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Sum of an infinite series: $1 + frac12 - frac13- frac14 + frac15 + frac16 - cdots$?
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I've been doing a lot of stuff with infinite series and Python lately, and have been getting some very interesting results. For example, the series 1 + 1/16 - 1/256 - 1/4096 + ... (two plus signs, then two minus signs, etc.) converges to 272/257 - wow! But the series 1 + 1/16 + 1/256 + 1/4096 - ... (four alternating plus and minus signs) converges to 69904/65537 - a very interesting result indeed.
Now I want to look at a series that doesn't involve powers of two. For example, I already know that the series 1 + 1/2 + 1/3 + 1/4 + ... does not converge, but the series 1 - 1/2 + 1/3 - 1/4 + ... converges to ln(2), and that's not too hard to prove.
But what about the series 1 + 1/2 - 1/3 - 1/4 + ... ? That has two alternating plus and minus signs. When I ran 1,000,000 iterations of the sequence in Python, I obtained a result approximately equal to 1.131971. I've checked a few things, and it's not ln(3), ln(4), ln(pi), or anything else (it's about ln(3.1017), but as far as I am concerned, the number 3.1017 does not have any significance here). Can someone please help me figure this out? (I don't have any experience with calculus)
Also, I tried a series involving triangular numbers. The series 1 + 1/3 + 1/6 + 1/10 + ... does not seem to converge. What about the series 1 - 1/3 + 1/6 - 1/10 + ... ? There HAS to be some sort of special result to that one, and after 10,000,000 iterations the result stabilized around 0.77258872.
Thanks everyone!
calculus sequences-and-series
edited Aug 6 at 18:08
Clayton
17.9k22882
asked Aug 6 at 17:50
Anonymous
152
add a comment |Â
up vote
1
down vote
favorite
I've been doing a lot of stuff with infinite series and Python lately, and have been getting some very interesting results. For example, the series 1 + 1/16 - 1/256 - 1/4096 + ... (two plus signs, then two minus signs, etc.) converges to 272/257 - wow! But the series 1 + 1/16 + 1/256 + 1/4096 - ... (four alternating plus and minus signs) converges to 69904/65537 - a very interesting result indeed.
Now I want to look at a series that doesn't involve powers of two. For example, I already know that the series 1 + 1/2 + 1/3 + 1/4 + ... does not converge, but the series 1 - 1/2 + 1/3 - 1/4 + ... converges to ln(2), and that's not too hard to prove.
But what about the series 1 + 1/2 - 1/3 - 1/4 + ... ? That has two alternating plus and minus signs. When I ran 1,000,000 iterations of the sequence in Python, I obtained a result approximately equal to 1.131971. I've checked a few things, and it's not ln(3), ln(4), ln(pi), or anything else (it's about ln(3.1017), but as far as I am concerned, the number 3.1017 does not have any significance here). Can someone please help me figure this out? (I don't have any experience with calculus)
Also, I tried a series involving triangular numbers. The series 1 + 1/3 + 1/6 + 1/10 + ... does not seem to converge. What about the series 1 - 1/3 + 1/6 - 1/10 + ... ? There HAS to be some sort of special result to that one, and after 10,000,000 iterations the result stabilized around 0.77258872.
Thanks everyone!
calculus sequences-and-series
edited Aug 6 at 18:08
Clayton
17.9k22882
asked Aug 6 at 17:50
Anonymous
152
3
If you have multiple questions, you should generally make one post per question -- i.e. your question about the series for triangular numbers should probably go in another post, not this one.
– Hurkyl
Aug 6 at 18:00
1
The sum of the reciprocals of the triangular numbers does converge, to a very simple value.
– saulspatz
Aug 6 at 18:19
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I've been doing a lot of stuff with infinite series and Python lately, and have been getting some very interesting results. For example, the series 1 + 1/16 - 1/256 - 1/4096 + ... (two plus signs, then two minus signs, etc.) converges to 272/257 - wow! But the series 1 + 1/16 + 1/256 + 1/4096 - ... (four alternating plus and minus signs) converges to 69904/65537 - a very interesting result indeed.
Now I want to look at a series that doesn't involve powers of two. For example, I already know that the series 1 + 1/2 + 1/3 + 1/4 + ... does not converge, but the series 1 - 1/2 + 1/3 - 1/4 + ... converges to ln(2), and that's not too hard to prove.
But what about the series 1 + 1/2 - 1/3 - 1/4 + ... ? That has two alternating plus and minus signs. When I ran 1,000,000 iterations of the sequence in Python, I obtained a result approximately equal to 1.131971. I've checked a few things, and it's not ln(3), ln(4), ln(pi), or anything else (it's about ln(3.1017), but as far as I am concerned, the number 3.1017 does not have any significance here). Can someone please help me figure this out? (I don't have any experience with calculus)
Also, I tried a series involving triangular numbers. The series 1 + 1/3 + 1/6 + 1/10 + ... does not seem to converge. What about the series 1 - 1/3 + 1/6 - 1/10 + ... ? There HAS to be some sort of special result to that one, and after 10,000,000 iterations the result stabilized around 0.77258872.
Thanks everyone!
calculus sequences-and-series
edited Aug 6 at 18:08
Clayton
17.9k22882
asked Aug 6 at 17:50
Anonymous
152
I've been doing a lot of stuff with infinite series and Python lately, and have been getting some very interesting results. For example, the series 1 + 1/16 - 1/256 - 1/4096 + ... (two plus signs, then two minus signs, etc.) converges to 272/257 - wow! But the series 1 + 1/16 + 1/256 + 1/4096 - ... (four alternating plus and minus signs) converges to 69904/65537 - a very interesting result indeed.
Now I want to look at a series that doesn't involve powers of two. For example, I already know that the series 1 + 1/2 + 1/3 + 1/4 + ... does not converge, but the series 1 - 1/2 + 1/3 - 1/4 + ... converges to ln(2), and that's not too hard to prove.
But what about the series 1 + 1/2 - 1/3 - 1/4 + ... ? That has two alternating plus and minus signs. When I ran 1,000,000 iterations of the sequence in Python, I obtained a result approximately equal to 1.131971. I've checked a few things, and it's not ln(3), ln(4), ln(pi), or anything else (it's about ln(3.1017), but as far as I am concerned, the number 3.1017 does not have any significance here). Can someone please help me figure this out? (I don't have any experience with calculus)
Also, I tried a series involving triangular numbers. The series 1 + 1/3 + 1/6 + 1/10 + ... does not seem to converge. What about the series 1 - 1/3 + 1/6 - 1/10 + ... ? There HAS to be some sort of special result to that one, and after 10,000,000 iterations the result stabilized around 0.77258872.
Thanks everyone!
calculus sequences-and-series
edited Aug 6 at 18:08
Clayton
17.9k22882
asked Aug 6 at 17:50
Anonymous
152
edited Aug 6 at 18:08
Clayton
17.9k22882
edited Aug 6 at 18:08
Clayton
17.9k22882
edited Aug 6 at 18:08
Clayton
17.9k22882
17.9k22882
asked Aug 6 at 17:50
Anonymous
152
asked Aug 6 at 17:50
Anonymous
152
asked Aug 6 at 17:50
Anonymous
152
152
3
If you have multiple questions, you should generally make one post per question -- i.e. your question about the series for triangular numbers should probably go in another post, not this one.
– Hurkyl
Aug 6 at 18:00
1
The sum of the reciprocals of the triangular numbers does converge, to a very simple value.
– saulspatz
Aug 6 at 18:19
add a comment |Â
3
If you have multiple questions, you should generally make one post per question -- i.e. your question about the series for triangular numbers should probably go in another post, not this one.
– Hurkyl
Aug 6 at 18:00
1
The sum of the reciprocals of the triangular numbers does converge, to a very simple value.
– saulspatz
Aug 6 at 18:19
3
3
If you have multiple questions, you should generally make one post per question -- i.e. your question about the series for triangular numbers should probably go in another post, not this one.
– Hurkyl
Aug 6 at 18:00
If you have multiple questions, you should generally make one post per question -- i.e. your question about the series for triangular numbers should probably go in another post, not this one.
– Hurkyl
Aug 6 at 18:00
1
1
The sum of the reciprocals of the triangular numbers does converge, to a very simple value.
– saulspatz
Aug 6 at 18:19
The sum of the reciprocals of the triangular numbers does converge, to a very simple value.
– saulspatz
Aug 6 at 18:19
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
10
down vote
If the pattern of signs is $++--$, we are looking for
$$ sum_kgeq 0left(frac14k+1+frac14k+2-frac14k+3-frac14k+4right)=sum_kgeq 0int_0^1x^4k(1+x-x^2-x^3),dx $$
or
$$ int_0^1frac(1+x)(1-x^2)1-x^4,dx = int_0^1frac1+x1+x^2,dx = colorredfracpi4+frac12log 2.$$
About reciprocal triangular numbers, the series $sum_ngeq 1frac2n(n+1)$ does converge by the $p$-test, and it converges to $colorred2$ since $frac2n(n+1)$ telescopes as $frac2n-frac2n+1$. By the same approach as before,
$$ sum_ngeq 1frac2(-1)^n+1n(n+1) = 2sum_ngeq 1int_0^1(-1)^n-1x^n-1(1-x),dx=2int_0^1frac1-x1+x,dx $$
equals $colorred4log 2-2$.
answered Aug 6 at 17:53
Jack D'Aurizio♦
270k31266630
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up vote
5
down vote
The given series is equal to $$
left(1-frac13+frac15-frac17+dotsright)+frac12left(1-frac12+frac13-frac14right)=piover4+frac12log2$$
as both sums are familiar.
answered Aug 6 at 17:58
saulspatz
10.7k21323
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up vote
3
down vote
You want
$$
sum_n=1^infty (-1)^n-1 left(frac12n-1+frac12nright)
=
sum_n=1^infty (-1)^n-1 frac12n-1+sum_n=1^infty (-1)^n-1frac12n tag$dagger$
$$
(You can check breaking the sum in two is fine, as both series do converge (conditionally).
Now,
$$
sum_n=1^infty frac12n = frac12log 2,.
$$
For the other term, note that for $|x|< 1$,
$$
sum_n=1^infty (-1)^n-1 fracx^2(n-1)2n-1
= sum_n=0^infty (-1)^n fracx^2n2n+1
= fracarctan xx
$$
recognizing the power series, from which, by Abel's theorem,
$$
sum_n=1^infty (-1)^n-1 frac12n-1
= fracarctan 11 = fracpi4,.
$$
Combining the two, we get from $(dagger)$ that
$$
sum_n=1^infty (-1)^n-1 left(frac12n-1+frac12nright)
= boxedfrac12log 2 + fracpi4,. tag$ddagger$
$$
edited Aug 6 at 18:01
Clayton
17.9k22882
answered Aug 6 at 18:00
Clement C.
47.2k33682
add a comment |Â
up vote
2
down vote
Period four patterns can often be studied by plugging in the fourth roots of unity; i.e. $1, i, -1, -i$. In this case, we can get everything we need from just $i$:
$$ ln(1 + i) = i + frac12 - fraci3 - frac14 + ldots
\ = i left(1 - frac13 + frac15 - ldots right)
+ left(frac12 - frac14 + frac16 - ldots right)$$
So we can use the fact that
$$ ln(1 + i) = ln |1+i| + i arg(1+i)
= ln sqrt2 + i fracpi4 $$
to identify
$$1 - frac13 + frac15 - ldots = fracpi4 $$
$$frac12 - frac14 + frac16 - ldots = ln sqrt2 = frac12 ln 2 $$
to get
$$ 1 + frac12 - frac13 - frac14 + ldots = fracpi4 + frac12 ln 2 $$
answered Aug 6 at 17:57
Hurkyl
108k9112253
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
10
down vote
If the pattern of signs is $++--$, we are looking for
$$ sum_kgeq 0left(frac14k+1+frac14k+2-frac14k+3-frac14k+4right)=sum_kgeq 0int_0^1x^4k(1+x-x^2-x^3),dx $$
or
$$ int_0^1frac(1+x)(1-x^2)1-x^4,dx = int_0^1frac1+x1+x^2,dx = colorredfracpi4+frac12log 2.$$
About reciprocal triangular numbers, the series $sum_ngeq 1frac2n(n+1)$ does converge by the $p$-test, and it converges to $colorred2$ since $frac2n(n+1)$ telescopes as $frac2n-frac2n+1$. By the same approach as before,
$$ sum_ngeq 1frac2(-1)^n+1n(n+1) = 2sum_ngeq 1int_0^1(-1)^n-1x^n-1(1-x),dx=2int_0^1frac1-x1+x,dx $$
equals $colorred4log 2-2$.
answered Aug 6 at 17:53
Jack D'Aurizio♦
270k31266630
add a comment |Â
up vote
10
down vote
If the pattern of signs is $++--$, we are looking for
$$ sum_kgeq 0left(frac14k+1+frac14k+2-frac14k+3-frac14k+4right)=sum_kgeq 0int_0^1x^4k(1+x-x^2-x^3),dx $$
or
$$ int_0^1frac(1+x)(1-x^2)1-x^4,dx = int_0^1frac1+x1+x^2,dx = colorredfracpi4+frac12log 2.$$
About reciprocal triangular numbers, the series $sum_ngeq 1frac2n(n+1)$ does converge by the $p$-test, and it converges to $colorred2$ since $frac2n(n+1)$ telescopes as $frac2n-frac2n+1$. By the same approach as before,
$$ sum_ngeq 1frac2(-1)^n+1n(n+1) = 2sum_ngeq 1int_0^1(-1)^n-1x^n-1(1-x),dx=2int_0^1frac1-x1+x,dx $$
equals $colorred4log 2-2$.
answered Aug 6 at 17:53
Jack D'Aurizio♦
270k31266630
add a comment |Â
up vote
10
down vote
up vote
10
down vote
If the pattern of signs is $++--$, we are looking for
$$ sum_kgeq 0left(frac14k+1+frac14k+2-frac14k+3-frac14k+4right)=sum_kgeq 0int_0^1x^4k(1+x-x^2-x^3),dx $$
or
$$ int_0^1frac(1+x)(1-x^2)1-x^4,dx = int_0^1frac1+x1+x^2,dx = colorredfracpi4+frac12log 2.$$
About reciprocal triangular numbers, the series $sum_ngeq 1frac2n(n+1)$ does converge by the $p$-test, and it converges to $colorred2$ since $frac2n(n+1)$ telescopes as $frac2n-frac2n+1$. By the same approach as before,
$$ sum_ngeq 1frac2(-1)^n+1n(n+1) = 2sum_ngeq 1int_0^1(-1)^n-1x^n-1(1-x),dx=2int_0^1frac1-x1+x,dx $$
equals $colorred4log 2-2$.
answered Aug 6 at 17:53
Jack D'Aurizio♦
270k31266630
If the pattern of signs is $++--$, we are looking for
$$ sum_kgeq 0left(frac14k+1+frac14k+2-frac14k+3-frac14k+4right)=sum_kgeq 0int_0^1x^4k(1+x-x^2-x^3),dx $$
or
$$ int_0^1frac(1+x)(1-x^2)1-x^4,dx = int_0^1frac1+x1+x^2,dx = colorredfracpi4+frac12log 2.$$
About reciprocal triangular numbers, the series $sum_ngeq 1frac2n(n+1)$ does converge by the $p$-test, and it converges to $colorred2$ since $frac2n(n+1)$ telescopes as $frac2n-frac2n+1$. By the same approach as before,
$$ sum_ngeq 1frac2(-1)^n+1n(n+1) = 2sum_ngeq 1int_0^1(-1)^n-1x^n-1(1-x),dx=2int_0^1frac1-x1+x,dx $$
equals $colorred4log 2-2$.
answered Aug 6 at 17:53
Jack D'Aurizio♦
270k31266630
edited Aug 6 at 18:07
answered Aug 6 at 17:53
Jack D'Aurizio♦
270k31266630
answered Aug 6 at 17:53
Jack D'Aurizio♦
270k31266630
answered Aug 6 at 17:53
Jack D'Aurizio♦
270k31266630
270k31266630
add a comment |Â
add a comment |Â
up vote
5
down vote
The given series is equal to $$
left(1-frac13+frac15-frac17+dotsright)+frac12left(1-frac12+frac13-frac14right)=piover4+frac12log2$$
as both sums are familiar.
answered Aug 6 at 17:58
saulspatz
10.7k21323
add a comment |Â
up vote
5
down vote
The given series is equal to $$
left(1-frac13+frac15-frac17+dotsright)+frac12left(1-frac12+frac13-frac14right)=piover4+frac12log2$$
as both sums are familiar.
answered Aug 6 at 17:58
saulspatz
10.7k21323
add a comment |Â
up vote
5
down vote
up vote
5
down vote
The given series is equal to $$
left(1-frac13+frac15-frac17+dotsright)+frac12left(1-frac12+frac13-frac14right)=piover4+frac12log2$$
as both sums are familiar.
answered Aug 6 at 17:58
saulspatz
10.7k21323
The given series is equal to $$
left(1-frac13+frac15-frac17+dotsright)+frac12left(1-frac12+frac13-frac14right)=piover4+frac12log2$$
as both sums are familiar.
answered Aug 6 at 17:58
saulspatz
10.7k21323
answered Aug 6 at 17:58
saulspatz
10.7k21323
answered Aug 6 at 17:58
saulspatz
10.7k21323
answered Aug 6 at 17:58
saulspatz
10.7k21323
10.7k21323
add a comment |Â
add a comment |Â
up vote
3
down vote
You want
$$
sum_n=1^infty (-1)^n-1 left(frac12n-1+frac12nright)
=
sum_n=1^infty (-1)^n-1 frac12n-1+sum_n=1^infty (-1)^n-1frac12n tag$dagger$
$$
(You can check breaking the sum in two is fine, as both series do converge (conditionally).
Now,
$$
sum_n=1^infty frac12n = frac12log 2,.
$$
For the other term, note that for $|x|< 1$,
$$
sum_n=1^infty (-1)^n-1 fracx^2(n-1)2n-1
= sum_n=0^infty (-1)^n fracx^2n2n+1
= fracarctan xx
$$
recognizing the power series, from which, by Abel's theorem,
$$
sum_n=1^infty (-1)^n-1 frac12n-1
= fracarctan 11 = fracpi4,.
$$
Combining the two, we get from $(dagger)$ that
$$
sum_n=1^infty (-1)^n-1 left(frac12n-1+frac12nright)
= boxedfrac12log 2 + fracpi4,. tag$ddagger$
$$
edited Aug 6 at 18:01
Clayton
17.9k22882
answered Aug 6 at 18:00
Clement C.
47.2k33682
add a comment |Â
up vote
3
down vote
You want
$$
sum_n=1^infty (-1)^n-1 left(frac12n-1+frac12nright)
=
sum_n=1^infty (-1)^n-1 frac12n-1+sum_n=1^infty (-1)^n-1frac12n tag$dagger$
$$
(You can check breaking the sum in two is fine, as both series do converge (conditionally).
Now,
$$
sum_n=1^infty frac12n = frac12log 2,.
$$
For the other term, note that for $|x|< 1$,
$$
sum_n=1^infty (-1)^n-1 fracx^2(n-1)2n-1
= sum_n=0^infty (-1)^n fracx^2n2n+1
= fracarctan xx
$$
recognizing the power series, from which, by Abel's theorem,
$$
sum_n=1^infty (-1)^n-1 frac12n-1
= fracarctan 11 = fracpi4,.
$$
Combining the two, we get from $(dagger)$ that
$$
sum_n=1^infty (-1)^n-1 left(frac12n-1+frac12nright)
= boxedfrac12log 2 + fracpi4,. tag$ddagger$
$$
edited Aug 6 at 18:01
Clayton
17.9k22882
answered Aug 6 at 18:00
Clement C.
47.2k33682
add a comment |Â
up vote
3
down vote
up vote
3
down vote
You want
$$
sum_n=1^infty (-1)^n-1 left(frac12n-1+frac12nright)
=
sum_n=1^infty (-1)^n-1 frac12n-1+sum_n=1^infty (-1)^n-1frac12n tag$dagger$
$$
(You can check breaking the sum in two is fine, as both series do converge (conditionally).
Now,
$$
sum_n=1^infty frac12n = frac12log 2,.
$$
For the other term, note that for $|x|< 1$,
$$
sum_n=1^infty (-1)^n-1 fracx^2(n-1)2n-1
= sum_n=0^infty (-1)^n fracx^2n2n+1
= fracarctan xx
$$
recognizing the power series, from which, by Abel's theorem,
$$
sum_n=1^infty (-1)^n-1 frac12n-1
= fracarctan 11 = fracpi4,.
$$
Combining the two, we get from $(dagger)$ that
$$
sum_n=1^infty (-1)^n-1 left(frac12n-1+frac12nright)
= boxedfrac12log 2 + fracpi4,. tag$ddagger$
$$
edited Aug 6 at 18:01
Clayton
17.9k22882
answered Aug 6 at 18:00
Clement C.
47.2k33682
You want
$$
sum_n=1^infty (-1)^n-1 left(frac12n-1+frac12nright)
=
sum_n=1^infty (-1)^n-1 frac12n-1+sum_n=1^infty (-1)^n-1frac12n tag$dagger$
$$
(You can check breaking the sum in two is fine, as both series do converge (conditionally).
Now,
$$
sum_n=1^infty frac12n = frac12log 2,.
$$
For the other term, note that for $|x|< 1$,
$$
sum_n=1^infty (-1)^n-1 fracx^2(n-1)2n-1
= sum_n=0^infty (-1)^n fracx^2n2n+1
= fracarctan xx
$$
recognizing the power series, from which, by Abel's theorem,
$$
sum_n=1^infty (-1)^n-1 frac12n-1
= fracarctan 11 = fracpi4,.
$$
Combining the two, we get from $(dagger)$ that
$$
sum_n=1^infty (-1)^n-1 left(frac12n-1+frac12nright)
= boxedfrac12log 2 + fracpi4,. tag$ddagger$
$$
edited Aug 6 at 18:01
Clayton
17.9k22882
answered Aug 6 at 18:00
Clement C.
47.2k33682
edited Aug 6 at 18:01
Clayton
17.9k22882
edited Aug 6 at 18:01
Clayton
17.9k22882
edited Aug 6 at 18:01
Clayton
17.9k22882
17.9k22882
answered Aug 6 at 18:00
Clement C.
47.2k33682
answered Aug 6 at 18:00
Clement C.
47.2k33682
answered Aug 6 at 18:00
Clement C.
47.2k33682
47.2k33682
add a comment |Â
add a comment |Â
up vote
2
down vote
Period four patterns can often be studied by plugging in the fourth roots of unity; i.e. $1, i, -1, -i$. In this case, we can get everything we need from just $i$:
$$ ln(1 + i) = i + frac12 - fraci3 - frac14 + ldots
\ = i left(1 - frac13 + frac15 - ldots right)
+ left(frac12 - frac14 + frac16 - ldots right)$$
So we can use the fact that
$$ ln(1 + i) = ln |1+i| + i arg(1+i)
= ln sqrt2 + i fracpi4 $$
to identify
$$1 - frac13 + frac15 - ldots = fracpi4 $$
$$frac12 - frac14 + frac16 - ldots = ln sqrt2 = frac12 ln 2 $$
to get
$$ 1 + frac12 - frac13 - frac14 + ldots = fracpi4 + frac12 ln 2 $$
answered Aug 6 at 17:57
Hurkyl
108k9112253
add a comment |Â
up vote
2
down vote
Period four patterns can often be studied by plugging in the fourth roots of unity; i.e. $1, i, -1, -i$. In this case, we can get everything we need from just $i$:
$$ ln(1 + i) = i + frac12 - fraci3 - frac14 + ldots
\ = i left(1 - frac13 + frac15 - ldots right)
+ left(frac12 - frac14 + frac16 - ldots right)$$
So we can use the fact that
$$ ln(1 + i) = ln |1+i| + i arg(1+i)
= ln sqrt2 + i fracpi4 $$
to identify
$$1 - frac13 + frac15 - ldots = fracpi4 $$
$$frac12 - frac14 + frac16 - ldots = ln sqrt2 = frac12 ln 2 $$
to get
$$ 1 + frac12 - frac13 - frac14 + ldots = fracpi4 + frac12 ln 2 $$
answered Aug 6 at 17:57
Hurkyl
108k9112253
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Period four patterns can often be studied by plugging in the fourth roots of unity; i.e. $1, i, -1, -i$. In this case, we can get everything we need from just $i$:
$$ ln(1 + i) = i + frac12 - fraci3 - frac14 + ldots
\ = i left(1 - frac13 + frac15 - ldots right)
+ left(frac12 - frac14 + frac16 - ldots right)$$
So we can use the fact that
$$ ln(1 + i) = ln |1+i| + i arg(1+i)
= ln sqrt2 + i fracpi4 $$
to identify
$$1 - frac13 + frac15 - ldots = fracpi4 $$
$$frac12 - frac14 + frac16 - ldots = ln sqrt2 = frac12 ln 2 $$
to get
$$ 1 + frac12 - frac13 - frac14 + ldots = fracpi4 + frac12 ln 2 $$
answered Aug 6 at 17:57
Hurkyl
108k9112253
Period four patterns can often be studied by plugging in the fourth roots of unity; i.e. $1, i, -1, -i$. In this case, we can get everything we need from just $i$:
$$ ln(1 + i) = i + frac12 - fraci3 - frac14 + ldots
\ = i left(1 - frac13 + frac15 - ldots right)
+ left(frac12 - frac14 + frac16 - ldots right)$$
So we can use the fact that
$$ ln(1 + i) = ln |1+i| + i arg(1+i)
= ln sqrt2 + i fracpi4 $$
to identify
$$1 - frac13 + frac15 - ldots = fracpi4 $$
$$frac12 - frac14 + frac16 - ldots = ln sqrt2 = frac12 ln 2 $$
to get
$$ 1 + frac12 - frac13 - frac14 + ldots = fracpi4 + frac12 ln 2 $$
answered Aug 6 at 17:57
Hurkyl
108k9112253
edited Aug 6 at 18:15
answered Aug 6 at 17:57
Hurkyl
108k9112253
answered Aug 6 at 17:57
Hurkyl
108k9112253
answered Aug 6 at 17:57
Hurkyl
108k9112253
108k9112253
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3
If you have multiple questions, you should generally make one post per question -- i.e. your question about the series for triangular numbers should probably go in another post, not this one.
– Hurkyl
Aug 6 at 18:00
1
The sum of the reciprocals of the triangular numbers does converge, to a very simple value.
– saulspatz
Aug 6 at 18:19