Then can the all terms of sequence of the partial sums of the series be strictly greater than zero?
up vote
-1
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It is given that $sum _{n=1} ^ infty a_n = 0$ but $sum _{n=1} ^ infty |a_n| $ is not convergent .
Then can we get a series whose all the terms of sequence of the partial sums of the series are strictly greater than zero?
I could not find an example.
Can anyone please give me a hint?
real-analysis sequences-and-series analysis
asked Nov 18 at 17:37
cmi
1,046212
add a comment |
up vote
-1
down vote
favorite
It is given that $sum _{n=1} ^ infty a_n = 0$ but $sum _{n=1} ^ infty |a_n| $ is not convergent .
Then can we get a series whose all the terms of sequence of the partial sums of the series are strictly greater than zero?
I could not find an example.
Can anyone please give me a hint?
real-analysis sequences-and-series analysis
asked Nov 18 at 17:37
cmi
1,046212
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
It is given that $sum _{n=1} ^ infty a_n = 0$ but $sum _{n=1} ^ infty |a_n| $ is not convergent .
Then can we get a series whose all the terms of sequence of the partial sums of the series are strictly greater than zero?
I could not find an example.
Can anyone please give me a hint?
real-analysis sequences-and-series analysis
asked Nov 18 at 17:37
cmi
1,046212
It is given that $sum _{n=1} ^ infty a_n = 0$ but $sum _{n=1} ^ infty |a_n| $ is not convergent .
Then can we get a series whose all the terms of sequence of the partial sums of the series are strictly greater than zero?
I could not find an example.
Can anyone please give me a hint?
real-analysis sequences-and-series analysis
real-analysis sequences-and-series analysis
asked Nov 18 at 17:37
cmi
1,046212
asked Nov 18 at 17:37
cmi
1,046212
edited Nov 18 at 17:43
asked Nov 18 at 17:37
cmi
1,046212
asked Nov 18 at 17:37
cmi
1,046212
asked Nov 18 at 17:37
cmi
1,046212
1,046212
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
If you define the sequence of partial sums $A_n=sum_{k=1}^{n}a_k$ as
$$ A_n = left{begin{array}{rcl}frac{1}{n^2}+frac{1}{n}&text{if }ntext{ is odd}\ frac{1}{n^2} & text{if }ntext{ is even}end{array} right.$$
then ${a_n}_{ngeq 1}$ is implicitly defined and it fulfills the wanted constraints.
answered Nov 18 at 17:45
Jack D'Aurizio
283k33275653
How can you prove the absolute divergence of the series?@Jack D'Aurizio
– cmi
Nov 18 at 18:18
@cmi: by comparison with the harmonic series.
– Jack D'Aurizio
Nov 18 at 18:21
add a comment |
up vote
1
down vote
If
$$
begin{array}{l}
a_0=1\
a_n=frac{(-1)^{n+1}}{leftlfloorfrac{n+1}2rightrfloor}-frac1{n(n+1)}
end{array}
$$
then
$$
sum_{k=0}^na_k=frac1{n+1}+frac{1-(-1)^n}{2leftlfloorfrac{n+1}2rightrfloor}
$$
answered Nov 18 at 18:49
robjohn♦
263k27301622
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
If you define the sequence of partial sums $A_n=sum_{k=1}^{n}a_k$ as
$$ A_n = left{begin{array}{rcl}frac{1}{n^2}+frac{1}{n}&text{if }ntext{ is odd}\ frac{1}{n^2} & text{if }ntext{ is even}end{array} right.$$
then ${a_n}_{ngeq 1}$ is implicitly defined and it fulfills the wanted constraints.
answered Nov 18 at 17:45
Jack D'Aurizio
283k33275653
How can you prove the absolute divergence of the series?@Jack D'Aurizio
– cmi
Nov 18 at 18:18
@cmi: by comparison with the harmonic series.
– Jack D'Aurizio
Nov 18 at 18:21
add a comment |
up vote
2
down vote
accepted
If you define the sequence of partial sums $A_n=sum_{k=1}^{n}a_k$ as
$$ A_n = left{begin{array}{rcl}frac{1}{n^2}+frac{1}{n}&text{if }ntext{ is odd}\ frac{1}{n^2} & text{if }ntext{ is even}end{array} right.$$
then ${a_n}_{ngeq 1}$ is implicitly defined and it fulfills the wanted constraints.
answered Nov 18 at 17:45
Jack D'Aurizio
283k33275653
How can you prove the absolute divergence of the series?@Jack D'Aurizio
– cmi
Nov 18 at 18:18
@cmi: by comparison with the harmonic series.
– Jack D'Aurizio
Nov 18 at 18:21
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
If you define the sequence of partial sums $A_n=sum_{k=1}^{n}a_k$ as
$$ A_n = left{begin{array}{rcl}frac{1}{n^2}+frac{1}{n}&text{if }ntext{ is odd}\ frac{1}{n^2} & text{if }ntext{ is even}end{array} right.$$
then ${a_n}_{ngeq 1}$ is implicitly defined and it fulfills the wanted constraints.
answered Nov 18 at 17:45
Jack D'Aurizio
283k33275653
If you define the sequence of partial sums $A_n=sum_{k=1}^{n}a_k$ as
$$ A_n = left{begin{array}{rcl}frac{1}{n^2}+frac{1}{n}&text{if }ntext{ is odd}\ frac{1}{n^2} & text{if }ntext{ is even}end{array} right.$$
then ${a_n}_{ngeq 1}$ is implicitly defined and it fulfills the wanted constraints.
answered Nov 18 at 17:45
Jack D'Aurizio
283k33275653
edited Nov 18 at 17:51
answered Nov 18 at 17:45
Jack D'Aurizio
283k33275653
answered Nov 18 at 17:45
Jack D'Aurizio
283k33275653
answered Nov 18 at 17:45
Jack D'Aurizio
283k33275653
283k33275653
How can you prove the absolute divergence of the series?@Jack D'Aurizio
– cmi
Nov 18 at 18:18
@cmi: by comparison with the harmonic series.
– Jack D'Aurizio
Nov 18 at 18:21
add a comment |
How can you prove the absolute divergence of the series?@Jack D'Aurizio
– cmi
Nov 18 at 18:18
@cmi: by comparison with the harmonic series.
– Jack D'Aurizio
Nov 18 at 18:21
How can you prove the absolute divergence of the series?@Jack D'Aurizio
– cmi
Nov 18 at 18:18
How can you prove the absolute divergence of the series?@Jack D'Aurizio
– cmi
Nov 18 at 18:18
@cmi: by comparison with the harmonic series.
– Jack D'Aurizio
Nov 18 at 18:21
@cmi: by comparison with the harmonic series.
– Jack D'Aurizio
Nov 18 at 18:21
add a comment |
up vote
1
down vote
If
$$
begin{array}{l}
a_0=1\
a_n=frac{(-1)^{n+1}}{leftlfloorfrac{n+1}2rightrfloor}-frac1{n(n+1)}
end{array}
$$
then
$$
sum_{k=0}^na_k=frac1{n+1}+frac{1-(-1)^n}{2leftlfloorfrac{n+1}2rightrfloor}
$$
answered Nov 18 at 18:49
robjohn♦
263k27301622
add a comment |
up vote
1
down vote
If
$$
begin{array}{l}
a_0=1\
a_n=frac{(-1)^{n+1}}{leftlfloorfrac{n+1}2rightrfloor}-frac1{n(n+1)}
end{array}
$$
then
$$
sum_{k=0}^na_k=frac1{n+1}+frac{1-(-1)^n}{2leftlfloorfrac{n+1}2rightrfloor}
$$
answered Nov 18 at 18:49
robjohn♦
263k27301622
add a comment |
up vote
1
down vote
up vote
1
down vote
If
$$
begin{array}{l}
a_0=1\
a_n=frac{(-1)^{n+1}}{leftlfloorfrac{n+1}2rightrfloor}-frac1{n(n+1)}
end{array}
$$
then
$$
sum_{k=0}^na_k=frac1{n+1}+frac{1-(-1)^n}{2leftlfloorfrac{n+1}2rightrfloor}
$$
answered Nov 18 at 18:49
robjohn♦
263k27301622
If
$$
begin{array}{l}
a_0=1\
a_n=frac{(-1)^{n+1}}{leftlfloorfrac{n+1}2rightrfloor}-frac1{n(n+1)}
end{array}
$$
then
$$
sum_{k=0}^na_k=frac1{n+1}+frac{1-(-1)^n}{2leftlfloorfrac{n+1}2rightrfloor}
$$
answered Nov 18 at 18:49
robjohn♦
263k27301622
answered Nov 18 at 18:49
robjohn♦
263k27301622
answered Nov 18 at 18:49
robjohn♦
263k27301622
answered Nov 18 at 18:49
robjohn♦
263k27301622
263k27301622
add a comment |
add a comment |
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