Mixing
Limit of product of infinite sequences and infinite series
$begingroup$
If the limit $$lim_{nto infty} n^p u_n =A tag{1}\ pgt 1;A lt infty$$ then I have to show that the limit of the series $sum_{n=0} ^{infty}u_n $ converges. I know that since $n^p$ blows up at infinity $u_n$ must somehow go to zero. I have tried the following:
$$lim_{nto infty} n^p u_n =A tag{2} = lim_{nto infty} n^p cdot lim_{nto infty} u_n \
therefore$$ $$lim_{nto infty}u_n = frac{lim_{nto infty}n^p}{A} tag{3}$$
From there I concluded that $u_n = frac{n^p}{A}$. Then applying the ratio test:
$$lim_{nto infty}frac{u_{n+1}}{u_n} = lim_{nto infty} {}frac{(n+1)^p}{n^p} lt 1 tag{4}$$
but obviously this doesn't work and is wrong. I am not convinced that $(2)$ is correct because I am not sure if the limits ($lim_{n to infty }n^p $ is $infty$ ???) exist. Also $(4)$ doesn't really work I guess because the limit is probably $=1$.
real-analysis sequences-and-series limits
asked Jan 25 at 12:08
daljit97daljit97
178111
$endgroup$
add a comment |
$begingroup$
If the limit $$lim_{nto infty} n^p u_n =A tag{1}\ pgt 1;A lt infty$$ then I have to show that the limit of the series $sum_{n=0} ^{infty}u_n $ converges. I know that since $n^p$ blows up at infinity $u_n$ must somehow go to zero. I have tried the following:
$$lim_{nto infty} n^p u_n =A tag{2} = lim_{nto infty} n^p cdot lim_{nto infty} u_n \
therefore$$ $$lim_{nto infty}u_n = frac{lim_{nto infty}n^p}{A} tag{3}$$
From there I concluded that $u_n = frac{n^p}{A}$. Then applying the ratio test:
$$lim_{nto infty}frac{u_{n+1}}{u_n} = lim_{nto infty} {}frac{(n+1)^p}{n^p} lt 1 tag{4}$$
but obviously this doesn't work and is wrong. I am not convinced that $(2)$ is correct because I am not sure if the limits ($lim_{n to infty }n^p $ is $infty$ ???) exist. Also $(4)$ doesn't really work I guess because the limit is probably $=1$.
real-analysis sequences-and-series limits
asked Jan 25 at 12:08
daljit97daljit97
178111
$endgroup$
add a comment |
$begingroup$
If the limit $$lim_{nto infty} n^p u_n =A tag{1}\ pgt 1;A lt infty$$ then I have to show that the limit of the series $sum_{n=0} ^{infty}u_n $ converges. I know that since $n^p$ blows up at infinity $u_n$ must somehow go to zero. I have tried the following:
$$lim_{nto infty} n^p u_n =A tag{2} = lim_{nto infty} n^p cdot lim_{nto infty} u_n \
therefore$$ $$lim_{nto infty}u_n = frac{lim_{nto infty}n^p}{A} tag{3}$$
From there I concluded that $u_n = frac{n^p}{A}$. Then applying the ratio test:
$$lim_{nto infty}frac{u_{n+1}}{u_n} = lim_{nto infty} {}frac{(n+1)^p}{n^p} lt 1 tag{4}$$
but obviously this doesn't work and is wrong. I am not convinced that $(2)$ is correct because I am not sure if the limits ($lim_{n to infty }n^p $ is $infty$ ???) exist. Also $(4)$ doesn't really work I guess because the limit is probably $=1$.
real-analysis sequences-and-series limits
asked Jan 25 at 12:08
daljit97daljit97
178111
$endgroup$
If the limit $$lim_{nto infty} n^p u_n =A tag{1}\ pgt 1;A lt infty$$ then I have to show that the limit of the series $sum_{n=0} ^{infty}u_n $ converges. I know that since $n^p$ blows up at infinity $u_n$ must somehow go to zero. I have tried the following:
$$lim_{nto infty} n^p u_n =A tag{2} = lim_{nto infty} n^p cdot lim_{nto infty} u_n \
therefore$$ $$lim_{nto infty}u_n = frac{lim_{nto infty}n^p}{A} tag{3}$$
From there I concluded that $u_n = frac{n^p}{A}$. Then applying the ratio test:
$$lim_{nto infty}frac{u_{n+1}}{u_n} = lim_{nto infty} {}frac{(n+1)^p}{n^p} lt 1 tag{4}$$
but obviously this doesn't work and is wrong. I am not convinced that $(2)$ is correct because I am not sure if the limits ($lim_{n to infty }n^p $ is $infty$ ???) exist. Also $(4)$ doesn't really work I guess because the limit is probably $=1$.
real-analysis sequences-and-series limits
real-analysis sequences-and-series limits
asked Jan 25 at 12:08
daljit97daljit97
178111
asked Jan 25 at 12:08
daljit97daljit97
178111
edited Jan 25 at 12:20
daljit97
asked Jan 25 at 12:08
daljit97daljit97
178111
asked Jan 25 at 12:08
daljit97daljit97
178111
asked Jan 25 at 12:08
daljit97daljit97
178111
178111
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
First of all, you are doing a lot of illegal stuff there. (2), as you already pointed out, does not work as $n^p$ is not convergent. Then you are essentially dividing by $0$ from (2) to (3) (among another elementary mistake) and after that, you conclude $u_n = frac{n^p}{A}$ from $limlimits_{nto infty}u_n = frac{limlimits_{nto infty}n^p}{A}$, which is obviously not correct, either.
Now for your actual question. Write $$sum_{n=0}^{infty}u_n = sum_{n=0}^{infty} frac{1}{n^p}(n^pu_n).$$
Can you figure it out now?
answered Jan 25 at 12:17
KlausKlaus
2,56713
$endgroup$
$begingroup$
I know that $sum_{n=0} ^ {infty} {frac{1}{n^p}}$ is convergent and I suppose that $sum_{n=0} ^ {infty} {n^p u_n}$ is also convergent?
$endgroup$
– daljit97
Jan 25 at 15:32
$begingroup$
No, $(n^p u_n)$ is just bounded, but that's sufficient.
$endgroup$
– Klaus
Jan 25 at 15:38
$begingroup$
Ok so I can say $sum_{n=0}^{infty}u_n lt sum_{n=0}^{infty} frac{1}{n^p}(M).$ for some $M in mathbb{R}$?
$endgroup$
– daljit97
Jan 25 at 15:43
$begingroup$
Better use an absolute value (i.e. $|u_n|$), but otherwise yes.
$endgroup$
– Klaus
Jan 25 at 15:45
$begingroup$
Ok since $sum_{n=0} ^ {infty} {frac{1}{n^p}}$ converges obviously $M sum_{n=0} ^ {infty} {frac{1}{n^p}}$ converges right?
$endgroup$
– daljit97
Jan 26 at 2:11
add a comment |
$begingroup$
As $lim_{nto infty} n^p u_n =A $ is finite, $(n^p u_n)$ is a bounded sequence (a sequence that converges is bounded). Let's say
$vert n^p u_n vert le B$ with $B >0$.
then you have $$vert u_n vert le frac{B}{n^p}$$ proving that $sum u_n$ is absolutely converging as for $p > 1$, $sum 1/n^p$ converges.
answered Jan 25 at 12:17
mathcounterexamples.netmathcounterexamples.net
27k22158
$endgroup$
add a comment |
$begingroup$
$(2)$ and $(3)$ are meaningless. $(4)$ is actually right (except that the limit is equal to $1$), but your derivation isn’t.
So, here, the ratio test is useless. The actual useful assumption is that $n^pu_n$ is convergent. Therefore it is bounded: so $|u_n| leq Mn^{-p}$ for each $n$. Can you take it from there?
answered Jan 25 at 12:21
MindlackMindlack
4,920211
$endgroup$
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
$begingroup$
First of all, you are doing a lot of illegal stuff there. (2), as you already pointed out, does not work as $n^p$ is not convergent. Then you are essentially dividing by $0$ from (2) to (3) (among another elementary mistake) and after that, you conclude $u_n = frac{n^p}{A}$ from $limlimits_{nto infty}u_n = frac{limlimits_{nto infty}n^p}{A}$, which is obviously not correct, either.
Now for your actual question. Write $$sum_{n=0}^{infty}u_n = sum_{n=0}^{infty} frac{1}{n^p}(n^pu_n).$$
Can you figure it out now?
answered Jan 25 at 12:17
KlausKlaus
2,56713
$endgroup$
$begingroup$
I know that $sum_{n=0} ^ {infty} {frac{1}{n^p}}$ is convergent and I suppose that $sum_{n=0} ^ {infty} {n^p u_n}$ is also convergent?
$endgroup$
– daljit97
Jan 25 at 15:32
$begingroup$
No, $(n^p u_n)$ is just bounded, but that's sufficient.
$endgroup$
– Klaus
Jan 25 at 15:38
$begingroup$
Ok so I can say $sum_{n=0}^{infty}u_n lt sum_{n=0}^{infty} frac{1}{n^p}(M).$ for some $M in mathbb{R}$?
$endgroup$
– daljit97
Jan 25 at 15:43
$begingroup$
Better use an absolute value (i.e. $|u_n|$), but otherwise yes.
$endgroup$
– Klaus
Jan 25 at 15:45
$begingroup$
Ok since $sum_{n=0} ^ {infty} {frac{1}{n^p}}$ converges obviously $M sum_{n=0} ^ {infty} {frac{1}{n^p}}$ converges right?
$endgroup$
– daljit97
Jan 26 at 2:11
add a comment |
$begingroup$
First of all, you are doing a lot of illegal stuff there. (2), as you already pointed out, does not work as $n^p$ is not convergent. Then you are essentially dividing by $0$ from (2) to (3) (among another elementary mistake) and after that, you conclude $u_n = frac{n^p}{A}$ from $limlimits_{nto infty}u_n = frac{limlimits_{nto infty}n^p}{A}$, which is obviously not correct, either.
Now for your actual question. Write $$sum_{n=0}^{infty}u_n = sum_{n=0}^{infty} frac{1}{n^p}(n^pu_n).$$
Can you figure it out now?
answered Jan 25 at 12:17
KlausKlaus
2,56713
$endgroup$
$begingroup$
I know that $sum_{n=0} ^ {infty} {frac{1}{n^p}}$ is convergent and I suppose that $sum_{n=0} ^ {infty} {n^p u_n}$ is also convergent?
$endgroup$
– daljit97
Jan 25 at 15:32
$begingroup$
No, $(n^p u_n)$ is just bounded, but that's sufficient.
$endgroup$
– Klaus
Jan 25 at 15:38
$begingroup$
Ok so I can say $sum_{n=0}^{infty}u_n lt sum_{n=0}^{infty} frac{1}{n^p}(M).$ for some $M in mathbb{R}$?
$endgroup$
– daljit97
Jan 25 at 15:43
$begingroup$
Better use an absolute value (i.e. $|u_n|$), but otherwise yes.
$endgroup$
– Klaus
Jan 25 at 15:45
$begingroup$
Ok since $sum_{n=0} ^ {infty} {frac{1}{n^p}}$ converges obviously $M sum_{n=0} ^ {infty} {frac{1}{n^p}}$ converges right?
$endgroup$
– daljit97
Jan 26 at 2:11
add a comment |
$begingroup$
First of all, you are doing a lot of illegal stuff there. (2), as you already pointed out, does not work as $n^p$ is not convergent. Then you are essentially dividing by $0$ from (2) to (3) (among another elementary mistake) and after that, you conclude $u_n = frac{n^p}{A}$ from $limlimits_{nto infty}u_n = frac{limlimits_{nto infty}n^p}{A}$, which is obviously not correct, either.
Now for your actual question. Write $$sum_{n=0}^{infty}u_n = sum_{n=0}^{infty} frac{1}{n^p}(n^pu_n).$$
Can you figure it out now?
answered Jan 25 at 12:17
KlausKlaus
2,56713
$endgroup$
First of all, you are doing a lot of illegal stuff there. (2), as you already pointed out, does not work as $n^p$ is not convergent. Then you are essentially dividing by $0$ from (2) to (3) (among another elementary mistake) and after that, you conclude $u_n = frac{n^p}{A}$ from $limlimits_{nto infty}u_n = frac{limlimits_{nto infty}n^p}{A}$, which is obviously not correct, either.
Now for your actual question. Write $$sum_{n=0}^{infty}u_n = sum_{n=0}^{infty} frac{1}{n^p}(n^pu_n).$$
Can you figure it out now?
answered Jan 25 at 12:17
KlausKlaus
2,56713
edited Jan 25 at 12:23
answered Jan 25 at 12:17
KlausKlaus
2,56713
answered Jan 25 at 12:17
KlausKlaus
2,56713
answered Jan 25 at 12:17
KlausKlaus
2,56713
2,56713
$begingroup$
I know that $sum_{n=0} ^ {infty} {frac{1}{n^p}}$ is convergent and I suppose that $sum_{n=0} ^ {infty} {n^p u_n}$ is also convergent?
$endgroup$
– daljit97
Jan 25 at 15:32
$begingroup$
No, $(n^p u_n)$ is just bounded, but that's sufficient.
$endgroup$
– Klaus
Jan 25 at 15:38
$begingroup$
Ok so I can say $sum_{n=0}^{infty}u_n lt sum_{n=0}^{infty} frac{1}{n^p}(M).$ for some $M in mathbb{R}$?
$endgroup$
– daljit97
Jan 25 at 15:43
$begingroup$
Better use an absolute value (i.e. $|u_n|$), but otherwise yes.
$endgroup$
– Klaus
Jan 25 at 15:45
$begingroup$
Ok since $sum_{n=0} ^ {infty} {frac{1}{n^p}}$ converges obviously $M sum_{n=0} ^ {infty} {frac{1}{n^p}}$ converges right?
$endgroup$
– daljit97
Jan 26 at 2:11
add a comment |
$begingroup$
I know that $sum_{n=0} ^ {infty} {frac{1}{n^p}}$ is convergent and I suppose that $sum_{n=0} ^ {infty} {n^p u_n}$ is also convergent?
$endgroup$
– daljit97
Jan 25 at 15:32
$begingroup$
No, $(n^p u_n)$ is just bounded, but that's sufficient.
$endgroup$
– Klaus
Jan 25 at 15:38
$begingroup$
Ok so I can say $sum_{n=0}^{infty}u_n lt sum_{n=0}^{infty} frac{1}{n^p}(M).$ for some $M in mathbb{R}$?
$endgroup$
– daljit97
Jan 25 at 15:43
$begingroup$
Better use an absolute value (i.e. $|u_n|$), but otherwise yes.
$endgroup$
– Klaus
Jan 25 at 15:45
$begingroup$
Ok since $sum_{n=0} ^ {infty} {frac{1}{n^p}}$ converges obviously $M sum_{n=0} ^ {infty} {frac{1}{n^p}}$ converges right?
$endgroup$
– daljit97
Jan 26 at 2:11
$begingroup$
I know that $sum_{n=0} ^ {infty} {frac{1}{n^p}}$ is convergent and I suppose that $sum_{n=0} ^ {infty} {n^p u_n}$ is also convergent?
$endgroup$
– daljit97
Jan 25 at 15:32
$begingroup$
I know that $sum_{n=0} ^ {infty} {frac{1}{n^p}}$ is convergent and I suppose that $sum_{n=0} ^ {infty} {n^p u_n}$ is also convergent?
$endgroup$
– daljit97
Jan 25 at 15:32
$begingroup$
No, $(n^p u_n)$ is just bounded, but that's sufficient.
$endgroup$
– Klaus
Jan 25 at 15:38
$begingroup$
No, $(n^p u_n)$ is just bounded, but that's sufficient.
$endgroup$
– Klaus
Jan 25 at 15:38
$begingroup$
Ok so I can say $sum_{n=0}^{infty}u_n lt sum_{n=0}^{infty} frac{1}{n^p}(M).$ for some $M in mathbb{R}$?
$endgroup$
– daljit97
Jan 25 at 15:43
$begingroup$
Ok so I can say $sum_{n=0}^{infty}u_n lt sum_{n=0}^{infty} frac{1}{n^p}(M).$ for some $M in mathbb{R}$?
$endgroup$
– daljit97
Jan 25 at 15:43
$begingroup$
Better use an absolute value (i.e. $|u_n|$), but otherwise yes.
$endgroup$
– Klaus
Jan 25 at 15:45
$begingroup$
Better use an absolute value (i.e. $|u_n|$), but otherwise yes.
$endgroup$
– Klaus
Jan 25 at 15:45
$begingroup$
Ok since $sum_{n=0} ^ {infty} {frac{1}{n^p}}$ converges obviously $M sum_{n=0} ^ {infty} {frac{1}{n^p}}$ converges right?
$endgroup$
– daljit97
Jan 26 at 2:11
$begingroup$
Ok since $sum_{n=0} ^ {infty} {frac{1}{n^p}}$ converges obviously $M sum_{n=0} ^ {infty} {frac{1}{n^p}}$ converges right?
$endgroup$
– daljit97
Jan 26 at 2:11
add a comment |
$begingroup$
As $lim_{nto infty} n^p u_n =A $ is finite, $(n^p u_n)$ is a bounded sequence (a sequence that converges is bounded). Let's say
$vert n^p u_n vert le B$ with $B >0$.
then you have $$vert u_n vert le frac{B}{n^p}$$ proving that $sum u_n$ is absolutely converging as for $p > 1$, $sum 1/n^p$ converges.
answered Jan 25 at 12:17
mathcounterexamples.netmathcounterexamples.net
27k22158
$endgroup$
add a comment |
$begingroup$
As $lim_{nto infty} n^p u_n =A $ is finite, $(n^p u_n)$ is a bounded sequence (a sequence that converges is bounded). Let's say
$vert n^p u_n vert le B$ with $B >0$.
then you have $$vert u_n vert le frac{B}{n^p}$$ proving that $sum u_n$ is absolutely converging as for $p > 1$, $sum 1/n^p$ converges.
answered Jan 25 at 12:17
mathcounterexamples.netmathcounterexamples.net
27k22158
$endgroup$
add a comment |
$begingroup$
As $lim_{nto infty} n^p u_n =A $ is finite, $(n^p u_n)$ is a bounded sequence (a sequence that converges is bounded). Let's say
$vert n^p u_n vert le B$ with $B >0$.
then you have $$vert u_n vert le frac{B}{n^p}$$ proving that $sum u_n$ is absolutely converging as for $p > 1$, $sum 1/n^p$ converges.
answered Jan 25 at 12:17
mathcounterexamples.netmathcounterexamples.net
27k22158
$endgroup$
As $lim_{nto infty} n^p u_n =A $ is finite, $(n^p u_n)$ is a bounded sequence (a sequence that converges is bounded). Let's say
$vert n^p u_n vert le B$ with $B >0$.
then you have $$vert u_n vert le frac{B}{n^p}$$ proving that $sum u_n$ is absolutely converging as for $p > 1$, $sum 1/n^p$ converges.
answered Jan 25 at 12:17
mathcounterexamples.netmathcounterexamples.net
27k22158
answered Jan 25 at 12:17
mathcounterexamples.netmathcounterexamples.net
27k22158
answered Jan 25 at 12:17
mathcounterexamples.netmathcounterexamples.net
27k22158
answered Jan 25 at 12:17
mathcounterexamples.netmathcounterexamples.net
27k22158
27k22158
add a comment |
add a comment |
$begingroup$
$(2)$ and $(3)$ are meaningless. $(4)$ is actually right (except that the limit is equal to $1$), but your derivation isn’t.
So, here, the ratio test is useless. The actual useful assumption is that $n^pu_n$ is convergent. Therefore it is bounded: so $|u_n| leq Mn^{-p}$ for each $n$. Can you take it from there?
answered Jan 25 at 12:21
MindlackMindlack
4,920211
$endgroup$
add a comment |
$begingroup$
$(2)$ and $(3)$ are meaningless. $(4)$ is actually right (except that the limit is equal to $1$), but your derivation isn’t.
So, here, the ratio test is useless. The actual useful assumption is that $n^pu_n$ is convergent. Therefore it is bounded: so $|u_n| leq Mn^{-p}$ for each $n$. Can you take it from there?
answered Jan 25 at 12:21
MindlackMindlack
4,920211
$endgroup$
add a comment |
$begingroup$
$(2)$ and $(3)$ are meaningless. $(4)$ is actually right (except that the limit is equal to $1$), but your derivation isn’t.
So, here, the ratio test is useless. The actual useful assumption is that $n^pu_n$ is convergent. Therefore it is bounded: so $|u_n| leq Mn^{-p}$ for each $n$. Can you take it from there?
answered Jan 25 at 12:21
MindlackMindlack
4,920211
$endgroup$
$(2)$ and $(3)$ are meaningless. $(4)$ is actually right (except that the limit is equal to $1$), but your derivation isn’t.
So, here, the ratio test is useless. The actual useful assumption is that $n^pu_n$ is convergent. Therefore it is bounded: so $|u_n| leq Mn^{-p}$ for each $n$. Can you take it from there?
answered Jan 25 at 12:21
MindlackMindlack
4,920211
answered Jan 25 at 12:21
MindlackMindlack
4,920211
answered Jan 25 at 12:21
MindlackMindlack
4,920211
answered Jan 25 at 12:21
MindlackMindlack
4,920211
4,920211
add a comment |
add a comment |
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