$sum_{n=1}^infty a_n^{b_n}$ converges
Show that if $sum_{n=1}^infty a_n$ converges and $a_ngeq0$ and $b_ngeq 1$ for each $n$ then $sum_{n=1}^infty a_n^{b_n}$ converges.
I know I need to use absolutely convergent but I am unsure how to do it.
real-analysis
asked Nov 29 '18 at 1:30
Vicky
412
add a comment |
Show that if $sum_{n=1}^infty a_n$ converges and $a_ngeq0$ and $b_ngeq 1$ for each $n$ then $sum_{n=1}^infty a_n^{b_n}$ converges.
I know I need to use absolutely convergent but I am unsure how to do it.
real-analysis
asked Nov 29 '18 at 1:30
Vicky
412
4
You could start by noting that eventually $a_n lt 1$ (why?), and use a comparison test.
– Tob Ernack
Nov 29 '18 at 1:32
@TobErnack You beat me to it! Who cares that it was by 7 minutes... I coulda been a contender!
– JDMan4444
Nov 29 '18 at 1:40
add a comment |
Show that if $sum_{n=1}^infty a_n$ converges and $a_ngeq0$ and $b_ngeq 1$ for each $n$ then $sum_{n=1}^infty a_n^{b_n}$ converges.
I know I need to use absolutely convergent but I am unsure how to do it.
real-analysis
asked Nov 29 '18 at 1:30
Vicky
412
Show that if $sum_{n=1}^infty a_n$ converges and $a_ngeq0$ and $b_ngeq 1$ for each $n$ then $sum_{n=1}^infty a_n^{b_n}$ converges.
I know I need to use absolutely convergent but I am unsure how to do it.
real-analysis
real-analysis
asked Nov 29 '18 at 1:30
Vicky
412
asked Nov 29 '18 at 1:30
Vicky
412
asked Nov 29 '18 at 1:30
Vicky
412
asked Nov 29 '18 at 1:30
Vicky
412
asked Nov 29 '18 at 1:30
Vicky
412
412
4
You could start by noting that eventually $a_n lt 1$ (why?), and use a comparison test.
– Tob Ernack
Nov 29 '18 at 1:32
@TobErnack You beat me to it! Who cares that it was by 7 minutes... I coulda been a contender!
– JDMan4444
Nov 29 '18 at 1:40
add a comment |
4
You could start by noting that eventually $a_n lt 1$ (why?), and use a comparison test.
– Tob Ernack
Nov 29 '18 at 1:32
@TobErnack You beat me to it! Who cares that it was by 7 minutes... I coulda been a contender!
– JDMan4444
Nov 29 '18 at 1:40
4
4
You could start by noting that eventually $a_n lt 1$ (why?), and use a comparison test.
– Tob Ernack
Nov 29 '18 at 1:32
You could start by noting that eventually $a_n lt 1$ (why?), and use a comparison test.
– Tob Ernack
Nov 29 '18 at 1:32
@TobErnack You beat me to it! Who cares that it was by 7 minutes... I coulda been a contender!
– JDMan4444
Nov 29 '18 at 1:40
@TobErnack You beat me to it! Who cares that it was by 7 minutes... I coulda been a contender!
– JDMan4444
Nov 29 '18 at 1:40
add a comment |
2 Answers
2
active
oldest
votes
Since $sum_{n=1}^infty a_n < infty$, hence we have $underset{n to infty}{lim} a_n to 0$.
Thus, $exists kin mathbb{N}$ such that $forall n > k, a_n < 1$
Define:
$$
begin{align}
A_k := sum_{n=1}^k a_n^{b_n} < infty
end{align}
$$
Then, since we have:
$$
begin{align}
A_{k'} :=
sum_{n=k+1}^infty a_n^{b_n} leq sum_{n=k+1}^infty a_n
< infty
end{align}
$$
Thus we obtain:
$$
sum_{n=1}^infty a_n^{b_n} = A_k + A_{k'} < infty
$$
edited Nov 29 '18 at 4:04
Mark Viola
130k1274170
answered Nov 29 '18 at 1:57
Sean Lee
1578
Your solution is correct. I just want to make notice of the following: the condition $a_n to 0$ is true for any convergent series.
– Will M.
Nov 29 '18 at 2:04
Ah yes, thank you!
– Sean Lee
Nov 29 '18 at 2:09
Removed the part about absolute convergence - its unnecessary
– Sean Lee
Nov 29 '18 at 2:17
(+1) Well written.
– Mark Viola
Nov 29 '18 at 4:04
add a comment |
Abridged proof. The convergence of the series $(a_n)$ guarantees $a_n to 0,$ hence $(a_n^{b_n})$ is also a convergent series given the assumptions $b_n geq 1.$ Q.E.D.
answered Nov 29 '18 at 1:42
Will M.
2,442314
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Since $sum_{n=1}^infty a_n < infty$, hence we have $underset{n to infty}{lim} a_n to 0$.
Thus, $exists kin mathbb{N}$ such that $forall n > k, a_n < 1$
Define:
$$
begin{align}
A_k := sum_{n=1}^k a_n^{b_n} < infty
end{align}
$$
Then, since we have:
$$
begin{align}
A_{k'} :=
sum_{n=k+1}^infty a_n^{b_n} leq sum_{n=k+1}^infty a_n
< infty
end{align}
$$
Thus we obtain:
$$
sum_{n=1}^infty a_n^{b_n} = A_k + A_{k'} < infty
$$
edited Nov 29 '18 at 4:04
Mark Viola
130k1274170
answered Nov 29 '18 at 1:57
Sean Lee
1578
Your solution is correct. I just want to make notice of the following: the condition $a_n to 0$ is true for any convergent series.
– Will M.
Nov 29 '18 at 2:04
Ah yes, thank you!
– Sean Lee
Nov 29 '18 at 2:09
Removed the part about absolute convergence - its unnecessary
– Sean Lee
Nov 29 '18 at 2:17
(+1) Well written.
– Mark Viola
Nov 29 '18 at 4:04
add a comment |
Since $sum_{n=1}^infty a_n < infty$, hence we have $underset{n to infty}{lim} a_n to 0$.
Thus, $exists kin mathbb{N}$ such that $forall n > k, a_n < 1$
Define:
$$
begin{align}
A_k := sum_{n=1}^k a_n^{b_n} < infty
end{align}
$$
Then, since we have:
$$
begin{align}
A_{k'} :=
sum_{n=k+1}^infty a_n^{b_n} leq sum_{n=k+1}^infty a_n
< infty
end{align}
$$
Thus we obtain:
$$
sum_{n=1}^infty a_n^{b_n} = A_k + A_{k'} < infty
$$
edited Nov 29 '18 at 4:04
Mark Viola
130k1274170
answered Nov 29 '18 at 1:57
Sean Lee
1578
Your solution is correct. I just want to make notice of the following: the condition $a_n to 0$ is true for any convergent series.
– Will M.
Nov 29 '18 at 2:04
Ah yes, thank you!
– Sean Lee
Nov 29 '18 at 2:09
Removed the part about absolute convergence - its unnecessary
– Sean Lee
Nov 29 '18 at 2:17
(+1) Well written.
– Mark Viola
Nov 29 '18 at 4:04
add a comment |
Since $sum_{n=1}^infty a_n < infty$, hence we have $underset{n to infty}{lim} a_n to 0$.
Thus, $exists kin mathbb{N}$ such that $forall n > k, a_n < 1$
Define:
$$
begin{align}
A_k := sum_{n=1}^k a_n^{b_n} < infty
end{align}
$$
Then, since we have:
$$
begin{align}
A_{k'} :=
sum_{n=k+1}^infty a_n^{b_n} leq sum_{n=k+1}^infty a_n
< infty
end{align}
$$
Thus we obtain:
$$
sum_{n=1}^infty a_n^{b_n} = A_k + A_{k'} < infty
$$
edited Nov 29 '18 at 4:04
Mark Viola
130k1274170
answered Nov 29 '18 at 1:57
Sean Lee
1578
Since $sum_{n=1}^infty a_n < infty$, hence we have $underset{n to infty}{lim} a_n to 0$.
Thus, $exists kin mathbb{N}$ such that $forall n > k, a_n < 1$
Define:
$$
begin{align}
A_k := sum_{n=1}^k a_n^{b_n} < infty
end{align}
$$
Then, since we have:
$$
begin{align}
A_{k'} :=
sum_{n=k+1}^infty a_n^{b_n} leq sum_{n=k+1}^infty a_n
< infty
end{align}
$$
Thus we obtain:
$$
sum_{n=1}^infty a_n^{b_n} = A_k + A_{k'} < infty
$$
edited Nov 29 '18 at 4:04
Mark Viola
130k1274170
answered Nov 29 '18 at 1:57
Sean Lee
1578
edited Nov 29 '18 at 4:04
Mark Viola
130k1274170
edited Nov 29 '18 at 4:04
Mark Viola
130k1274170
edited Nov 29 '18 at 4:04
Mark Viola
130k1274170
130k1274170
answered Nov 29 '18 at 1:57
Sean Lee
1578
answered Nov 29 '18 at 1:57
Sean Lee
1578
answered Nov 29 '18 at 1:57
Sean Lee
1578
1578
Your solution is correct. I just want to make notice of the following: the condition $a_n to 0$ is true for any convergent series.
– Will M.
Nov 29 '18 at 2:04
Ah yes, thank you!
– Sean Lee
Nov 29 '18 at 2:09
Removed the part about absolute convergence - its unnecessary
– Sean Lee
Nov 29 '18 at 2:17
(+1) Well written.
– Mark Viola
Nov 29 '18 at 4:04
add a comment |
Your solution is correct. I just want to make notice of the following: the condition $a_n to 0$ is true for any convergent series.
– Will M.
Nov 29 '18 at 2:04
Ah yes, thank you!
– Sean Lee
Nov 29 '18 at 2:09
Removed the part about absolute convergence - its unnecessary
– Sean Lee
Nov 29 '18 at 2:17
(+1) Well written.
– Mark Viola
Nov 29 '18 at 4:04
Your solution is correct. I just want to make notice of the following: the condition $a_n to 0$ is true for any convergent series.
– Will M.
Nov 29 '18 at 2:04
Your solution is correct. I just want to make notice of the following: the condition $a_n to 0$ is true for any convergent series.
– Will M.
Nov 29 '18 at 2:04
Ah yes, thank you!
– Sean Lee
Nov 29 '18 at 2:09
Ah yes, thank you!
– Sean Lee
Nov 29 '18 at 2:09
Removed the part about absolute convergence - its unnecessary
– Sean Lee
Nov 29 '18 at 2:17
Removed the part about absolute convergence - its unnecessary
– Sean Lee
Nov 29 '18 at 2:17
(+1) Well written.
– Mark Viola
Nov 29 '18 at 4:04
(+1) Well written.
– Mark Viola
Nov 29 '18 at 4:04
add a comment |
Abridged proof. The convergence of the series $(a_n)$ guarantees $a_n to 0,$ hence $(a_n^{b_n})$ is also a convergent series given the assumptions $b_n geq 1.$ Q.E.D.
answered Nov 29 '18 at 1:42
Will M.
2,442314
add a comment |
Abridged proof. The convergence of the series $(a_n)$ guarantees $a_n to 0,$ hence $(a_n^{b_n})$ is also a convergent series given the assumptions $b_n geq 1.$ Q.E.D.
answered Nov 29 '18 at 1:42
Will M.
2,442314
add a comment |
Abridged proof. The convergence of the series $(a_n)$ guarantees $a_n to 0,$ hence $(a_n^{b_n})$ is also a convergent series given the assumptions $b_n geq 1.$ Q.E.D.
answered Nov 29 '18 at 1:42
Will M.
2,442314
Abridged proof. The convergence of the series $(a_n)$ guarantees $a_n to 0,$ hence $(a_n^{b_n})$ is also a convergent series given the assumptions $b_n geq 1.$ Q.E.D.
answered Nov 29 '18 at 1:42
Will M.
2,442314
answered Nov 29 '18 at 1:42
Will M.
2,442314
answered Nov 29 '18 at 1:42
Will M.
2,442314
answered Nov 29 '18 at 1:42
Will M.
2,442314
2,442314
add a comment |
add a comment |
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4
You could start by noting that eventually $a_n lt 1$ (why?), and use a comparison test.
– Tob Ernack
Nov 29 '18 at 1:32
@TobErnack You beat me to it! Who cares that it was by 7 minutes... I coulda been a contender!
– JDMan4444
Nov 29 '18 at 1:40