Convergence of this particular series
$begingroup$
Let $a_1=1 $ and $ a_n=2-frac 1n$ for $ngeq 2$. Then $$sum_{n=1}^inftybigg(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}bigg)$$
converges to?
I started by expanding the sum:
$sum_{n=1}^inftybigg(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}bigg)=bigg(frac{1}{a_1^2}-frac{1}{a_2^2}bigg)+bigg(frac{1}{a_2^2}-frac{1}{a_3^2}bigg)+bigg(frac{1}{a_3^2}-frac{1}{a_4^2}bigg)+ldots$
$=frac{1}{a_1^2}+bigg(-frac{1}{a_2^2}+frac{1}{a_2^2}bigg)+bigg(-frac{1}{a_3^2}+frac{1}{a_3^2}bigg)+bigg(-frac{1}{a_4^2}+frac{1}{a_4^2}bigg)+ldots$
$=frac{1}{a_1^2}$
$=1$
Now, the strange part is that the answer to this question is known to me as $0.75$. Is there something wrong with what I did? How does one arrive at $0.75$ as a solution?
PS: I am aware that infinite sums can behave in strange ways, and sometimes more than one solution may follow logically. Is this one such case?
sequences-and-series convergence
$endgroup$
add a comment |
$begingroup$
Let $a_1=1 $ and $ a_n=2-frac 1n$ for $ngeq 2$. Then $$sum_{n=1}^inftybigg(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}bigg)$$
converges to?
I started by expanding the sum:
$sum_{n=1}^inftybigg(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}bigg)=bigg(frac{1}{a_1^2}-frac{1}{a_2^2}bigg)+bigg(frac{1}{a_2^2}-frac{1}{a_3^2}bigg)+bigg(frac{1}{a_3^2}-frac{1}{a_4^2}bigg)+ldots$
$=frac{1}{a_1^2}+bigg(-frac{1}{a_2^2}+frac{1}{a_2^2}bigg)+bigg(-frac{1}{a_3^2}+frac{1}{a_3^2}bigg)+bigg(-frac{1}{a_4^2}+frac{1}{a_4^2}bigg)+ldots$
$=frac{1}{a_1^2}$
$=1$
Now, the strange part is that the answer to this question is known to me as $0.75$. Is there something wrong with what I did? How does one arrive at $0.75$ as a solution?
PS: I am aware that infinite sums can behave in strange ways, and sometimes more than one solution may follow logically. Is this one such case?
sequences-and-series convergence
$endgroup$
add a comment |
$begingroup$
Let $a_1=1 $ and $ a_n=2-frac 1n$ for $ngeq 2$. Then $$sum_{n=1}^inftybigg(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}bigg)$$
converges to?
I started by expanding the sum:
$sum_{n=1}^inftybigg(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}bigg)=bigg(frac{1}{a_1^2}-frac{1}{a_2^2}bigg)+bigg(frac{1}{a_2^2}-frac{1}{a_3^2}bigg)+bigg(frac{1}{a_3^2}-frac{1}{a_4^2}bigg)+ldots$
$=frac{1}{a_1^2}+bigg(-frac{1}{a_2^2}+frac{1}{a_2^2}bigg)+bigg(-frac{1}{a_3^2}+frac{1}{a_3^2}bigg)+bigg(-frac{1}{a_4^2}+frac{1}{a_4^2}bigg)+ldots$
$=frac{1}{a_1^2}$
$=1$
Now, the strange part is that the answer to this question is known to me as $0.75$. Is there something wrong with what I did? How does one arrive at $0.75$ as a solution?
PS: I am aware that infinite sums can behave in strange ways, and sometimes more than one solution may follow logically. Is this one such case?
sequences-and-series convergence
$endgroup$
Let $a_1=1 $ and $ a_n=2-frac 1n$ for $ngeq 2$. Then $$sum_{n=1}^inftybigg(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}bigg)$$
converges to?
I started by expanding the sum:
$sum_{n=1}^inftybigg(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}bigg)=bigg(frac{1}{a_1^2}-frac{1}{a_2^2}bigg)+bigg(frac{1}{a_2^2}-frac{1}{a_3^2}bigg)+bigg(frac{1}{a_3^2}-frac{1}{a_4^2}bigg)+ldots$
$=frac{1}{a_1^2}+bigg(-frac{1}{a_2^2}+frac{1}{a_2^2}bigg)+bigg(-frac{1}{a_3^2}+frac{1}{a_3^2}bigg)+bigg(-frac{1}{a_4^2}+frac{1}{a_4^2}bigg)+ldots$
$=frac{1}{a_1^2}$
$=1$
Now, the strange part is that the answer to this question is known to me as $0.75$. Is there something wrong with what I did? How does one arrive at $0.75$ as a solution?
PS: I am aware that infinite sums can behave in strange ways, and sometimes more than one solution may follow logically. Is this one such case?
sequences-and-series convergence
sequences-and-series convergence
asked 1 hour ago
s0ulr3aper07s0ulr3aper07
3069
3069
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Let $$S_N=sum_{n=1}^Nleft(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}right)=frac{1}{a_1^2}-frac{1}{a_{N+1}^2}.$$
Since $a_1=1$ and $a_nto 2$, we have
$$lim_{Ntoinfty}S_N=1-frac{1}{2^2}=frac{3}{4}.$$
$endgroup$
add a comment |
$begingroup$
$a_n$ does not approach zero,
so the end term can not
be disregarded.
In other words,
the sum up to $n$ is
$frac1{a_1^2}
-frac1{a_n^2}
to 1-frac14
=frac34
$.
$endgroup$
add a comment |
$begingroup$
Infinite sums can behave in strange ways
True!
Sometimes more than one solution may follow logically
Definitely not true. There is one answer in every case (as long as you allow "the sum diverges" as an answer). Sure, there are subtle errors which may make it appear that a sum has more than one value, but that's exactly what they are - errors.
To avoid these errors, don't manipulate infinite sums unless you have a conclusive proof that the manipulation is justified. If you don't, go back to the definition. In your case this says
$$eqalign{sum_{n=1}^inftybigg(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}bigg)
&=lim_{Ntoinfty}sum_{n=1}^Nbigg(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}bigg)cr
&=lim_{Ntoinfty}biggl(frac{1}{a_1^2}-frac{1}{a_{N+1}^2}biggr) .cr}$$
See if you can finish the job from here.
$endgroup$
$begingroup$
I stated that more than one solution may follow logically because of Grandi's series where 0, 1, and half are possible answers. Although, I suppose you could simply call the series divergent and completely avoid the controversy altogether.
$endgroup$
– s0ulr3aper07
28 mins ago
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $$S_N=sum_{n=1}^Nleft(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}right)=frac{1}{a_1^2}-frac{1}{a_{N+1}^2}.$$
Since $a_1=1$ and $a_nto 2$, we have
$$lim_{Ntoinfty}S_N=1-frac{1}{2^2}=frac{3}{4}.$$
$endgroup$
add a comment |
$begingroup$
Let $$S_N=sum_{n=1}^Nleft(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}right)=frac{1}{a_1^2}-frac{1}{a_{N+1}^2}.$$
Since $a_1=1$ and $a_nto 2$, we have
$$lim_{Ntoinfty}S_N=1-frac{1}{2^2}=frac{3}{4}.$$
$endgroup$
add a comment |
$begingroup$
Let $$S_N=sum_{n=1}^Nleft(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}right)=frac{1}{a_1^2}-frac{1}{a_{N+1}^2}.$$
Since $a_1=1$ and $a_nto 2$, we have
$$lim_{Ntoinfty}S_N=1-frac{1}{2^2}=frac{3}{4}.$$
$endgroup$
Let $$S_N=sum_{n=1}^Nleft(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}right)=frac{1}{a_1^2}-frac{1}{a_{N+1}^2}.$$
Since $a_1=1$ and $a_nto 2$, we have
$$lim_{Ntoinfty}S_N=1-frac{1}{2^2}=frac{3}{4}.$$
answered 58 mins ago
Eclipse SunEclipse Sun
7,3841437
7,3841437
add a comment |
add a comment |
$begingroup$
$a_n$ does not approach zero,
so the end term can not
be disregarded.
In other words,
the sum up to $n$ is
$frac1{a_1^2}
-frac1{a_n^2}
to 1-frac14
=frac34
$.
$endgroup$
add a comment |
$begingroup$
$a_n$ does not approach zero,
so the end term can not
be disregarded.
In other words,
the sum up to $n$ is
$frac1{a_1^2}
-frac1{a_n^2}
to 1-frac14
=frac34
$.
$endgroup$
add a comment |
$begingroup$
$a_n$ does not approach zero,
so the end term can not
be disregarded.
In other words,
the sum up to $n$ is
$frac1{a_1^2}
-frac1{a_n^2}
to 1-frac14
=frac34
$.
$endgroup$
$a_n$ does not approach zero,
so the end term can not
be disregarded.
In other words,
the sum up to $n$ is
$frac1{a_1^2}
-frac1{a_n^2}
to 1-frac14
=frac34
$.
answered 55 mins ago
marty cohenmarty cohen
73.4k549128
73.4k549128
add a comment |
add a comment |
$begingroup$
Infinite sums can behave in strange ways
True!
Sometimes more than one solution may follow logically
Definitely not true. There is one answer in every case (as long as you allow "the sum diverges" as an answer). Sure, there are subtle errors which may make it appear that a sum has more than one value, but that's exactly what they are - errors.
To avoid these errors, don't manipulate infinite sums unless you have a conclusive proof that the manipulation is justified. If you don't, go back to the definition. In your case this says
$$eqalign{sum_{n=1}^inftybigg(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}bigg)
&=lim_{Ntoinfty}sum_{n=1}^Nbigg(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}bigg)cr
&=lim_{Ntoinfty}biggl(frac{1}{a_1^2}-frac{1}{a_{N+1}^2}biggr) .cr}$$
See if you can finish the job from here.
$endgroup$
$begingroup$
I stated that more than one solution may follow logically because of Grandi's series where 0, 1, and half are possible answers. Although, I suppose you could simply call the series divergent and completely avoid the controversy altogether.
$endgroup$
– s0ulr3aper07
28 mins ago
add a comment |
$begingroup$
Infinite sums can behave in strange ways
True!
Sometimes more than one solution may follow logically
Definitely not true. There is one answer in every case (as long as you allow "the sum diverges" as an answer). Sure, there are subtle errors which may make it appear that a sum has more than one value, but that's exactly what they are - errors.
To avoid these errors, don't manipulate infinite sums unless you have a conclusive proof that the manipulation is justified. If you don't, go back to the definition. In your case this says
$$eqalign{sum_{n=1}^inftybigg(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}bigg)
&=lim_{Ntoinfty}sum_{n=1}^Nbigg(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}bigg)cr
&=lim_{Ntoinfty}biggl(frac{1}{a_1^2}-frac{1}{a_{N+1}^2}biggr) .cr}$$
See if you can finish the job from here.
$endgroup$
$begingroup$
I stated that more than one solution may follow logically because of Grandi's series where 0, 1, and half are possible answers. Although, I suppose you could simply call the series divergent and completely avoid the controversy altogether.
$endgroup$
– s0ulr3aper07
28 mins ago
add a comment |
$begingroup$
Infinite sums can behave in strange ways
True!
Sometimes more than one solution may follow logically
Definitely not true. There is one answer in every case (as long as you allow "the sum diverges" as an answer). Sure, there are subtle errors which may make it appear that a sum has more than one value, but that's exactly what they are - errors.
To avoid these errors, don't manipulate infinite sums unless you have a conclusive proof that the manipulation is justified. If you don't, go back to the definition. In your case this says
$$eqalign{sum_{n=1}^inftybigg(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}bigg)
&=lim_{Ntoinfty}sum_{n=1}^Nbigg(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}bigg)cr
&=lim_{Ntoinfty}biggl(frac{1}{a_1^2}-frac{1}{a_{N+1}^2}biggr) .cr}$$
See if you can finish the job from here.
$endgroup$
Infinite sums can behave in strange ways
True!
Sometimes more than one solution may follow logically
Definitely not true. There is one answer in every case (as long as you allow "the sum diverges" as an answer). Sure, there are subtle errors which may make it appear that a sum has more than one value, but that's exactly what they are - errors.
To avoid these errors, don't manipulate infinite sums unless you have a conclusive proof that the manipulation is justified. If you don't, go back to the definition. In your case this says
$$eqalign{sum_{n=1}^inftybigg(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}bigg)
&=lim_{Ntoinfty}sum_{n=1}^Nbigg(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}bigg)cr
&=lim_{Ntoinfty}biggl(frac{1}{a_1^2}-frac{1}{a_{N+1}^2}biggr) .cr}$$
See if you can finish the job from here.
answered 55 mins ago
DavidDavid
68.1k664126
68.1k664126
$begingroup$
I stated that more than one solution may follow logically because of Grandi's series where 0, 1, and half are possible answers. Although, I suppose you could simply call the series divergent and completely avoid the controversy altogether.
$endgroup$
– s0ulr3aper07
28 mins ago
add a comment |
$begingroup$
I stated that more than one solution may follow logically because of Grandi's series where 0, 1, and half are possible answers. Although, I suppose you could simply call the series divergent and completely avoid the controversy altogether.
$endgroup$
– s0ulr3aper07
28 mins ago
$begingroup$
I stated that more than one solution may follow logically because of Grandi's series where 0, 1, and half are possible answers. Although, I suppose you could simply call the series divergent and completely avoid the controversy altogether.
$endgroup$
– s0ulr3aper07
28 mins ago
$begingroup$
I stated that more than one solution may follow logically because of Grandi's series where 0, 1, and half are possible answers. Although, I suppose you could simply call the series divergent and completely avoid the controversy altogether.
$endgroup$
– s0ulr3aper07
28 mins ago
add a comment |
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