Convergence test from Demidovich: $sum_{n=1}^{infty} frac{n^{n + frac{1}{n}}}{(n + frac{1}{n})^n}$
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I'm just learning how to test series for convergence and have encountered this series from the Demidovich's book and I can't really decide what criteria should I use. Could you please give me some hint(s)?
$$sum_{n=1}^{infty} frac{n^{n + frac{1}{n}}}{(n + frac{1}{n})^n}$$
calculus sequences-and-series convergence divergent-series
edited Dec 9 '18 at 13:24
Martin Sleziak
44.9k10121274
asked May 17 '16 at 23:12
GogisGogis
103113
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add a comment |
$begingroup$
I'm just learning how to test series for convergence and have encountered this series from the Demidovich's book and I can't really decide what criteria should I use. Could you please give me some hint(s)?
$$sum_{n=1}^{infty} frac{n^{n + frac{1}{n}}}{(n + frac{1}{n})^n}$$
calculus sequences-and-series convergence divergent-series
edited Dec 9 '18 at 13:24
Martin Sleziak
44.9k10121274
asked May 17 '16 at 23:12
GogisGogis
103113
$endgroup$
add a comment |
$begingroup$
I'm just learning how to test series for convergence and have encountered this series from the Demidovich's book and I can't really decide what criteria should I use. Could you please give me some hint(s)?
$$sum_{n=1}^{infty} frac{n^{n + frac{1}{n}}}{(n + frac{1}{n})^n}$$
calculus sequences-and-series convergence divergent-series
edited Dec 9 '18 at 13:24
Martin Sleziak
44.9k10121274
asked May 17 '16 at 23:12
GogisGogis
103113
$endgroup$
I'm just learning how to test series for convergence and have encountered this series from the Demidovich's book and I can't really decide what criteria should I use. Could you please give me some hint(s)?
$$sum_{n=1}^{infty} frac{n^{n + frac{1}{n}}}{(n + frac{1}{n})^n}$$
calculus sequences-and-series convergence divergent-series
calculus sequences-and-series convergence divergent-series
edited Dec 9 '18 at 13:24
Martin Sleziak
44.9k10121274
asked May 17 '16 at 23:12
GogisGogis
103113
edited Dec 9 '18 at 13:24
Martin Sleziak
44.9k10121274
asked May 17 '16 at 23:12
GogisGogis
103113
edited Dec 9 '18 at 13:24
Martin Sleziak
44.9k10121274
edited Dec 9 '18 at 13:24
Martin Sleziak
44.9k10121274
edited Dec 9 '18 at 13:24
Martin Sleziak
44.9k10121274
44.9k10121274
asked May 17 '16 at 23:12
GogisGogis
103113
asked May 17 '16 at 23:12
GogisGogis
103113
asked May 17 '16 at 23:12
GogisGogis
103113
103113
add a comment |
add a comment |
2 Answers
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Hint: Since $n^{1/n} to 1,$ you can forget about it. We're left with an $n$th term equal to
$$frac{n^n}{(n+1/n)^n} = left ( frac{1}{1 + 1/n^2} right )^n.$$
answered May 17 '16 at 23:32
zhw.zhw.
74.5k43175
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add a comment |
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The $n^{th}$ root test looks tempting because of all the $n^{th}$ powers, but it only works when the rate of increase or decrease of the terms of the series is exponential, which is not the case here. Let's work out a couple of limits instead:
$$lim_{nrightarrowinfty}n^{frac1n}=lim_{nrightarrowinfty}expleft(frac{ln n}nright)=expleft(frac 00right)=lim_{nrightarrowinfty}expleft(frac{frac1n}1right)=expleft(frac01right)=e^0=1$$
And
$$begin{align}lim_{nrightarrowinfty}left(1+frac1{n^2}right)^n&=lim_{nrightarrowinfty}expleft(nlnleft(1+frac1{n^2}right)right)\
&=lim_{nrightarrowinfty}expleft(frac{lnleft(1+frac1{n^2}right)}{frac1n}right)=expleft(frac00right)\
&=lim_{nrightarrowinfty}expleft(frac{frac{-frac2{n^3}}{1+frac1{n^2}}}{-frac1{n^2}}right)=lim_{nrightarrowinfty}expleft(frac{frac2n}{1+frac1{n^2}}right)\
&=expleft(frac0{1+0}right)=e^0=1end{align}$$
Then
$$lim_{nrightarrowinfty}frac{n^{n+frac1n}}{left(n+frac1nright)^n}=lim_{nrightarrowinfty}frac{n^{frac1n}}{left(1+frac1{n^2}right)^n}=frac11=1ne0$$
So the sum diverges by the divergence test.
answered May 17 '16 at 23:37
user5713492user5713492
11.1k2919
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2 Answers
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2 Answers
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$begingroup$
Hint: Since $n^{1/n} to 1,$ you can forget about it. We're left with an $n$th term equal to
$$frac{n^n}{(n+1/n)^n} = left ( frac{1}{1 + 1/n^2} right )^n.$$
answered May 17 '16 at 23:32
zhw.zhw.
74.5k43175
$endgroup$
add a comment |
$begingroup$
Hint: Since $n^{1/n} to 1,$ you can forget about it. We're left with an $n$th term equal to
$$frac{n^n}{(n+1/n)^n} = left ( frac{1}{1 + 1/n^2} right )^n.$$
answered May 17 '16 at 23:32
zhw.zhw.
74.5k43175
$endgroup$
add a comment |
$begingroup$
Hint: Since $n^{1/n} to 1,$ you can forget about it. We're left with an $n$th term equal to
$$frac{n^n}{(n+1/n)^n} = left ( frac{1}{1 + 1/n^2} right )^n.$$
answered May 17 '16 at 23:32
zhw.zhw.
74.5k43175
$endgroup$
Hint: Since $n^{1/n} to 1,$ you can forget about it. We're left with an $n$th term equal to
$$frac{n^n}{(n+1/n)^n} = left ( frac{1}{1 + 1/n^2} right )^n.$$
answered May 17 '16 at 23:32
zhw.zhw.
74.5k43175
answered May 17 '16 at 23:32
zhw.zhw.
74.5k43175
answered May 17 '16 at 23:32
zhw.zhw.
74.5k43175
answered May 17 '16 at 23:32
zhw.zhw.
74.5k43175
74.5k43175
add a comment |
add a comment |
$begingroup$
The $n^{th}$ root test looks tempting because of all the $n^{th}$ powers, but it only works when the rate of increase or decrease of the terms of the series is exponential, which is not the case here. Let's work out a couple of limits instead:
$$lim_{nrightarrowinfty}n^{frac1n}=lim_{nrightarrowinfty}expleft(frac{ln n}nright)=expleft(frac 00right)=lim_{nrightarrowinfty}expleft(frac{frac1n}1right)=expleft(frac01right)=e^0=1$$
And
$$begin{align}lim_{nrightarrowinfty}left(1+frac1{n^2}right)^n&=lim_{nrightarrowinfty}expleft(nlnleft(1+frac1{n^2}right)right)\
&=lim_{nrightarrowinfty}expleft(frac{lnleft(1+frac1{n^2}right)}{frac1n}right)=expleft(frac00right)\
&=lim_{nrightarrowinfty}expleft(frac{frac{-frac2{n^3}}{1+frac1{n^2}}}{-frac1{n^2}}right)=lim_{nrightarrowinfty}expleft(frac{frac2n}{1+frac1{n^2}}right)\
&=expleft(frac0{1+0}right)=e^0=1end{align}$$
Then
$$lim_{nrightarrowinfty}frac{n^{n+frac1n}}{left(n+frac1nright)^n}=lim_{nrightarrowinfty}frac{n^{frac1n}}{left(1+frac1{n^2}right)^n}=frac11=1ne0$$
So the sum diverges by the divergence test.
answered May 17 '16 at 23:37
user5713492user5713492
11.1k2919
$endgroup$
add a comment |
$begingroup$
The $n^{th}$ root test looks tempting because of all the $n^{th}$ powers, but it only works when the rate of increase or decrease of the terms of the series is exponential, which is not the case here. Let's work out a couple of limits instead:
$$lim_{nrightarrowinfty}n^{frac1n}=lim_{nrightarrowinfty}expleft(frac{ln n}nright)=expleft(frac 00right)=lim_{nrightarrowinfty}expleft(frac{frac1n}1right)=expleft(frac01right)=e^0=1$$
And
$$begin{align}lim_{nrightarrowinfty}left(1+frac1{n^2}right)^n&=lim_{nrightarrowinfty}expleft(nlnleft(1+frac1{n^2}right)right)\
&=lim_{nrightarrowinfty}expleft(frac{lnleft(1+frac1{n^2}right)}{frac1n}right)=expleft(frac00right)\
&=lim_{nrightarrowinfty}expleft(frac{frac{-frac2{n^3}}{1+frac1{n^2}}}{-frac1{n^2}}right)=lim_{nrightarrowinfty}expleft(frac{frac2n}{1+frac1{n^2}}right)\
&=expleft(frac0{1+0}right)=e^0=1end{align}$$
Then
$$lim_{nrightarrowinfty}frac{n^{n+frac1n}}{left(n+frac1nright)^n}=lim_{nrightarrowinfty}frac{n^{frac1n}}{left(1+frac1{n^2}right)^n}=frac11=1ne0$$
So the sum diverges by the divergence test.
answered May 17 '16 at 23:37
user5713492user5713492
11.1k2919
$endgroup$
add a comment |
$begingroup$
The $n^{th}$ root test looks tempting because of all the $n^{th}$ powers, but it only works when the rate of increase or decrease of the terms of the series is exponential, which is not the case here. Let's work out a couple of limits instead:
$$lim_{nrightarrowinfty}n^{frac1n}=lim_{nrightarrowinfty}expleft(frac{ln n}nright)=expleft(frac 00right)=lim_{nrightarrowinfty}expleft(frac{frac1n}1right)=expleft(frac01right)=e^0=1$$
And
$$begin{align}lim_{nrightarrowinfty}left(1+frac1{n^2}right)^n&=lim_{nrightarrowinfty}expleft(nlnleft(1+frac1{n^2}right)right)\
&=lim_{nrightarrowinfty}expleft(frac{lnleft(1+frac1{n^2}right)}{frac1n}right)=expleft(frac00right)\
&=lim_{nrightarrowinfty}expleft(frac{frac{-frac2{n^3}}{1+frac1{n^2}}}{-frac1{n^2}}right)=lim_{nrightarrowinfty}expleft(frac{frac2n}{1+frac1{n^2}}right)\
&=expleft(frac0{1+0}right)=e^0=1end{align}$$
Then
$$lim_{nrightarrowinfty}frac{n^{n+frac1n}}{left(n+frac1nright)^n}=lim_{nrightarrowinfty}frac{n^{frac1n}}{left(1+frac1{n^2}right)^n}=frac11=1ne0$$
So the sum diverges by the divergence test.
answered May 17 '16 at 23:37
user5713492user5713492
11.1k2919
$endgroup$
The $n^{th}$ root test looks tempting because of all the $n^{th}$ powers, but it only works when the rate of increase or decrease of the terms of the series is exponential, which is not the case here. Let's work out a couple of limits instead:
$$lim_{nrightarrowinfty}n^{frac1n}=lim_{nrightarrowinfty}expleft(frac{ln n}nright)=expleft(frac 00right)=lim_{nrightarrowinfty}expleft(frac{frac1n}1right)=expleft(frac01right)=e^0=1$$
And
$$begin{align}lim_{nrightarrowinfty}left(1+frac1{n^2}right)^n&=lim_{nrightarrowinfty}expleft(nlnleft(1+frac1{n^2}right)right)\
&=lim_{nrightarrowinfty}expleft(frac{lnleft(1+frac1{n^2}right)}{frac1n}right)=expleft(frac00right)\
&=lim_{nrightarrowinfty}expleft(frac{frac{-frac2{n^3}}{1+frac1{n^2}}}{-frac1{n^2}}right)=lim_{nrightarrowinfty}expleft(frac{frac2n}{1+frac1{n^2}}right)\
&=expleft(frac0{1+0}right)=e^0=1end{align}$$
Then
$$lim_{nrightarrowinfty}frac{n^{n+frac1n}}{left(n+frac1nright)^n}=lim_{nrightarrowinfty}frac{n^{frac1n}}{left(1+frac1{n^2}right)^n}=frac11=1ne0$$
So the sum diverges by the divergence test.
answered May 17 '16 at 23:37
user5713492user5713492
11.1k2919
answered May 17 '16 at 23:37
user5713492user5713492
11.1k2919
answered May 17 '16 at 23:37
user5713492user5713492
11.1k2919
answered May 17 '16 at 23:37
user5713492user5713492
11.1k2919
11.1k2919
add a comment |
add a comment |
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