Methods to show the convergence/divergence of following series:
$begingroup$
This is the series I have a question about: $sum_{n=1}^{infty} frac{(ln n)^2 }{2^{ln n}}$.
The root and ratio test are both inconclusive, so I tried Raabe's test of convergence. It seems too difficult to find the limit because the term $frac{x_n}{x_{n+1}}$ becomes huge (also when applying de l'Hopital's rule). Could someone help me please? Are there any rules/methods on how to approach proving the convergence of series? How can they be applied on this given series?
real-analysis
asked Jan 17 at 13:17
ZacharyZachary
3229
$endgroup$
add a comment |
$begingroup$
This is the series I have a question about: $sum_{n=1}^{infty} frac{(ln n)^2 }{2^{ln n}}$.
The root and ratio test are both inconclusive, so I tried Raabe's test of convergence. It seems too difficult to find the limit because the term $frac{x_n}{x_{n+1}}$ becomes huge (also when applying de l'Hopital's rule). Could someone help me please? Are there any rules/methods on how to approach proving the convergence of series? How can they be applied on this given series?
real-analysis
asked Jan 17 at 13:17
ZacharyZachary
3229
$endgroup$
$begingroup$
Use Cauchy-Condensation test...
$endgroup$
– RAM_3R
Jan 17 at 13:50
add a comment |
$begingroup$
This is the series I have a question about: $sum_{n=1}^{infty} frac{(ln n)^2 }{2^{ln n}}$.
The root and ratio test are both inconclusive, so I tried Raabe's test of convergence. It seems too difficult to find the limit because the term $frac{x_n}{x_{n+1}}$ becomes huge (also when applying de l'Hopital's rule). Could someone help me please? Are there any rules/methods on how to approach proving the convergence of series? How can they be applied on this given series?
real-analysis
asked Jan 17 at 13:17
ZacharyZachary
3229
$endgroup$
This is the series I have a question about: $sum_{n=1}^{infty} frac{(ln n)^2 }{2^{ln n}}$.
The root and ratio test are both inconclusive, so I tried Raabe's test of convergence. It seems too difficult to find the limit because the term $frac{x_n}{x_{n+1}}$ becomes huge (also when applying de l'Hopital's rule). Could someone help me please? Are there any rules/methods on how to approach proving the convergence of series? How can they be applied on this given series?
real-analysis
real-analysis
asked Jan 17 at 13:17
ZacharyZachary
3229
asked Jan 17 at 13:17
ZacharyZachary
3229
asked Jan 17 at 13:17
ZacharyZachary
3229
asked Jan 17 at 13:17
ZacharyZachary
3229
asked Jan 17 at 13:17
ZacharyZachary
3229
3229
$begingroup$
Use Cauchy-Condensation test...
$endgroup$
– RAM_3R
Jan 17 at 13:50
add a comment |
$begingroup$
Use Cauchy-Condensation test...
$endgroup$
– RAM_3R
Jan 17 at 13:50
$begingroup$
Use Cauchy-Condensation test...
$endgroup$
– RAM_3R
Jan 17 at 13:50
$begingroup$
Use Cauchy-Condensation test...
$endgroup$
– RAM_3R
Jan 17 at 13:50
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
$a^{ln n} = e^{ln a ln n} =n^{ln a}$
so the sum
$sum (ln n)^b/a^{ln n}$
converges if
$ln a > 1$,
diverges if
$ln a < 1$,
and depends on the value of $b$
if $ln a=1$.
In this case,
$ln a =ln 2 < 1$
so the sum diverges.
answered Jan 17 at 13:39
marty cohenmarty cohen
75.5k549130
$endgroup$
add a comment |
$begingroup$
Let $a_n=sum(ln n)^2/2^{ln n}=sum(ln n)^2/n^{ln 2}$
Now $s_n=sum frac{2^n .(ln 2^n)^2}{2^{nln 2}}=sum 2^{n(1-ln 2)}n^2(ln 2)^2$
Since $1-ln 2>0$, So $s_n$ diverges $Rightarrow$ $a_n$ diverges
answered Jan 17 at 13:57
RAM_3RRAM_3R
602216
$endgroup$
add a comment |
$begingroup$
There is this useful Cauchy condensation test, implying that $sum_{n=1}^{infty}a_n$ is convergent iff $sum_{n=1}^{infty}2^na_{2^n}$ is convergent, usable if $a_n$ is nonincreasing sequence of nonnegative numbers. Since $2^{ln n}$ is faster than $(ln$ $n)^2$ one can assume that this sequence is nonincreasing ( as it is nonincreasing for $n$ large enough ), and of course $a_n$ are nonnegative, so one can use this test. It follows that our sum converges iff sum $sum_{n=1}^{infty}2^n(ln(2^n))^2/2^{ln(2^n)}$ = $sum_{n=1}^{infty}2^n(nln2)^2/2^{nln2}$ = $sum_{n=1}^{infty}n^2(ln2)^2/2^{n(ln2 - 1)}$ = $sum_{n=1}^{infty}n^2(ln2)^22^{n(1-ln2)}$ converges, but this one obviously diverges, as $ln2$ $< 1$.
answered Jan 17 at 16:42
Jakub AndruszkiewiczJakub Andruszkiewicz
2116
$endgroup$
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
$begingroup$
$a^{ln n} = e^{ln a ln n} =n^{ln a}$
so the sum
$sum (ln n)^b/a^{ln n}$
converges if
$ln a > 1$,
diverges if
$ln a < 1$,
and depends on the value of $b$
if $ln a=1$.
In this case,
$ln a =ln 2 < 1$
so the sum diverges.
answered Jan 17 at 13:39
marty cohenmarty cohen
75.5k549130
$endgroup$
add a comment |
$begingroup$
$a^{ln n} = e^{ln a ln n} =n^{ln a}$
so the sum
$sum (ln n)^b/a^{ln n}$
converges if
$ln a > 1$,
diverges if
$ln a < 1$,
and depends on the value of $b$
if $ln a=1$.
In this case,
$ln a =ln 2 < 1$
so the sum diverges.
answered Jan 17 at 13:39
marty cohenmarty cohen
75.5k549130
$endgroup$
add a comment |
$begingroup$
$a^{ln n} = e^{ln a ln n} =n^{ln a}$
so the sum
$sum (ln n)^b/a^{ln n}$
converges if
$ln a > 1$,
diverges if
$ln a < 1$,
and depends on the value of $b$
if $ln a=1$.
In this case,
$ln a =ln 2 < 1$
so the sum diverges.
answered Jan 17 at 13:39
marty cohenmarty cohen
75.5k549130
$endgroup$
$a^{ln n} = e^{ln a ln n} =n^{ln a}$
so the sum
$sum (ln n)^b/a^{ln n}$
converges if
$ln a > 1$,
diverges if
$ln a < 1$,
and depends on the value of $b$
if $ln a=1$.
In this case,
$ln a =ln 2 < 1$
so the sum diverges.
answered Jan 17 at 13:39
marty cohenmarty cohen
75.5k549130
answered Jan 17 at 13:39
marty cohenmarty cohen
75.5k549130
answered Jan 17 at 13:39
marty cohenmarty cohen
75.5k549130
answered Jan 17 at 13:39
marty cohenmarty cohen
75.5k549130
75.5k549130
add a comment |
add a comment |
$begingroup$
Let $a_n=sum(ln n)^2/2^{ln n}=sum(ln n)^2/n^{ln 2}$
Now $s_n=sum frac{2^n .(ln 2^n)^2}{2^{nln 2}}=sum 2^{n(1-ln 2)}n^2(ln 2)^2$
Since $1-ln 2>0$, So $s_n$ diverges $Rightarrow$ $a_n$ diverges
answered Jan 17 at 13:57
RAM_3RRAM_3R
602216
$endgroup$
add a comment |
$begingroup$
Let $a_n=sum(ln n)^2/2^{ln n}=sum(ln n)^2/n^{ln 2}$
Now $s_n=sum frac{2^n .(ln 2^n)^2}{2^{nln 2}}=sum 2^{n(1-ln 2)}n^2(ln 2)^2$
Since $1-ln 2>0$, So $s_n$ diverges $Rightarrow$ $a_n$ diverges
answered Jan 17 at 13:57
RAM_3RRAM_3R
602216
$endgroup$
add a comment |
$begingroup$
Let $a_n=sum(ln n)^2/2^{ln n}=sum(ln n)^2/n^{ln 2}$
Now $s_n=sum frac{2^n .(ln 2^n)^2}{2^{nln 2}}=sum 2^{n(1-ln 2)}n^2(ln 2)^2$
Since $1-ln 2>0$, So $s_n$ diverges $Rightarrow$ $a_n$ diverges
answered Jan 17 at 13:57
RAM_3RRAM_3R
602216
$endgroup$
Let $a_n=sum(ln n)^2/2^{ln n}=sum(ln n)^2/n^{ln 2}$
Now $s_n=sum frac{2^n .(ln 2^n)^2}{2^{nln 2}}=sum 2^{n(1-ln 2)}n^2(ln 2)^2$
Since $1-ln 2>0$, So $s_n$ diverges $Rightarrow$ $a_n$ diverges
answered Jan 17 at 13:57
RAM_3RRAM_3R
602216
edited Jan 17 at 16:17
answered Jan 17 at 13:57
RAM_3RRAM_3R
602216
answered Jan 17 at 13:57
RAM_3RRAM_3R
602216
answered Jan 17 at 13:57
RAM_3RRAM_3R
602216
602216
add a comment |
add a comment |
$begingroup$
There is this useful Cauchy condensation test, implying that $sum_{n=1}^{infty}a_n$ is convergent iff $sum_{n=1}^{infty}2^na_{2^n}$ is convergent, usable if $a_n$ is nonincreasing sequence of nonnegative numbers. Since $2^{ln n}$ is faster than $(ln$ $n)^2$ one can assume that this sequence is nonincreasing ( as it is nonincreasing for $n$ large enough ), and of course $a_n$ are nonnegative, so one can use this test. It follows that our sum converges iff sum $sum_{n=1}^{infty}2^n(ln(2^n))^2/2^{ln(2^n)}$ = $sum_{n=1}^{infty}2^n(nln2)^2/2^{nln2}$ = $sum_{n=1}^{infty}n^2(ln2)^2/2^{n(ln2 - 1)}$ = $sum_{n=1}^{infty}n^2(ln2)^22^{n(1-ln2)}$ converges, but this one obviously diverges, as $ln2$ $< 1$.
answered Jan 17 at 16:42
Jakub AndruszkiewiczJakub Andruszkiewicz
2116
$endgroup$
add a comment |
$begingroup$
There is this useful Cauchy condensation test, implying that $sum_{n=1}^{infty}a_n$ is convergent iff $sum_{n=1}^{infty}2^na_{2^n}$ is convergent, usable if $a_n$ is nonincreasing sequence of nonnegative numbers. Since $2^{ln n}$ is faster than $(ln$ $n)^2$ one can assume that this sequence is nonincreasing ( as it is nonincreasing for $n$ large enough ), and of course $a_n$ are nonnegative, so one can use this test. It follows that our sum converges iff sum $sum_{n=1}^{infty}2^n(ln(2^n))^2/2^{ln(2^n)}$ = $sum_{n=1}^{infty}2^n(nln2)^2/2^{nln2}$ = $sum_{n=1}^{infty}n^2(ln2)^2/2^{n(ln2 - 1)}$ = $sum_{n=1}^{infty}n^2(ln2)^22^{n(1-ln2)}$ converges, but this one obviously diverges, as $ln2$ $< 1$.
answered Jan 17 at 16:42
Jakub AndruszkiewiczJakub Andruszkiewicz
2116
$endgroup$
add a comment |
$begingroup$
There is this useful Cauchy condensation test, implying that $sum_{n=1}^{infty}a_n$ is convergent iff $sum_{n=1}^{infty}2^na_{2^n}$ is convergent, usable if $a_n$ is nonincreasing sequence of nonnegative numbers. Since $2^{ln n}$ is faster than $(ln$ $n)^2$ one can assume that this sequence is nonincreasing ( as it is nonincreasing for $n$ large enough ), and of course $a_n$ are nonnegative, so one can use this test. It follows that our sum converges iff sum $sum_{n=1}^{infty}2^n(ln(2^n))^2/2^{ln(2^n)}$ = $sum_{n=1}^{infty}2^n(nln2)^2/2^{nln2}$ = $sum_{n=1}^{infty}n^2(ln2)^2/2^{n(ln2 - 1)}$ = $sum_{n=1}^{infty}n^2(ln2)^22^{n(1-ln2)}$ converges, but this one obviously diverges, as $ln2$ $< 1$.
answered Jan 17 at 16:42
Jakub AndruszkiewiczJakub Andruszkiewicz
2116
$endgroup$
There is this useful Cauchy condensation test, implying that $sum_{n=1}^{infty}a_n$ is convergent iff $sum_{n=1}^{infty}2^na_{2^n}$ is convergent, usable if $a_n$ is nonincreasing sequence of nonnegative numbers. Since $2^{ln n}$ is faster than $(ln$ $n)^2$ one can assume that this sequence is nonincreasing ( as it is nonincreasing for $n$ large enough ), and of course $a_n$ are nonnegative, so one can use this test. It follows that our sum converges iff sum $sum_{n=1}^{infty}2^n(ln(2^n))^2/2^{ln(2^n)}$ = $sum_{n=1}^{infty}2^n(nln2)^2/2^{nln2}$ = $sum_{n=1}^{infty}n^2(ln2)^2/2^{n(ln2 - 1)}$ = $sum_{n=1}^{infty}n^2(ln2)^22^{n(1-ln2)}$ converges, but this one obviously diverges, as $ln2$ $< 1$.
answered Jan 17 at 16:42
Jakub AndruszkiewiczJakub Andruszkiewicz
2116
answered Jan 17 at 16:42
Jakub AndruszkiewiczJakub Andruszkiewicz
2116
answered Jan 17 at 16:42
Jakub AndruszkiewiczJakub Andruszkiewicz
2116
answered Jan 17 at 16:42
Jakub AndruszkiewiczJakub Andruszkiewicz
2116
2116
add a comment |
add a comment |
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$begingroup$
Use Cauchy-Condensation test...
$endgroup$
– RAM_3R
Jan 17 at 13:50