Mixing
Does $Sigma_{n=1}^{infty} frac {sqrt {2n-1}~ log (4n+1)} {n(n+1)}$ converge?
$begingroup$
Does $Sigma_{n=1}^{infty} dfrac {sqrt {2n-1}~ log (4n+1)} {n(n+1)}$ converge?
Attempt: I have been trying to use the comparison test for a while, but I can't find a suitable comparator.
For Example, since, $Sigma_{n=1}^{infty} dfrac {1}{n(n+1)}$ converges, I was thinking of an expression involving the above to use as a comparator. But can not find one.
Please guide me on how to move ahead.
Thank you very much for your help.
calculus real-analysis sequences-and-series
edited Jan 27 '15 at 21:09
dustin
6,72892969
asked Jan 22 '15 at 12:08
MathManMathMan
3,64841970
$endgroup$
add a comment |
$begingroup$
Does $Sigma_{n=1}^{infty} dfrac {sqrt {2n-1}~ log (4n+1)} {n(n+1)}$ converge?
Attempt: I have been trying to use the comparison test for a while, but I can't find a suitable comparator.
For Example, since, $Sigma_{n=1}^{infty} dfrac {1}{n(n+1)}$ converges, I was thinking of an expression involving the above to use as a comparator. But can not find one.
Please guide me on how to move ahead.
Thank you very much for your help.
calculus real-analysis sequences-and-series
edited Jan 27 '15 at 21:09
dustin
6,72892969
asked Jan 22 '15 at 12:08
MathManMathMan
3,64841970
$endgroup$
add a comment |
$begingroup$
Does $Sigma_{n=1}^{infty} dfrac {sqrt {2n-1}~ log (4n+1)} {n(n+1)}$ converge?
Attempt: I have been trying to use the comparison test for a while, but I can't find a suitable comparator.
For Example, since, $Sigma_{n=1}^{infty} dfrac {1}{n(n+1)}$ converges, I was thinking of an expression involving the above to use as a comparator. But can not find one.
Please guide me on how to move ahead.
Thank you very much for your help.
calculus real-analysis sequences-and-series
edited Jan 27 '15 at 21:09
dustin
6,72892969
asked Jan 22 '15 at 12:08
MathManMathMan
3,64841970
$endgroup$
Does $Sigma_{n=1}^{infty} dfrac {sqrt {2n-1}~ log (4n+1)} {n(n+1)}$ converge?
Attempt: I have been trying to use the comparison test for a while, but I can't find a suitable comparator.
For Example, since, $Sigma_{n=1}^{infty} dfrac {1}{n(n+1)}$ converges, I was thinking of an expression involving the above to use as a comparator. But can not find one.
Please guide me on how to move ahead.
Thank you very much for your help.
calculus real-analysis sequences-and-series
calculus real-analysis sequences-and-series
edited Jan 27 '15 at 21:09
dustin
6,72892969
asked Jan 22 '15 at 12:08
MathManMathMan
3,64841970
edited Jan 27 '15 at 21:09
dustin
6,72892969
asked Jan 22 '15 at 12:08
MathManMathMan
3,64841970
edited Jan 27 '15 at 21:09
dustin
6,72892969
edited Jan 27 '15 at 21:09
dustin
6,72892969
edited Jan 27 '15 at 21:09
dustin
6,72892969
6,72892969
asked Jan 22 '15 at 12:08
MathManMathMan
3,64841970
asked Jan 22 '15 at 12:08
MathManMathMan
3,64841970
asked Jan 22 '15 at 12:08
MathManMathMan
3,64841970
3,64841970
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
First using the L'Hôpital's rule we see that
$$ln(n)=_infty o(n^alpha),quad forall alpha>0$$
Now we have
$$frac {sqrt {2n-1}~ log (4n+1)} {n(n+1)}sim_infty sqrt 2frac{ln(4n)}{n^{3/2}}=oleft(frac1{n^{3/2-alpha}}right)$$
so it suffices to choose $alpha>0$ such that $frac32-alpha>1$ to conclude that the given series is convergent.
$endgroup$
$begingroup$
$sqrt 2frac{ln(4n)}{n^{3/2}}=oleft(frac1{n^{3/2-alpha}}right)$. Could you please explain the steps after that?
$endgroup$
– MathMan
Jan 22 '15 at 13:34
$begingroup$
If you choose $alpha>0$ st $frac{3}{2}-alpha>1$ then $sum frac{1}{n^{3/2-alpha}}$ converges so your series will also converge.
$endgroup$
– Bhauryal
Jan 22 '15 at 18:54
add a comment |
$begingroup$
For large $n$ $$frac{sqrt{2n-1} hspace{1mm}ln(4n+1) }{ n(n+1) }approx frac{sqrt{2n} hspace{1mm}ln(4n) }{ n(n+1) }=sqrt{2}frac{ln4n}{sqrt{n}(n+1)}=sqrt{2} hspace{1mm} ln4frac{1}{sqrt{n}(n+1)}+sqrt{2}frac{ln n}{sqrt{n}(n+1)}$$ The first term is the term from a convergent series.(Hint:- It is dominated by a convergent $p$-series). Now consider the series $sum_{n=1}^{infty} frac{ln n}{sqrt{n}n}$. We are done if we could show that this series converges. Integral test seems to be efficient in this case:- begin{equation} begin{aligned}
lim_{n to infty}int_1^{n}frac{ln x}{xsqrt{x}} dx&=lim_{n to infty}int_0^{ln n}ue^{-frac{u}{2}}du qquad (mbox{where $u=ln x$} )\
&=int_0^{infty}ue^{-frac{u}{2}}du \
&=4
end{aligned}
end{equation}Thus $sum_{n=1}^{infty}frac{ln n}{nsqrt{n}}$ converges. Since $frac{ln n}{sqrt{n} (n+1)}<frac{ln n}{nsqrt{n}}$, so by Comparison test the series $sum_{n=1}^{infty} frac{ln n}{sqrt{n} (n+1)}$ converges. Ultimately, your series converges.
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
First using the L'Hôpital's rule we see that
$$ln(n)=_infty o(n^alpha),quad forall alpha>0$$
Now we have
$$frac {sqrt {2n-1}~ log (4n+1)} {n(n+1)}sim_infty sqrt 2frac{ln(4n)}{n^{3/2}}=oleft(frac1{n^{3/2-alpha}}right)$$
so it suffices to choose $alpha>0$ such that $frac32-alpha>1$ to conclude that the given series is convergent.
$endgroup$
$begingroup$
$sqrt 2frac{ln(4n)}{n^{3/2}}=oleft(frac1{n^{3/2-alpha}}right)$. Could you please explain the steps after that?
$endgroup$
– MathMan
Jan 22 '15 at 13:34
$begingroup$
If you choose $alpha>0$ st $frac{3}{2}-alpha>1$ then $sum frac{1}{n^{3/2-alpha}}$ converges so your series will also converge.
$endgroup$
– Bhauryal
Jan 22 '15 at 18:54
add a comment |
$begingroup$
First using the L'Hôpital's rule we see that
$$ln(n)=_infty o(n^alpha),quad forall alpha>0$$
Now we have
$$frac {sqrt {2n-1}~ log (4n+1)} {n(n+1)}sim_infty sqrt 2frac{ln(4n)}{n^{3/2}}=oleft(frac1{n^{3/2-alpha}}right)$$
so it suffices to choose $alpha>0$ such that $frac32-alpha>1$ to conclude that the given series is convergent.
$endgroup$
$begingroup$
$sqrt 2frac{ln(4n)}{n^{3/2}}=oleft(frac1{n^{3/2-alpha}}right)$. Could you please explain the steps after that?
$endgroup$
– MathMan
Jan 22 '15 at 13:34
$begingroup$
If you choose $alpha>0$ st $frac{3}{2}-alpha>1$ then $sum frac{1}{n^{3/2-alpha}}$ converges so your series will also converge.
$endgroup$
– Bhauryal
Jan 22 '15 at 18:54
add a comment |
$begingroup$
First using the L'Hôpital's rule we see that
$$ln(n)=_infty o(n^alpha),quad forall alpha>0$$
Now we have
$$frac {sqrt {2n-1}~ log (4n+1)} {n(n+1)}sim_infty sqrt 2frac{ln(4n)}{n^{3/2}}=oleft(frac1{n^{3/2-alpha}}right)$$
so it suffices to choose $alpha>0$ such that $frac32-alpha>1$ to conclude that the given series is convergent.
$endgroup$
First using the L'Hôpital's rule we see that
$$ln(n)=_infty o(n^alpha),quad forall alpha>0$$
Now we have
$$frac {sqrt {2n-1}~ log (4n+1)} {n(n+1)}sim_infty sqrt 2frac{ln(4n)}{n^{3/2}}=oleft(frac1{n^{3/2-alpha}}right)$$
so it suffices to choose $alpha>0$ such that $frac32-alpha>1$ to conclude that the given series is convergent.
answered Jan 22 '15 at 12:15
user63181
$begingroup$
$sqrt 2frac{ln(4n)}{n^{3/2}}=oleft(frac1{n^{3/2-alpha}}right)$. Could you please explain the steps after that?
$endgroup$
– MathMan
Jan 22 '15 at 13:34
$begingroup$
If you choose $alpha>0$ st $frac{3}{2}-alpha>1$ then $sum frac{1}{n^{3/2-alpha}}$ converges so your series will also converge.
$endgroup$
– Bhauryal
Jan 22 '15 at 18:54
add a comment |
$begingroup$
$sqrt 2frac{ln(4n)}{n^{3/2}}=oleft(frac1{n^{3/2-alpha}}right)$. Could you please explain the steps after that?
$endgroup$
– MathMan
Jan 22 '15 at 13:34
$begingroup$
If you choose $alpha>0$ st $frac{3}{2}-alpha>1$ then $sum frac{1}{n^{3/2-alpha}}$ converges so your series will also converge.
$endgroup$
– Bhauryal
Jan 22 '15 at 18:54
$begingroup$
$sqrt 2frac{ln(4n)}{n^{3/2}}=oleft(frac1{n^{3/2-alpha}}right)$. Could you please explain the steps after that?
$endgroup$
– MathMan
Jan 22 '15 at 13:34
$begingroup$
$sqrt 2frac{ln(4n)}{n^{3/2}}=oleft(frac1{n^{3/2-alpha}}right)$. Could you please explain the steps after that?
$endgroup$
– MathMan
Jan 22 '15 at 13:34
$begingroup$
If you choose $alpha>0$ st $frac{3}{2}-alpha>1$ then $sum frac{1}{n^{3/2-alpha}}$ converges so your series will also converge.
$endgroup$
– Bhauryal
Jan 22 '15 at 18:54
$begingroup$
If you choose $alpha>0$ st $frac{3}{2}-alpha>1$ then $sum frac{1}{n^{3/2-alpha}}$ converges so your series will also converge.
$endgroup$
– Bhauryal
Jan 22 '15 at 18:54
add a comment |
$begingroup$
For large $n$ $$frac{sqrt{2n-1} hspace{1mm}ln(4n+1) }{ n(n+1) }approx frac{sqrt{2n} hspace{1mm}ln(4n) }{ n(n+1) }=sqrt{2}frac{ln4n}{sqrt{n}(n+1)}=sqrt{2} hspace{1mm} ln4frac{1}{sqrt{n}(n+1)}+sqrt{2}frac{ln n}{sqrt{n}(n+1)}$$ The first term is the term from a convergent series.(Hint:- It is dominated by a convergent $p$-series). Now consider the series $sum_{n=1}^{infty} frac{ln n}{sqrt{n}n}$. We are done if we could show that this series converges. Integral test seems to be efficient in this case:- begin{equation} begin{aligned}
lim_{n to infty}int_1^{n}frac{ln x}{xsqrt{x}} dx&=lim_{n to infty}int_0^{ln n}ue^{-frac{u}{2}}du qquad (mbox{where $u=ln x$} )\
&=int_0^{infty}ue^{-frac{u}{2}}du \
&=4
end{aligned}
end{equation}Thus $sum_{n=1}^{infty}frac{ln n}{nsqrt{n}}$ converges. Since $frac{ln n}{sqrt{n} (n+1)}<frac{ln n}{nsqrt{n}}$, so by Comparison test the series $sum_{n=1}^{infty} frac{ln n}{sqrt{n} (n+1)}$ converges. Ultimately, your series converges.
$endgroup$
add a comment |
$begingroup$
For large $n$ $$frac{sqrt{2n-1} hspace{1mm}ln(4n+1) }{ n(n+1) }approx frac{sqrt{2n} hspace{1mm}ln(4n) }{ n(n+1) }=sqrt{2}frac{ln4n}{sqrt{n}(n+1)}=sqrt{2} hspace{1mm} ln4frac{1}{sqrt{n}(n+1)}+sqrt{2}frac{ln n}{sqrt{n}(n+1)}$$ The first term is the term from a convergent series.(Hint:- It is dominated by a convergent $p$-series). Now consider the series $sum_{n=1}^{infty} frac{ln n}{sqrt{n}n}$. We are done if we could show that this series converges. Integral test seems to be efficient in this case:- begin{equation} begin{aligned}
lim_{n to infty}int_1^{n}frac{ln x}{xsqrt{x}} dx&=lim_{n to infty}int_0^{ln n}ue^{-frac{u}{2}}du qquad (mbox{where $u=ln x$} )\
&=int_0^{infty}ue^{-frac{u}{2}}du \
&=4
end{aligned}
end{equation}Thus $sum_{n=1}^{infty}frac{ln n}{nsqrt{n}}$ converges. Since $frac{ln n}{sqrt{n} (n+1)}<frac{ln n}{nsqrt{n}}$, so by Comparison test the series $sum_{n=1}^{infty} frac{ln n}{sqrt{n} (n+1)}$ converges. Ultimately, your series converges.
$endgroup$
add a comment |
$begingroup$
For large $n$ $$frac{sqrt{2n-1} hspace{1mm}ln(4n+1) }{ n(n+1) }approx frac{sqrt{2n} hspace{1mm}ln(4n) }{ n(n+1) }=sqrt{2}frac{ln4n}{sqrt{n}(n+1)}=sqrt{2} hspace{1mm} ln4frac{1}{sqrt{n}(n+1)}+sqrt{2}frac{ln n}{sqrt{n}(n+1)}$$ The first term is the term from a convergent series.(Hint:- It is dominated by a convergent $p$-series). Now consider the series $sum_{n=1}^{infty} frac{ln n}{sqrt{n}n}$. We are done if we could show that this series converges. Integral test seems to be efficient in this case:- begin{equation} begin{aligned}
lim_{n to infty}int_1^{n}frac{ln x}{xsqrt{x}} dx&=lim_{n to infty}int_0^{ln n}ue^{-frac{u}{2}}du qquad (mbox{where $u=ln x$} )\
&=int_0^{infty}ue^{-frac{u}{2}}du \
&=4
end{aligned}
end{equation}Thus $sum_{n=1}^{infty}frac{ln n}{nsqrt{n}}$ converges. Since $frac{ln n}{sqrt{n} (n+1)}<frac{ln n}{nsqrt{n}}$, so by Comparison test the series $sum_{n=1}^{infty} frac{ln n}{sqrt{n} (n+1)}$ converges. Ultimately, your series converges.
$endgroup$
For large $n$ $$frac{sqrt{2n-1} hspace{1mm}ln(4n+1) }{ n(n+1) }approx frac{sqrt{2n} hspace{1mm}ln(4n) }{ n(n+1) }=sqrt{2}frac{ln4n}{sqrt{n}(n+1)}=sqrt{2} hspace{1mm} ln4frac{1}{sqrt{n}(n+1)}+sqrt{2}frac{ln n}{sqrt{n}(n+1)}$$ The first term is the term from a convergent series.(Hint:- It is dominated by a convergent $p$-series). Now consider the series $sum_{n=1}^{infty} frac{ln n}{sqrt{n}n}$. We are done if we could show that this series converges. Integral test seems to be efficient in this case:- begin{equation} begin{aligned}
lim_{n to infty}int_1^{n}frac{ln x}{xsqrt{x}} dx&=lim_{n to infty}int_0^{ln n}ue^{-frac{u}{2}}du qquad (mbox{where $u=ln x$} )\
&=int_0^{infty}ue^{-frac{u}{2}}du \
&=4
end{aligned}
end{equation}Thus $sum_{n=1}^{infty}frac{ln n}{nsqrt{n}}$ converges. Since $frac{ln n}{sqrt{n} (n+1)}<frac{ln n}{nsqrt{n}}$, so by Comparison test the series $sum_{n=1}^{infty} frac{ln n}{sqrt{n} (n+1)}$ converges. Ultimately, your series converges.
answered Jan 11 at 18:15
user440191
add a comment |
add a comment |
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