Mixing
Interchanging a limit and an infinite alternate series
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I am having troubles to explain if the following equality holds or not
$$lim_ktoinftysum_n=1^infty(-1)^n(n+k)^-1=sum_n=1^infty(-1)^nlim_ktoinfty(n+k)^-1=0.$$
As far as I see, I can't apply the dominated convergence theorem since $|f_k(n)|=|(-1)^n(n+k)^-1|=(n+k)^-1$ can't be dominated by summable sequence over $n$. How could I proceed?
sequences-and-series convergence lebesgue-measure
asked Sep 1 at 16:59
Dubglass
978
add a comment |Â
up vote
3
down vote
favorite
I am having troubles to explain if the following equality holds or not
$$lim_ktoinftysum_n=1^infty(-1)^n(n+k)^-1=sum_n=1^infty(-1)^nlim_ktoinfty(n+k)^-1=0.$$
As far as I see, I can't apply the dominated convergence theorem since $|f_k(n)|=|(-1)^n(n+k)^-1|=(n+k)^-1$ can't be dominated by summable sequence over $n$. How could I proceed?
sequences-and-series convergence lebesgue-measure
asked Sep 1 at 16:59
Dubglass
978
The first of your equalities comes unjustified. But with a conditional convergence, you can easily apply whatever-you-want to, say, pairwise summation.
– metamorphy
Sep 1 at 17:04
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I am having troubles to explain if the following equality holds or not
$$lim_ktoinftysum_n=1^infty(-1)^n(n+k)^-1=sum_n=1^infty(-1)^nlim_ktoinfty(n+k)^-1=0.$$
As far as I see, I can't apply the dominated convergence theorem since $|f_k(n)|=|(-1)^n(n+k)^-1|=(n+k)^-1$ can't be dominated by summable sequence over $n$. How could I proceed?
sequences-and-series convergence lebesgue-measure
asked Sep 1 at 16:59
Dubglass
978
I am having troubles to explain if the following equality holds or not
$$lim_ktoinftysum_n=1^infty(-1)^n(n+k)^-1=sum_n=1^infty(-1)^nlim_ktoinfty(n+k)^-1=0.$$
As far as I see, I can't apply the dominated convergence theorem since $|f_k(n)|=|(-1)^n(n+k)^-1|=(n+k)^-1$ can't be dominated by summable sequence over $n$. How could I proceed?
sequences-and-series convergence lebesgue-measure
asked Sep 1 at 16:59
Dubglass
978
asked Sep 1 at 16:59
Dubglass
978
asked Sep 1 at 16:59
Dubglass
978
asked Sep 1 at 16:59
Dubglass
978
978
The first of your equalities comes unjustified. But with a conditional convergence, you can easily apply whatever-you-want to, say, pairwise summation.
– metamorphy
Sep 1 at 17:04
add a comment |Â
The first of your equalities comes unjustified. But with a conditional convergence, you can easily apply whatever-you-want to, say, pairwise summation.
– metamorphy
Sep 1 at 17:04
The first of your equalities comes unjustified. But with a conditional convergence, you can easily apply whatever-you-want to, say, pairwise summation.
– metamorphy
Sep 1 at 17:04
The first of your equalities comes unjustified. But with a conditional convergence, you can easily apply whatever-you-want to, say, pairwise summation.
– metamorphy
Sep 1 at 17:04
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
5
down vote
accepted
HINT:
Note that we can write
$$sum_n=1^infty frac(-1)^nn+k=sum_n=1^inftyleft(frac12n+k-frac12n-1+kright)$$
Can you proceed now?
answered Sep 1 at 17:07
Mark Viola
126k1172167
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up vote
4
down vote
Or:
$$ sum_ngeq 1frac(-1)^nn+k = sum_ngeq 1int_0^+infty(-1)^n e^-nx e^-kx,dx =int_0^+inftyfracdx(1+e^x)e^kxleq int_0^+inftyfracdx2e^kx=frac12k.$$
answered Sep 1 at 18:40
Jack D'Aurizio♦
273k32268637
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
HINT:
Note that we can write
$$sum_n=1^infty frac(-1)^nn+k=sum_n=1^inftyleft(frac12n+k-frac12n-1+kright)$$
Can you proceed now?
answered Sep 1 at 17:07
Mark Viola
126k1172167
add a comment |Â
up vote
5
down vote
accepted
HINT:
Note that we can write
$$sum_n=1^infty frac(-1)^nn+k=sum_n=1^inftyleft(frac12n+k-frac12n-1+kright)$$
Can you proceed now?
answered Sep 1 at 17:07
Mark Viola
126k1172167
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
HINT:
Note that we can write
$$sum_n=1^infty frac(-1)^nn+k=sum_n=1^inftyleft(frac12n+k-frac12n-1+kright)$$
Can you proceed now?
answered Sep 1 at 17:07
Mark Viola
126k1172167
HINT:
Note that we can write
$$sum_n=1^infty frac(-1)^nn+k=sum_n=1^inftyleft(frac12n+k-frac12n-1+kright)$$
Can you proceed now?
answered Sep 1 at 17:07
Mark Viola
126k1172167
answered Sep 1 at 17:07
Mark Viola
126k1172167
answered Sep 1 at 17:07
Mark Viola
126k1172167
answered Sep 1 at 17:07
Mark Viola
126k1172167
126k1172167
add a comment |Â
add a comment |Â
up vote
4
down vote
Or:
$$ sum_ngeq 1frac(-1)^nn+k = sum_ngeq 1int_0^+infty(-1)^n e^-nx e^-kx,dx =int_0^+inftyfracdx(1+e^x)e^kxleq int_0^+inftyfracdx2e^kx=frac12k.$$
answered Sep 1 at 18:40
Jack D'Aurizio♦
273k32268637
add a comment |Â
up vote
4
down vote
Or:
$$ sum_ngeq 1frac(-1)^nn+k = sum_ngeq 1int_0^+infty(-1)^n e^-nx e^-kx,dx =int_0^+inftyfracdx(1+e^x)e^kxleq int_0^+inftyfracdx2e^kx=frac12k.$$
answered Sep 1 at 18:40
Jack D'Aurizio♦
273k32268637
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Or:
$$ sum_ngeq 1frac(-1)^nn+k = sum_ngeq 1int_0^+infty(-1)^n e^-nx e^-kx,dx =int_0^+inftyfracdx(1+e^x)e^kxleq int_0^+inftyfracdx2e^kx=frac12k.$$
answered Sep 1 at 18:40
Jack D'Aurizio♦
273k32268637
Or:
$$ sum_ngeq 1frac(-1)^nn+k = sum_ngeq 1int_0^+infty(-1)^n e^-nx e^-kx,dx =int_0^+inftyfracdx(1+e^x)e^kxleq int_0^+inftyfracdx2e^kx=frac12k.$$
answered Sep 1 at 18:40
Jack D'Aurizio♦
273k32268637
answered Sep 1 at 18:40
Jack D'Aurizio♦
273k32268637
answered Sep 1 at 18:40
Jack D'Aurizio♦
273k32268637
answered Sep 1 at 18:40
Jack D'Aurizio♦
273k32268637
273k32268637
add a comment |Â
add a comment |Â
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The first of your equalities comes unjustified. But with a conditional convergence, you can easily apply whatever-you-want to, say, pairwise summation.
– metamorphy
Sep 1 at 17:04