
    – Yves Daoust
    Jul 18 at 9:36












up vote
2
down vote










up vote
2
down vote









Also, from
$$fracx1+xleq ln(1+x)leq x, forall x>-1$$
we have
$$sqrtfrac1n+1leq sqrtlnleft(1+frac1nright)leq frac1sqrtn$$
or
$$0leq frac1sqrtn-sqrtlnleft(1+frac1nright) leq frac1sqrtn-sqrtfrac1n+1$$
where
$$0leq frac1sqrtn-sqrtfrac1n+1=fracsqrtn+1-sqrtnsqrtn+1sqrtn< fracsqrtn+1-sqrtnn=frac1n(sqrtn+1+sqrtn)< frac12sqrtn^3$$
As a result
$$0leqsumlimits_n=1^infty left( frac1sqrtn - sqrtlnfracn+1n right)<
frac12sumlimits_n=1^inftyfrac1n^frac32$$
which converges, since $frac32>1$.






share|cite|improve this answer













Also, from
$$fracx1+xleq ln(1+x)leq x, forall x>-1$$
we have
$$sqrtfrac1n+1leq sqrtlnleft(1+frac1nright)leq frac1sqrtn$$
or
$$0leq frac1sqrtn-sqrtlnleft(1+frac1nright) leq frac1sqrtn-sqrtfrac1n+1$$
where
$$0leq frac1sqrtn-sqrtfrac1n+1=fracsqrtn+1-sqrtnsqrtn+1sqrtn< fracsqrtn+1-sqrtnn=frac1n(sqrtn+1+sqrtn)< frac12sqrtn^3$$
As a result
$$0leqsumlimits_n=1^infty left( frac1sqrtn - sqrtlnfracn+1n right)<
frac12sumlimits_n=1^inftyfrac1n^frac32$$
which converges, since $frac32>1$.







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 18 at 8:04









rtybase

8,86721433




8,86721433







  • 1




    Nice bracketing of the logarithm.
    – Yves Daoust
    Jul 18 at 9:36












  • 1




    Nice bracketing of the logarithm.
    – Yves Daoust
    Jul 18 at 9:36







1




1




Nice bracketing of the logarithm.
– Yves Daoust
Jul 18 at 9:36




Nice bracketing of the logarithm.
– Yves Daoust
Jul 18 at 9:36












 

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