Mixing
Limits: Writing a rigorous solution
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I'm going through some past papers, and I'm asked to determine whether $$sum_k=1^infty e^left(frac1k^2right)$$ converges or diverges. The question is fine, I'm just not sure about the level of rigour required. I cannot find any worked solutions that could give me an indication.
I have two variation of my solution:
1) As $k rightarrow infty$, it follows that $frac1k^2 rightarrow 0$. This in turn means that $e^left(frac1k^2right) rightarrow 1$, which means as $k$ gets sufficiently large, $$sum_k=1^infty e^left(frac1k^2right) = sum_k=1^N e^left(frac1k^2right) + sum_k=N+1^infty 1^left(frac1k^2right)$$ Since the tail diverges, so does the series. $square$
2) For all $k in mathbbN$, it follows that $$1^left(frac1k^2right) < e^left(frac1k^2right) iff sum_k=1^infty 1^left(frac1k^2right) < sum_k=1^infty e^left(frac1k^2right)$$ Since $$sum_k=1^infty 1^left(frac1k^2right) = infty$$ it follows by the comparison test that $$sum_k=1^infty e^left(frac1k^2right) = infty$$ So the series diverges. $square$
Is one solution better than the other? Is there an even better way to write a rigorous limit solution? Please give me some feedback.
limits
asked Jul 22 at 17:15
user9750060
13710
add a comment |Â
up vote
1
down vote
favorite
I'm going through some past papers, and I'm asked to determine whether $$sum_k=1^infty e^left(frac1k^2right)$$ converges or diverges. The question is fine, I'm just not sure about the level of rigour required. I cannot find any worked solutions that could give me an indication.
I have two variation of my solution:
1) As $k rightarrow infty$, it follows that $frac1k^2 rightarrow 0$. This in turn means that $e^left(frac1k^2right) rightarrow 1$, which means as $k$ gets sufficiently large, $$sum_k=1^infty e^left(frac1k^2right) = sum_k=1^N e^left(frac1k^2right) + sum_k=N+1^infty 1^left(frac1k^2right)$$ Since the tail diverges, so does the series. $square$
2) For all $k in mathbbN$, it follows that $$1^left(frac1k^2right) < e^left(frac1k^2right) iff sum_k=1^infty 1^left(frac1k^2right) < sum_k=1^infty e^left(frac1k^2right)$$ Since $$sum_k=1^infty 1^left(frac1k^2right) = infty$$ it follows by the comparison test that $$sum_k=1^infty e^left(frac1k^2right) = infty$$ So the series diverges. $square$
Is one solution better than the other? Is there an even better way to write a rigorous limit solution? Please give me some feedback.
limits
asked Jul 22 at 17:15
user9750060
13710
3
$e^1/k^2>1$ for all $kinBbb N$,
– Lord Shark the Unknown
Jul 22 at 17:17
1
It can be much shorter: the general term of a convergent series tends to $0$. This one doesn't.
– Bernard
Jul 22 at 17:18
1
If you wanted to make your existing strategies "more rigorous", you can use them to lower-bound the sum of the first $n$ terms. The sum is at least $n$, which becomes arbitrarily large.
– J.G.
Jul 22 at 19:30
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm going through some past papers, and I'm asked to determine whether $$sum_k=1^infty e^left(frac1k^2right)$$ converges or diverges. The question is fine, I'm just not sure about the level of rigour required. I cannot find any worked solutions that could give me an indication.
I have two variation of my solution:
1) As $k rightarrow infty$, it follows that $frac1k^2 rightarrow 0$. This in turn means that $e^left(frac1k^2right) rightarrow 1$, which means as $k$ gets sufficiently large, $$sum_k=1^infty e^left(frac1k^2right) = sum_k=1^N e^left(frac1k^2right) + sum_k=N+1^infty 1^left(frac1k^2right)$$ Since the tail diverges, so does the series. $square$
2) For all $k in mathbbN$, it follows that $$1^left(frac1k^2right) < e^left(frac1k^2right) iff sum_k=1^infty 1^left(frac1k^2right) < sum_k=1^infty e^left(frac1k^2right)$$ Since $$sum_k=1^infty 1^left(frac1k^2right) = infty$$ it follows by the comparison test that $$sum_k=1^infty e^left(frac1k^2right) = infty$$ So the series diverges. $square$
Is one solution better than the other? Is there an even better way to write a rigorous limit solution? Please give me some feedback.
limits
asked Jul 22 at 17:15
user9750060
13710
I'm going through some past papers, and I'm asked to determine whether $$sum_k=1^infty e^left(frac1k^2right)$$ converges or diverges. The question is fine, I'm just not sure about the level of rigour required. I cannot find any worked solutions that could give me an indication.
I have two variation of my solution:
1) As $k rightarrow infty$, it follows that $frac1k^2 rightarrow 0$. This in turn means that $e^left(frac1k^2right) rightarrow 1$, which means as $k$ gets sufficiently large, $$sum_k=1^infty e^left(frac1k^2right) = sum_k=1^N e^left(frac1k^2right) + sum_k=N+1^infty 1^left(frac1k^2right)$$ Since the tail diverges, so does the series. $square$
2) For all $k in mathbbN$, it follows that $$1^left(frac1k^2right) < e^left(frac1k^2right) iff sum_k=1^infty 1^left(frac1k^2right) < sum_k=1^infty e^left(frac1k^2right)$$ Since $$sum_k=1^infty 1^left(frac1k^2right) = infty$$ it follows by the comparison test that $$sum_k=1^infty e^left(frac1k^2right) = infty$$ So the series diverges. $square$
Is one solution better than the other? Is there an even better way to write a rigorous limit solution? Please give me some feedback.
limits
asked Jul 22 at 17:15
user9750060
13710
edited Jul 22 at 17:19
asked Jul 22 at 17:15
user9750060
13710
asked Jul 22 at 17:15
user9750060
13710
asked Jul 22 at 17:15
user9750060
13710
13710
3
$e^1/k^2>1$ for all $kinBbb N$,
– Lord Shark the Unknown
Jul 22 at 17:17
1
It can be much shorter: the general term of a convergent series tends to $0$. This one doesn't.
– Bernard
Jul 22 at 17:18
1
If you wanted to make your existing strategies "more rigorous", you can use them to lower-bound the sum of the first $n$ terms. The sum is at least $n$, which becomes arbitrarily large.
– J.G.
Jul 22 at 19:30
add a comment |Â
3
$e^1/k^2>1$ for all $kinBbb N$,
– Lord Shark the Unknown
Jul 22 at 17:17
1
It can be much shorter: the general term of a convergent series tends to $0$. This one doesn't.
– Bernard
Jul 22 at 17:18
1
If you wanted to make your existing strategies "more rigorous", you can use them to lower-bound the sum of the first $n$ terms. The sum is at least $n$, which becomes arbitrarily large.
– J.G.
Jul 22 at 19:30
3
3
$e^1/k^2>1$ for all $kinBbb N$,
– Lord Shark the Unknown
Jul 22 at 17:17
$e^1/k^2>1$ for all $kinBbb N$,
– Lord Shark the Unknown
Jul 22 at 17:17
1
1
It can be much shorter: the general term of a convergent series tends to $0$. This one doesn't.
– Bernard
Jul 22 at 17:18
It can be much shorter: the general term of a convergent series tends to $0$. This one doesn't.
– Bernard
Jul 22 at 17:18
1
1
If you wanted to make your existing strategies "more rigorous", you can use them to lower-bound the sum of the first $n$ terms. The sum is at least $n$, which becomes arbitrarily large.
– J.G.
Jul 22 at 19:30
If you wanted to make your existing strategies "more rigorous", you can use them to lower-bound the sum of the first $n$ terms. The sum is at least $n$, which becomes arbitrarily large.
– J.G.
Jul 22 at 19:30
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
It suffices to observe that
$$e^left(frac1k^2right) rightarrow 1$$
therefore the series doesn’t converge (it is a necessary condition that $a_k to 0$).
answered Jul 22 at 17:19
gimusi
65.2k73583
Great thanks, but can I really use that reasoning? What about the harmonic series? The terms tend to zero, but the series diverges... Or am I misunderstanding something?
– user9750060
Jul 22 at 17:22
1
"not $A$ implies not $B$" is not the same as "$A$ implies $B$". (in this example $A$ is "the series converges" and $B$ "its terms tend to zero"). @user9750060
– Lord Shark the Unknown
Jul 22 at 17:24
1
Good observation. The fact is that “$sum a_k$ converges $implies a_kto 0$†but the converse is not true. From here if $a_knot to 0$ we can conclude that the series doesn’t converge.
– gimusi
Jul 22 at 17:25
1
In other words we cannot find a convergent series with $a_knot to 0$.
– gimusi
Jul 22 at 17:27
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
It suffices to observe that
$$e^left(frac1k^2right) rightarrow 1$$
therefore the series doesn’t converge (it is a necessary condition that $a_k to 0$).
answered Jul 22 at 17:19
gimusi
65.2k73583
Great thanks, but can I really use that reasoning? What about the harmonic series? The terms tend to zero, but the series diverges... Or am I misunderstanding something?
– user9750060
Jul 22 at 17:22
1
"not $A$ implies not $B$" is not the same as "$A$ implies $B$". (in this example $A$ is "the series converges" and $B$ "its terms tend to zero"). @user9750060
– Lord Shark the Unknown
Jul 22 at 17:24
1
Good observation. The fact is that “$sum a_k$ converges $implies a_kto 0$†but the converse is not true. From here if $a_knot to 0$ we can conclude that the series doesn’t converge.
– gimusi
Jul 22 at 17:25
1
In other words we cannot find a convergent series with $a_knot to 0$.
– gimusi
Jul 22 at 17:27
add a comment |Â
up vote
2
down vote
accepted
It suffices to observe that
$$e^left(frac1k^2right) rightarrow 1$$
therefore the series doesn’t converge (it is a necessary condition that $a_k to 0$).
answered Jul 22 at 17:19
gimusi
65.2k73583
Great thanks, but can I really use that reasoning? What about the harmonic series? The terms tend to zero, but the series diverges... Or am I misunderstanding something?
– user9750060
Jul 22 at 17:22
1
"not $A$ implies not $B$" is not the same as "$A$ implies $B$". (in this example $A$ is "the series converges" and $B$ "its terms tend to zero"). @user9750060
– Lord Shark the Unknown
Jul 22 at 17:24
1
Good observation. The fact is that “$sum a_k$ converges $implies a_kto 0$†but the converse is not true. From here if $a_knot to 0$ we can conclude that the series doesn’t converge.
– gimusi
Jul 22 at 17:25
1
In other words we cannot find a convergent series with $a_knot to 0$.
– gimusi
Jul 22 at 17:27
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
It suffices to observe that
$$e^left(frac1k^2right) rightarrow 1$$
therefore the series doesn’t converge (it is a necessary condition that $a_k to 0$).
answered Jul 22 at 17:19
gimusi
65.2k73583
It suffices to observe that
$$e^left(frac1k^2right) rightarrow 1$$
therefore the series doesn’t converge (it is a necessary condition that $a_k to 0$).
answered Jul 22 at 17:19
gimusi
65.2k73583
answered Jul 22 at 17:19
gimusi
65.2k73583
answered Jul 22 at 17:19
gimusi
65.2k73583
answered Jul 22 at 17:19
gimusi
65.2k73583
65.2k73583
Great thanks, but can I really use that reasoning? What about the harmonic series? The terms tend to zero, but the series diverges... Or am I misunderstanding something?
– user9750060
Jul 22 at 17:22
1
"not $A$ implies not $B$" is not the same as "$A$ implies $B$". (in this example $A$ is "the series converges" and $B$ "its terms tend to zero"). @user9750060
– Lord Shark the Unknown
Jul 22 at 17:24
1
Good observation. The fact is that “$sum a_k$ converges $implies a_kto 0$†but the converse is not true. From here if $a_knot to 0$ we can conclude that the series doesn’t converge.
– gimusi
Jul 22 at 17:25
1
In other words we cannot find a convergent series with $a_knot to 0$.
– gimusi
Jul 22 at 17:27
add a comment |Â
Great thanks, but can I really use that reasoning? What about the harmonic series? The terms tend to zero, but the series diverges... Or am I misunderstanding something?
– user9750060
Jul 22 at 17:22
1
"not $A$ implies not $B$" is not the same as "$A$ implies $B$". (in this example $A$ is "the series converges" and $B$ "its terms tend to zero"). @user9750060
– Lord Shark the Unknown
Jul 22 at 17:24
1
Good observation. The fact is that “$sum a_k$ converges $implies a_kto 0$†but the converse is not true. From here if $a_knot to 0$ we can conclude that the series doesn’t converge.
– gimusi
Jul 22 at 17:25
1
In other words we cannot find a convergent series with $a_knot to 0$.
– gimusi
Jul 22 at 17:27
Great thanks, but can I really use that reasoning? What about the harmonic series? The terms tend to zero, but the series diverges... Or am I misunderstanding something?
– user9750060
Jul 22 at 17:22
Great thanks, but can I really use that reasoning? What about the harmonic series? The terms tend to zero, but the series diverges... Or am I misunderstanding something?
– user9750060
Jul 22 at 17:22
1
1
"not $A$ implies not $B$" is not the same as "$A$ implies $B$". (in this example $A$ is "the series converges" and $B$ "its terms tend to zero"). @user9750060
– Lord Shark the Unknown
Jul 22 at 17:24
"not $A$ implies not $B$" is not the same as "$A$ implies $B$". (in this example $A$ is "the series converges" and $B$ "its terms tend to zero"). @user9750060
– Lord Shark the Unknown
Jul 22 at 17:24
1
1
Good observation. The fact is that “$sum a_k$ converges $implies a_kto 0$†but the converse is not true. From here if $a_knot to 0$ we can conclude that the series doesn’t converge.
– gimusi
Jul 22 at 17:25
Good observation. The fact is that “$sum a_k$ converges $implies a_kto 0$†but the converse is not true. From here if $a_knot to 0$ we can conclude that the series doesn’t converge.
– gimusi
Jul 22 at 17:25
1
1
In other words we cannot find a convergent series with $a_knot to 0$.
– gimusi
Jul 22 at 17:27
In other words we cannot find a convergent series with $a_knot to 0$.
– gimusi
Jul 22 at 17:27
add a comment |Â
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3
$e^1/k^2>1$ for all $kinBbb N$,
– Lord Shark the Unknown
Jul 22 at 17:17
1
It can be much shorter: the general term of a convergent series tends to $0$. This one doesn't.
– Bernard
Jul 22 at 17:18
1
If you wanted to make your existing strategies "more rigorous", you can use them to lower-bound the sum of the first $n$ terms. The sum is at least $n$, which becomes arbitrarily large.
– J.G.
Jul 22 at 19:30