Mixing
Proof that $sum_n = 0^inftydfrac1n(n+1)$ converges to $1$
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
I'm looking for some verification of my proof. The problem is asking me to prove that the sequence $s_n$ defined by
$$s_n = dfrac1n(n + 1)$$
converges to $1$.I thought it goes to $0$, but maybe I'm missing something. Here's the problem word-for-word:
Problem
Define the sequence $s_n$ by
$$s_n = dfrac12 cdot 1 + dfrac13 cdot 2 + ldots + dfrac1(n + 1)(n)$$ for every index $n$.
Prove that the limit as $n$ approaches infinity of $s_n$ equals $1$.
My proof:
Let $epsilon > 0$ be arbitrarily chosen. Choose $M = dfrac1epsilon$, and let $n geq M$ be arbitrarily chosen. Then,
$$left| dfrac1n(n + 1) - 1 right| < left| dfrac1n(n + 1)right| < left|dfrac1nright| = dfrac1n < epsilon.$$
(I was able to remove the absolute value because $n > 0 ,$ always.)
Equivalently, this means that
$$n > dfrac1epsilon.$$
Thus, choosing $M = dfrac1e$ means that the above inequality always holds, which completes the proof.
Updated Proof
Ok, so I think I was actually supposed to prove
$$sum_n = 0^infty dfrac1n(n + 1) = 1.$$
By applying partial fraction decomposition, I found that this summation equals
$$sum_n = 0^inftydfrac1n - sum_n = 0^inftydfrac1n + 1.$$
From here, I can clearly see that everything cancels out except for the $1$ term. But is it possible to show this using the definition of convergence?
proof-verification epsilon-delta
asked Jul 17 at 21:04
Hat
789115
 |Â
show 1 more comment
up vote
2
down vote
favorite
I'm looking for some verification of my proof. The problem is asking me to prove that the sequence $s_n$ defined by
$$s_n = dfrac1n(n + 1)$$
converges to $1$.I thought it goes to $0$, but maybe I'm missing something. Here's the problem word-for-word:
Problem
Define the sequence $s_n$ by
$$s_n = dfrac12 cdot 1 + dfrac13 cdot 2 + ldots + dfrac1(n + 1)(n)$$ for every index $n$.
Prove that the limit as $n$ approaches infinity of $s_n$ equals $1$.
My proof:
Let $epsilon > 0$ be arbitrarily chosen. Choose $M = dfrac1epsilon$, and let $n geq M$ be arbitrarily chosen. Then,
$$left| dfrac1n(n + 1) - 1 right| < left| dfrac1n(n + 1)right| < left|dfrac1nright| = dfrac1n < epsilon.$$
(I was able to remove the absolute value because $n > 0 ,$ always.)
Equivalently, this means that
$$n > dfrac1epsilon.$$
Thus, choosing $M = dfrac1e$ means that the above inequality always holds, which completes the proof.
Updated Proof
Ok, so I think I was actually supposed to prove
$$sum_n = 0^infty dfrac1n(n + 1) = 1.$$
By applying partial fraction decomposition, I found that this summation equals
$$sum_n = 0^inftydfrac1n - sum_n = 0^inftydfrac1n + 1.$$
From here, I can clearly see that everything cancels out except for the $1$ term. But is it possible to show this using the definition of convergence?
proof-verification epsilon-delta
asked Jul 17 at 21:04
Hat
789115
But as $n$ grows, $s_nto 0$.
– lulu
Jul 17 at 21:05
Yeah, I was thinking the same. Maybe I'm missing something in my interpretation. Let me copy the problem word for word.
– Hat
Jul 17 at 21:06
1
Specifically, your claim that $big vert frac 1n(n+1)-1big vert <big vert frac 1n(n+1)big vert $ is false for large $n$.
– lulu
Jul 17 at 21:06
1
The sum, $sum_n=1^infty frac 1n(n+1)$ does indeed equal $1$. Hint: expand the summands by partial fractions.
– lulu
Jul 17 at 21:09
Be careful with distributing infinite sums. This is the sum of terms like $frac1n-frac1(n+1)$, but the infinite sums of the individual pieces both diverge to infinity, so you cannot write things as a difference of infinite sums as in your final expression.
– Mark S.
Jul 18 at 1:35
 |Â
show 1 more comment
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I'm looking for some verification of my proof. The problem is asking me to prove that the sequence $s_n$ defined by
$$s_n = dfrac1n(n + 1)$$
converges to $1$.I thought it goes to $0$, but maybe I'm missing something. Here's the problem word-for-word:
Problem
Define the sequence $s_n$ by
$$s_n = dfrac12 cdot 1 + dfrac13 cdot 2 + ldots + dfrac1(n + 1)(n)$$ for every index $n$.
Prove that the limit as $n$ approaches infinity of $s_n$ equals $1$.
My proof:
Let $epsilon > 0$ be arbitrarily chosen. Choose $M = dfrac1epsilon$, and let $n geq M$ be arbitrarily chosen. Then,
$$left| dfrac1n(n + 1) - 1 right| < left| dfrac1n(n + 1)right| < left|dfrac1nright| = dfrac1n < epsilon.$$
(I was able to remove the absolute value because $n > 0 ,$ always.)
Equivalently, this means that
$$n > dfrac1epsilon.$$
Thus, choosing $M = dfrac1e$ means that the above inequality always holds, which completes the proof.
Updated Proof
Ok, so I think I was actually supposed to prove
$$sum_n = 0^infty dfrac1n(n + 1) = 1.$$
By applying partial fraction decomposition, I found that this summation equals
$$sum_n = 0^inftydfrac1n - sum_n = 0^inftydfrac1n + 1.$$
From here, I can clearly see that everything cancels out except for the $1$ term. But is it possible to show this using the definition of convergence?
proof-verification epsilon-delta
asked Jul 17 at 21:04
Hat
789115
I'm looking for some verification of my proof. The problem is asking me to prove that the sequence $s_n$ defined by
$$s_n = dfrac1n(n + 1)$$
converges to $1$.I thought it goes to $0$, but maybe I'm missing something. Here's the problem word-for-word:
Problem
Define the sequence $s_n$ by
$$s_n = dfrac12 cdot 1 + dfrac13 cdot 2 + ldots + dfrac1(n + 1)(n)$$ for every index $n$.
Prove that the limit as $n$ approaches infinity of $s_n$ equals $1$.
My proof:
Let $epsilon > 0$ be arbitrarily chosen. Choose $M = dfrac1epsilon$, and let $n geq M$ be arbitrarily chosen. Then,
$$left| dfrac1n(n + 1) - 1 right| < left| dfrac1n(n + 1)right| < left|dfrac1nright| = dfrac1n < epsilon.$$
(I was able to remove the absolute value because $n > 0 ,$ always.)
Equivalently, this means that
$$n > dfrac1epsilon.$$
Thus, choosing $M = dfrac1e$ means that the above inequality always holds, which completes the proof.
Updated Proof
Ok, so I think I was actually supposed to prove
$$sum_n = 0^infty dfrac1n(n + 1) = 1.$$
By applying partial fraction decomposition, I found that this summation equals
$$sum_n = 0^inftydfrac1n - sum_n = 0^inftydfrac1n + 1.$$
From here, I can clearly see that everything cancels out except for the $1$ term. But is it possible to show this using the definition of convergence?
proof-verification epsilon-delta
asked Jul 17 at 21:04
Hat
789115
edited Jul 17 at 21:20
asked Jul 17 at 21:04
Hat
789115
asked Jul 17 at 21:04
Hat
789115
asked Jul 17 at 21:04
Hat
789115
789115
But as $n$ grows, $s_nto 0$.
– lulu
Jul 17 at 21:05
Yeah, I was thinking the same. Maybe I'm missing something in my interpretation. Let me copy the problem word for word.
– Hat
Jul 17 at 21:06
1
Specifically, your claim that $big vert frac 1n(n+1)-1big vert <big vert frac 1n(n+1)big vert $ is false for large $n$.
– lulu
Jul 17 at 21:06
1
The sum, $sum_n=1^infty frac 1n(n+1)$ does indeed equal $1$. Hint: expand the summands by partial fractions.
– lulu
Jul 17 at 21:09
Be careful with distributing infinite sums. This is the sum of terms like $frac1n-frac1(n+1)$, but the infinite sums of the individual pieces both diverge to infinity, so you cannot write things as a difference of infinite sums as in your final expression.
– Mark S.
Jul 18 at 1:35
 |Â
show 1 more comment
But as $n$ grows, $s_nto 0$.
– lulu
Jul 17 at 21:05
Yeah, I was thinking the same. Maybe I'm missing something in my interpretation. Let me copy the problem word for word.
– Hat
Jul 17 at 21:06
1
Specifically, your claim that $big vert frac 1n(n+1)-1big vert <big vert frac 1n(n+1)big vert $ is false for large $n$.
– lulu
Jul 17 at 21:06
1
The sum, $sum_n=1^infty frac 1n(n+1)$ does indeed equal $1$. Hint: expand the summands by partial fractions.
– lulu
Jul 17 at 21:09
Be careful with distributing infinite sums. This is the sum of terms like $frac1n-frac1(n+1)$, but the infinite sums of the individual pieces both diverge to infinity, so you cannot write things as a difference of infinite sums as in your final expression.
– Mark S.
Jul 18 at 1:35
But as $n$ grows, $s_nto 0$.
– lulu
Jul 17 at 21:05
But as $n$ grows, $s_nto 0$.
– lulu
Jul 17 at 21:05
Yeah, I was thinking the same. Maybe I'm missing something in my interpretation. Let me copy the problem word for word.
– Hat
Jul 17 at 21:06
Yeah, I was thinking the same. Maybe I'm missing something in my interpretation. Let me copy the problem word for word.
– Hat
Jul 17 at 21:06
1
1
Specifically, your claim that $big vert frac 1n(n+1)-1big vert <big vert frac 1n(n+1)big vert $ is false for large $n$.
– lulu
Jul 17 at 21:06
Specifically, your claim that $big vert frac 1n(n+1)-1big vert <big vert frac 1n(n+1)big vert $ is false for large $n$.
– lulu
Jul 17 at 21:06
1
1
The sum, $sum_n=1^infty frac 1n(n+1)$ does indeed equal $1$. Hint: expand the summands by partial fractions.
– lulu
Jul 17 at 21:09
The sum, $sum_n=1^infty frac 1n(n+1)$ does indeed equal $1$. Hint: expand the summands by partial fractions.
– lulu
Jul 17 at 21:09
Be careful with distributing infinite sums. This is the sum of terms like $frac1n-frac1(n+1)$, but the infinite sums of the individual pieces both diverge to infinity, so you cannot write things as a difference of infinite sums as in your final expression.
– Mark S.
Jul 18 at 1:35
Be careful with distributing infinite sums. This is the sum of terms like $frac1n-frac1(n+1)$, but the infinite sums of the individual pieces both diverge to infinity, so you cannot write things as a difference of infinite sums as in your final expression.
– Mark S.
Jul 18 at 1:35
 |Â
show 1 more comment
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
We have that
$$0le s_n = dfrac1n(n + 1)le frac1nto 0$$
and
$$sum_1^infty s_n = sum_1^infty dfrac1n(n + 1)=sum_1^infty left(dfrac1n-dfrac1n + 1right) =1colorred-frac12+frac12-frac13+frac13-frac14+frac14-ldots$$
note also that
$$S_k=sum_1^k s_n=1-frac1k+1 to 1$$
answered Jul 17 at 21:12
gimusi
65.4k73584
I also found this. Is there a way to do it using the definition of convergence?
– Hat
Jul 17 at 21:14
@Hat To do that you should express the partial sum $S_k$ and show that $S_kto 1$ as $kto infty$. But telescoping is a very good and rigorous argument.
– gimusi
Jul 17 at 21:16
@Hat Indeed we have $S_k=1-frac1k+1$ and it easy to show that $S_k to 1$.
– gimusi
Jul 17 at 21:20
How did you get that $S_k = 1 - dfrac1k + 1$?
– Hat
Jul 17 at 21:21
@Hat Always by telescoping argument but in that way we can make it more rigorous considering a partial sum and then taking the limit.
– gimusi
Jul 17 at 21:23
 |Â
show 2 more comments
up vote
3
down vote
It is wrong and it could no possibly be right since the limit of the sequence is $0$, not $1$. Perhaps that what you were asked to prove was that $$displaystylesum_n=1^inftyfrac1n(n+1)=1.$$
The error in your proof lies in the (false) inequality$$left|frac1n(n+1)-1right|<left|frac1n(n+1)right|.$$
answered Jul 17 at 21:07
José Carlos Santos
114k1698177
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
We have that
$$0le s_n = dfrac1n(n + 1)le frac1nto 0$$
and
$$sum_1^infty s_n = sum_1^infty dfrac1n(n + 1)=sum_1^infty left(dfrac1n-dfrac1n + 1right) =1colorred-frac12+frac12-frac13+frac13-frac14+frac14-ldots$$
note also that
$$S_k=sum_1^k s_n=1-frac1k+1 to 1$$
answered Jul 17 at 21:12
gimusi
65.4k73584
I also found this. Is there a way to do it using the definition of convergence?
– Hat
Jul 17 at 21:14
@Hat To do that you should express the partial sum $S_k$ and show that $S_kto 1$ as $kto infty$. But telescoping is a very good and rigorous argument.
– gimusi
Jul 17 at 21:16
@Hat Indeed we have $S_k=1-frac1k+1$ and it easy to show that $S_k to 1$.
– gimusi
Jul 17 at 21:20
How did you get that $S_k = 1 - dfrac1k + 1$?
– Hat
Jul 17 at 21:21
@Hat Always by telescoping argument but in that way we can make it more rigorous considering a partial sum and then taking the limit.
– gimusi
Jul 17 at 21:23
 |Â
show 2 more comments
up vote
4
down vote
accepted
We have that
$$0le s_n = dfrac1n(n + 1)le frac1nto 0$$
and
$$sum_1^infty s_n = sum_1^infty dfrac1n(n + 1)=sum_1^infty left(dfrac1n-dfrac1n + 1right) =1colorred-frac12+frac12-frac13+frac13-frac14+frac14-ldots$$
note also that
$$S_k=sum_1^k s_n=1-frac1k+1 to 1$$
answered Jul 17 at 21:12
gimusi
65.4k73584
I also found this. Is there a way to do it using the definition of convergence?
– Hat
Jul 17 at 21:14
@Hat To do that you should express the partial sum $S_k$ and show that $S_kto 1$ as $kto infty$. But telescoping is a very good and rigorous argument.
– gimusi
Jul 17 at 21:16
@Hat Indeed we have $S_k=1-frac1k+1$ and it easy to show that $S_k to 1$.
– gimusi
Jul 17 at 21:20
How did you get that $S_k = 1 - dfrac1k + 1$?
– Hat
Jul 17 at 21:21
@Hat Always by telescoping argument but in that way we can make it more rigorous considering a partial sum and then taking the limit.
– gimusi
Jul 17 at 21:23
 |Â
show 2 more comments
up vote
4
down vote
accepted
up vote
4
down vote
accepted
We have that
$$0le s_n = dfrac1n(n + 1)le frac1nto 0$$
and
$$sum_1^infty s_n = sum_1^infty dfrac1n(n + 1)=sum_1^infty left(dfrac1n-dfrac1n + 1right) =1colorred-frac12+frac12-frac13+frac13-frac14+frac14-ldots$$
note also that
$$S_k=sum_1^k s_n=1-frac1k+1 to 1$$
answered Jul 17 at 21:12
gimusi
65.4k73584
We have that
$$0le s_n = dfrac1n(n + 1)le frac1nto 0$$
and
$$sum_1^infty s_n = sum_1^infty dfrac1n(n + 1)=sum_1^infty left(dfrac1n-dfrac1n + 1right) =1colorred-frac12+frac12-frac13+frac13-frac14+frac14-ldots$$
note also that
$$S_k=sum_1^k s_n=1-frac1k+1 to 1$$
answered Jul 17 at 21:12
gimusi
65.4k73584
edited Jul 17 at 21:21
answered Jul 17 at 21:12
gimusi
65.4k73584
answered Jul 17 at 21:12
gimusi
65.4k73584
answered Jul 17 at 21:12
gimusi
65.4k73584
65.4k73584
I also found this. Is there a way to do it using the definition of convergence?
– Hat
Jul 17 at 21:14
@Hat To do that you should express the partial sum $S_k$ and show that $S_kto 1$ as $kto infty$. But telescoping is a very good and rigorous argument.
– gimusi
Jul 17 at 21:16
@Hat Indeed we have $S_k=1-frac1k+1$ and it easy to show that $S_k to 1$.
– gimusi
Jul 17 at 21:20
How did you get that $S_k = 1 - dfrac1k + 1$?
– Hat
Jul 17 at 21:21
@Hat Always by telescoping argument but in that way we can make it more rigorous considering a partial sum and then taking the limit.
– gimusi
Jul 17 at 21:23
 |Â
show 2 more comments
I also found this. Is there a way to do it using the definition of convergence?
– Hat
Jul 17 at 21:14
@Hat To do that you should express the partial sum $S_k$ and show that $S_kto 1$ as $kto infty$. But telescoping is a very good and rigorous argument.
– gimusi
Jul 17 at 21:16
@Hat Indeed we have $S_k=1-frac1k+1$ and it easy to show that $S_k to 1$.
– gimusi
Jul 17 at 21:20
How did you get that $S_k = 1 - dfrac1k + 1$?
– Hat
Jul 17 at 21:21
@Hat Always by telescoping argument but in that way we can make it more rigorous considering a partial sum and then taking the limit.
– gimusi
Jul 17 at 21:23
I also found this. Is there a way to do it using the definition of convergence?
– Hat
Jul 17 at 21:14
I also found this. Is there a way to do it using the definition of convergence?
– Hat
Jul 17 at 21:14
@Hat To do that you should express the partial sum $S_k$ and show that $S_kto 1$ as $kto infty$. But telescoping is a very good and rigorous argument.
– gimusi
Jul 17 at 21:16
@Hat To do that you should express the partial sum $S_k$ and show that $S_kto 1$ as $kto infty$. But telescoping is a very good and rigorous argument.
– gimusi
Jul 17 at 21:16
@Hat Indeed we have $S_k=1-frac1k+1$ and it easy to show that $S_k to 1$.
– gimusi
Jul 17 at 21:20
@Hat Indeed we have $S_k=1-frac1k+1$ and it easy to show that $S_k to 1$.
– gimusi
Jul 17 at 21:20
How did you get that $S_k = 1 - dfrac1k + 1$?
– Hat
Jul 17 at 21:21
How did you get that $S_k = 1 - dfrac1k + 1$?
– Hat
Jul 17 at 21:21
@Hat Always by telescoping argument but in that way we can make it more rigorous considering a partial sum and then taking the limit.
– gimusi
Jul 17 at 21:23
@Hat Always by telescoping argument but in that way we can make it more rigorous considering a partial sum and then taking the limit.
– gimusi
Jul 17 at 21:23
 |Â
show 2 more comments
up vote
3
down vote
It is wrong and it could no possibly be right since the limit of the sequence is $0$, not $1$. Perhaps that what you were asked to prove was that $$displaystylesum_n=1^inftyfrac1n(n+1)=1.$$
The error in your proof lies in the (false) inequality$$left|frac1n(n+1)-1right|<left|frac1n(n+1)right|.$$
answered Jul 17 at 21:07
José Carlos Santos
114k1698177
add a comment |Â
up vote
3
down vote
It is wrong and it could no possibly be right since the limit of the sequence is $0$, not $1$. Perhaps that what you were asked to prove was that $$displaystylesum_n=1^inftyfrac1n(n+1)=1.$$
The error in your proof lies in the (false) inequality$$left|frac1n(n+1)-1right|<left|frac1n(n+1)right|.$$
answered Jul 17 at 21:07
José Carlos Santos
114k1698177
add a comment |Â
up vote
3
down vote
up vote
3
down vote
It is wrong and it could no possibly be right since the limit of the sequence is $0$, not $1$. Perhaps that what you were asked to prove was that $$displaystylesum_n=1^inftyfrac1n(n+1)=1.$$
The error in your proof lies in the (false) inequality$$left|frac1n(n+1)-1right|<left|frac1n(n+1)right|.$$
answered Jul 17 at 21:07
José Carlos Santos
114k1698177
It is wrong and it could no possibly be right since the limit of the sequence is $0$, not $1$. Perhaps that what you were asked to prove was that $$displaystylesum_n=1^inftyfrac1n(n+1)=1.$$
The error in your proof lies in the (false) inequality$$left|frac1n(n+1)-1right|<left|frac1n(n+1)right|.$$
answered Jul 17 at 21:07
José Carlos Santos
114k1698177
answered Jul 17 at 21:07
José Carlos Santos
114k1698177
answered Jul 17 at 21:07
José Carlos Santos
114k1698177
answered Jul 17 at 21:07
José Carlos Santos
114k1698177
114k1698177
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2854919%2fproof-that-sum-n-0-infty-dfrac1nn1-converges-to-1%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
But as $n$ grows, $s_nto 0$.
– lulu
Jul 17 at 21:05
Yeah, I was thinking the same. Maybe I'm missing something in my interpretation. Let me copy the problem word for word.
– Hat
Jul 17 at 21:06
1
Specifically, your claim that $big vert frac 1n(n+1)-1big vert <big vert frac 1n(n+1)big vert $ is false for large $n$.
– lulu
Jul 17 at 21:06
1
The sum, $sum_n=1^infty frac 1n(n+1)$ does indeed equal $1$. Hint: expand the summands by partial fractions.
– lulu
Jul 17 at 21:09
Be careful with distributing infinite sums. This is the sum of terms like $frac1n-frac1(n+1)$, but the infinite sums of the individual pieces both diverge to infinity, so you cannot write things as a difference of infinite sums as in your final expression.
– Mark S.
Jul 18 at 1:35