Proving that the sum of the difference of square roots of partial sums diverges
Let $(a_n)_{n∈N}$ be a sequence of positive numbers which tends to zero but such that $sum_{n=1}^infty a_n$ diverges. Let $(A_n)_{n∈N}$ be the sequence of partial sums $A_n=sum_{k=1}^n a_k$,and let $b_{n+1}=sqrt{A_{n+1}}−sqrt{A_{n}}$. Show that $lim_{n→∞}frac{b_n}{a_n}= 0$ but that $sum b_n$ is still divergent.
Question 1:
I was able to answer the first part, since
$$frac{b_{n+1}}{a_{n+1}} = frac{sqrt{A_{n+1}}−sqrt{A_{n}}}{a_{n+1}} = frac{A_{n+1}-A_n}{a_{n+1}(sqrt{A_{n+1}}+sqrt{A_{n}})} = frac{1}{sqrt{A_{n+1}}+sqrt{A_{n}}}$$ which converges to $0$ since $sqrt{A_n}$ tends to infinity.
The second part, I have no idea about. Apparently $$sum b_k = sqrt{A_n}$$ Why is this true?
Question 2: It is also said that
In this sense there is no ‘smallest’ divergent series, and one can similarly show that there is no ‘largest’ convergent one.
I don't see why this follows. What about our findings conveys these ideas?
real-analysis sequences-and-series convergence
asked Oct 17 at 23:09
Tiwa Aina
2,676421
add a comment |
Let $(a_n)_{n∈N}$ be a sequence of positive numbers which tends to zero but such that $sum_{n=1}^infty a_n$ diverges. Let $(A_n)_{n∈N}$ be the sequence of partial sums $A_n=sum_{k=1}^n a_k$,and let $b_{n+1}=sqrt{A_{n+1}}−sqrt{A_{n}}$. Show that $lim_{n→∞}frac{b_n}{a_n}= 0$ but that $sum b_n$ is still divergent.
Question 1:
I was able to answer the first part, since
$$frac{b_{n+1}}{a_{n+1}} = frac{sqrt{A_{n+1}}−sqrt{A_{n}}}{a_{n+1}} = frac{A_{n+1}-A_n}{a_{n+1}(sqrt{A_{n+1}}+sqrt{A_{n}})} = frac{1}{sqrt{A_{n+1}}+sqrt{A_{n}}}$$ which converges to $0$ since $sqrt{A_n}$ tends to infinity.
The second part, I have no idea about. Apparently $$sum b_k = sqrt{A_n}$$ Why is this true?
Question 2: It is also said that
In this sense there is no ‘smallest’ divergent series, and one can similarly show that there is no ‘largest’ convergent one.
I don't see why this follows. What about our findings conveys these ideas?
real-analysis sequences-and-series convergence
asked Oct 17 at 23:09
Tiwa Aina
2,676421
1
For question 1) just write down $b_1+b_2+cdots +b_N$ and observe that lots of terms cancel out.
– Kavi Rama Murthy
Oct 17 at 23:21
add a comment |
Let $(a_n)_{n∈N}$ be a sequence of positive numbers which tends to zero but such that $sum_{n=1}^infty a_n$ diverges. Let $(A_n)_{n∈N}$ be the sequence of partial sums $A_n=sum_{k=1}^n a_k$,and let $b_{n+1}=sqrt{A_{n+1}}−sqrt{A_{n}}$. Show that $lim_{n→∞}frac{b_n}{a_n}= 0$ but that $sum b_n$ is still divergent.
Question 1:
I was able to answer the first part, since
$$frac{b_{n+1}}{a_{n+1}} = frac{sqrt{A_{n+1}}−sqrt{A_{n}}}{a_{n+1}} = frac{A_{n+1}-A_n}{a_{n+1}(sqrt{A_{n+1}}+sqrt{A_{n}})} = frac{1}{sqrt{A_{n+1}}+sqrt{A_{n}}}$$ which converges to $0$ since $sqrt{A_n}$ tends to infinity.
The second part, I have no idea about. Apparently $$sum b_k = sqrt{A_n}$$ Why is this true?
Question 2: It is also said that
In this sense there is no ‘smallest’ divergent series, and one can similarly show that there is no ‘largest’ convergent one.
I don't see why this follows. What about our findings conveys these ideas?
real-analysis sequences-and-series convergence
asked Oct 17 at 23:09
Tiwa Aina
2,676421
Let $(a_n)_{n∈N}$ be a sequence of positive numbers which tends to zero but such that $sum_{n=1}^infty a_n$ diverges. Let $(A_n)_{n∈N}$ be the sequence of partial sums $A_n=sum_{k=1}^n a_k$,and let $b_{n+1}=sqrt{A_{n+1}}−sqrt{A_{n}}$. Show that $lim_{n→∞}frac{b_n}{a_n}= 0$ but that $sum b_n$ is still divergent.
Question 1:
I was able to answer the first part, since
$$frac{b_{n+1}}{a_{n+1}} = frac{sqrt{A_{n+1}}−sqrt{A_{n}}}{a_{n+1}} = frac{A_{n+1}-A_n}{a_{n+1}(sqrt{A_{n+1}}+sqrt{A_{n}})} = frac{1}{sqrt{A_{n+1}}+sqrt{A_{n}}}$$ which converges to $0$ since $sqrt{A_n}$ tends to infinity.
The second part, I have no idea about. Apparently $$sum b_k = sqrt{A_n}$$ Why is this true?
Question 2: It is also said that
In this sense there is no ‘smallest’ divergent series, and one can similarly show that there is no ‘largest’ convergent one.
I don't see why this follows. What about our findings conveys these ideas?
real-analysis sequences-and-series convergence
real-analysis sequences-and-series convergence
asked Oct 17 at 23:09
Tiwa Aina
2,676421
asked Oct 17 at 23:09
Tiwa Aina
2,676421
asked Oct 17 at 23:09
Tiwa Aina
2,676421
asked Oct 17 at 23:09
Tiwa Aina
2,676421
asked Oct 17 at 23:09
Tiwa Aina
2,676421
2,676421
1
For question 1) just write down $b_1+b_2+cdots +b_N$ and observe that lots of terms cancel out.
– Kavi Rama Murthy
Oct 17 at 23:21
add a comment |
1
For question 1) just write down $b_1+b_2+cdots +b_N$ and observe that lots of terms cancel out.
– Kavi Rama Murthy
Oct 17 at 23:21
1
1
For question 1) just write down $b_1+b_2+cdots +b_N$ and observe that lots of terms cancel out.
– Kavi Rama Murthy
Oct 17 at 23:21
For question 1) just write down $b_1+b_2+cdots +b_N$ and observe that lots of terms cancel out.
– Kavi Rama Murthy
Oct 17 at 23:21
add a comment |
1 Answer
1
active
oldest
votes
We have$$sum_{k=1}^{n}b_{k+1}{=sum_{k=1}^{n}sqrt{A_{k+1}}-sqrt{A_k}\=sqrt{A_{2}}-sqrt{A_1}\+sqrt{A_{3}}-sqrt{2}\+sqrt{A_{4}}-sqrt{A_3}\+cdots\+sqrt{A_{n+1}}-sqrt{A_n}}\=sqrt{A_{n+1}}-sqrt{A_1}$$according to telescopic rule (https://en.wikipedia.org/wiki/Telescoping_series) therefore $sum_{k=1}^{n}b_{k+1}$ diverges since $sqrt{A_{n+1}}-sqrt{A_1}$ diverges.
answered Nov 30 at 19:26
Mostafa Ayaz
13.7k3836
add a comment |
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1 Answer
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1 Answer
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oldest
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We have$$sum_{k=1}^{n}b_{k+1}{=sum_{k=1}^{n}sqrt{A_{k+1}}-sqrt{A_k}\=sqrt{A_{2}}-sqrt{A_1}\+sqrt{A_{3}}-sqrt{2}\+sqrt{A_{4}}-sqrt{A_3}\+cdots\+sqrt{A_{n+1}}-sqrt{A_n}}\=sqrt{A_{n+1}}-sqrt{A_1}$$according to telescopic rule (https://en.wikipedia.org/wiki/Telescoping_series) therefore $sum_{k=1}^{n}b_{k+1}$ diverges since $sqrt{A_{n+1}}-sqrt{A_1}$ diverges.
answered Nov 30 at 19:26
Mostafa Ayaz
13.7k3836
add a comment |
We have$$sum_{k=1}^{n}b_{k+1}{=sum_{k=1}^{n}sqrt{A_{k+1}}-sqrt{A_k}\=sqrt{A_{2}}-sqrt{A_1}\+sqrt{A_{3}}-sqrt{2}\+sqrt{A_{4}}-sqrt{A_3}\+cdots\+sqrt{A_{n+1}}-sqrt{A_n}}\=sqrt{A_{n+1}}-sqrt{A_1}$$according to telescopic rule (https://en.wikipedia.org/wiki/Telescoping_series) therefore $sum_{k=1}^{n}b_{k+1}$ diverges since $sqrt{A_{n+1}}-sqrt{A_1}$ diverges.
answered Nov 30 at 19:26
Mostafa Ayaz
13.7k3836
add a comment |
We have$$sum_{k=1}^{n}b_{k+1}{=sum_{k=1}^{n}sqrt{A_{k+1}}-sqrt{A_k}\=sqrt{A_{2}}-sqrt{A_1}\+sqrt{A_{3}}-sqrt{2}\+sqrt{A_{4}}-sqrt{A_3}\+cdots\+sqrt{A_{n+1}}-sqrt{A_n}}\=sqrt{A_{n+1}}-sqrt{A_1}$$according to telescopic rule (https://en.wikipedia.org/wiki/Telescoping_series) therefore $sum_{k=1}^{n}b_{k+1}$ diverges since $sqrt{A_{n+1}}-sqrt{A_1}$ diverges.
answered Nov 30 at 19:26
Mostafa Ayaz
13.7k3836
We have$$sum_{k=1}^{n}b_{k+1}{=sum_{k=1}^{n}sqrt{A_{k+1}}-sqrt{A_k}\=sqrt{A_{2}}-sqrt{A_1}\+sqrt{A_{3}}-sqrt{2}\+sqrt{A_{4}}-sqrt{A_3}\+cdots\+sqrt{A_{n+1}}-sqrt{A_n}}\=sqrt{A_{n+1}}-sqrt{A_1}$$according to telescopic rule (https://en.wikipedia.org/wiki/Telescoping_series) therefore $sum_{k=1}^{n}b_{k+1}$ diverges since $sqrt{A_{n+1}}-sqrt{A_1}$ diverges.
answered Nov 30 at 19:26
Mostafa Ayaz
13.7k3836
answered Nov 30 at 19:26
Mostafa Ayaz
13.7k3836
answered Nov 30 at 19:26
Mostafa Ayaz
13.7k3836
answered Nov 30 at 19:26
Mostafa Ayaz
13.7k3836
13.7k3836
add a comment |
add a comment |
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For question 1) just write down $b_1+b_2+cdots +b_N$ and observe that lots of terms cancel out.
– Kavi Rama Murthy
Oct 17 at 23:21