Mixing
Evaluate $sum_{n=1}^infty arctanleft(frac{1}{8n^2} right)$
$begingroup$
I found this series in Jack D'Aurizio's Superior Mathematics from an Elementary Point of View on his user page. So I've seen similar series to this, so I figured I tried to make it telescope. I managed to write it using the difference formula for $arctanleft(xright)$ so,
begin{align}
sum_{n = 1}^{infty}
arctanleft(frac{1}{8n^{2}}right) & =
sum_{n = 1}^{infty}
left[vphantom{large A}arctanleft(4n + 1right) -arctanleft(4n - 1right)right]
\[1mm] & =
sum_{n = 1}^{infty}left(-1right)^{n}arctanleft(2n + 1right)
end{align}
Writing the series that though does not seem to help since none of the terms cancel with each other. What should I do find the answer ?.
sequences-and-series trigonometry
edited Jan 6 at 20:21
J.G.
24.8k22539
asked Jan 6 at 18:14
Tom HimlerTom Himler
933213
$endgroup$
add a comment |
$begingroup$
I found this series in Jack D'Aurizio's Superior Mathematics from an Elementary Point of View on his user page. So I've seen similar series to this, so I figured I tried to make it telescope. I managed to write it using the difference formula for $arctanleft(xright)$ so,
begin{align}
sum_{n = 1}^{infty}
arctanleft(frac{1}{8n^{2}}right) & =
sum_{n = 1}^{infty}
left[vphantom{large A}arctanleft(4n + 1right) -arctanleft(4n - 1right)right]
\[1mm] & =
sum_{n = 1}^{infty}left(-1right)^{n}arctanleft(2n + 1right)
end{align}
Writing the series that though does not seem to help since none of the terms cancel with each other. What should I do find the answer ?.
sequences-and-series trigonometry
edited Jan 6 at 20:21
J.G.
24.8k22539
asked Jan 6 at 18:14
Tom HimlerTom Himler
933213
$endgroup$
1
$begingroup$
That last sum doesn't converge since the absolute value of the terms tend to $pi/2$ not $0$. The first two sums do converge though.
$endgroup$
– JimmyK4542
Jan 6 at 18:24
$begingroup$
The last equality is not correct...
$endgroup$
– Fabian
Jan 6 at 18:25
$begingroup$
Try this approach for a sutable $x$.
$endgroup$
– A.Γ.
Jan 6 at 18:31
$begingroup$
I see you guy's point. I put that there since Desmos seemed to say it's equal to the original sum, but it definitely diverges.
$endgroup$
– Tom Himler
Jan 6 at 18:45
add a comment |
$begingroup$
I found this series in Jack D'Aurizio's Superior Mathematics from an Elementary Point of View on his user page. So I've seen similar series to this, so I figured I tried to make it telescope. I managed to write it using the difference formula for $arctanleft(xright)$ so,
begin{align}
sum_{n = 1}^{infty}
arctanleft(frac{1}{8n^{2}}right) & =
sum_{n = 1}^{infty}
left[vphantom{large A}arctanleft(4n + 1right) -arctanleft(4n - 1right)right]
\[1mm] & =
sum_{n = 1}^{infty}left(-1right)^{n}arctanleft(2n + 1right)
end{align}
Writing the series that though does not seem to help since none of the terms cancel with each other. What should I do find the answer ?.
sequences-and-series trigonometry
edited Jan 6 at 20:21
J.G.
24.8k22539
asked Jan 6 at 18:14
Tom HimlerTom Himler
933213
$endgroup$
I found this series in Jack D'Aurizio's Superior Mathematics from an Elementary Point of View on his user page. So I've seen similar series to this, so I figured I tried to make it telescope. I managed to write it using the difference formula for $arctanleft(xright)$ so,
begin{align}
sum_{n = 1}^{infty}
arctanleft(frac{1}{8n^{2}}right) & =
sum_{n = 1}^{infty}
left[vphantom{large A}arctanleft(4n + 1right) -arctanleft(4n - 1right)right]
\[1mm] & =
sum_{n = 1}^{infty}left(-1right)^{n}arctanleft(2n + 1right)
end{align}
Writing the series that though does not seem to help since none of the terms cancel with each other. What should I do find the answer ?.
sequences-and-series trigonometry
sequences-and-series trigonometry
edited Jan 6 at 20:21
J.G.
24.8k22539
asked Jan 6 at 18:14
Tom HimlerTom Himler
933213
edited Jan 6 at 20:21
J.G.
24.8k22539
asked Jan 6 at 18:14
Tom HimlerTom Himler
933213
edited Jan 6 at 20:21
J.G.
24.8k22539
edited Jan 6 at 20:21
J.G.
24.8k22539
edited Jan 6 at 20:21
J.G.
24.8k22539
24.8k22539
asked Jan 6 at 18:14
Tom HimlerTom Himler
933213
asked Jan 6 at 18:14
Tom HimlerTom Himler
933213
asked Jan 6 at 18:14
Tom HimlerTom Himler
933213
933213
1
$begingroup$
That last sum doesn't converge since the absolute value of the terms tend to $pi/2$ not $0$. The first two sums do converge though.
$endgroup$
– JimmyK4542
Jan 6 at 18:24
$begingroup$
The last equality is not correct...
$endgroup$
– Fabian
Jan 6 at 18:25
$begingroup$
Try this approach for a sutable $x$.
$endgroup$
– A.Γ.
Jan 6 at 18:31
$begingroup$
I see you guy's point. I put that there since Desmos seemed to say it's equal to the original sum, but it definitely diverges.
$endgroup$
– Tom Himler
Jan 6 at 18:45
add a comment |
1
$begingroup$
That last sum doesn't converge since the absolute value of the terms tend to $pi/2$ not $0$. The first two sums do converge though.
$endgroup$
– JimmyK4542
Jan 6 at 18:24
$begingroup$
The last equality is not correct...
$endgroup$
– Fabian
Jan 6 at 18:25
$begingroup$
Try this approach for a sutable $x$.
$endgroup$
– A.Γ.
Jan 6 at 18:31
$begingroup$
I see you guy's point. I put that there since Desmos seemed to say it's equal to the original sum, but it definitely diverges.
$endgroup$
– Tom Himler
Jan 6 at 18:45
1
1
$begingroup$
That last sum doesn't converge since the absolute value of the terms tend to $pi/2$ not $0$. The first two sums do converge though.
$endgroup$
– JimmyK4542
Jan 6 at 18:24
$begingroup$
That last sum doesn't converge since the absolute value of the terms tend to $pi/2$ not $0$. The first two sums do converge though.
$endgroup$
– JimmyK4542
Jan 6 at 18:24
$begingroup$
The last equality is not correct...
$endgroup$
– Fabian
Jan 6 at 18:25
$begingroup$
The last equality is not correct...
$endgroup$
– Fabian
Jan 6 at 18:25
$begingroup$
Try this approach for a sutable $x$.
$endgroup$
– A.Γ.
Jan 6 at 18:31
$begingroup$
Try this approach for a sutable $x$.
$endgroup$
– A.Γ.
Jan 6 at 18:31
$begingroup$
I see you guy's point. I put that there since Desmos seemed to say it's equal to the original sum, but it definitely diverges.
$endgroup$
– Tom Himler
Jan 6 at 18:45
$begingroup$
I see you guy's point. I put that there since Desmos seemed to say it's equal to the original sum, but it definitely diverges.
$endgroup$
– Tom Himler
Jan 6 at 18:45
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We can compute this sum by considering it as the argument of an infinite product. Here the argument of the nth factor of the product is $-text{arctan}(1/8n^2)$.
$$begin{align*}
sum_{n=1}^inftytext{arctan}left(frac{1}{8n^2}right) &=-text{arg}prod_{n=1}^inftyleft(1-frac{i}{8n^2}right) \
&= -text{arg}prod_{n=1}^inftyleft(1-frac{left(sqrt{i/8}right)^2}{n^2}right)\
&=-text{arg}frac{sinleft(pisqrt{i/8}right)}{pisqrt{i/8}},qquadtext{by Euler's product for the sine} \
&=-text{arg}left(frac{(1-i)sqrt{2}cosh(pi/4)}{pi}+frac{(1+i)sqrt{2}sinh(pi/4)}{pi}right) \
&=-text{arctan}left(frac{sinh(pi/4)-cosh(pi/4)}{sinh(pi/4)+cosh(pi/4)}right) \
&= text{arccot}left(e^{pi/2}right).
end{align*}$$
answered Jan 6 at 18:40
dxdydzdxdydz
4971411
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add a comment |
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$begingroup$
We can compute this sum by considering it as the argument of an infinite product. Here the argument of the nth factor of the product is $-text{arctan}(1/8n^2)$.
$$begin{align*}
sum_{n=1}^inftytext{arctan}left(frac{1}{8n^2}right) &=-text{arg}prod_{n=1}^inftyleft(1-frac{i}{8n^2}right) \
&= -text{arg}prod_{n=1}^inftyleft(1-frac{left(sqrt{i/8}right)^2}{n^2}right)\
&=-text{arg}frac{sinleft(pisqrt{i/8}right)}{pisqrt{i/8}},qquadtext{by Euler's product for the sine} \
&=-text{arg}left(frac{(1-i)sqrt{2}cosh(pi/4)}{pi}+frac{(1+i)sqrt{2}sinh(pi/4)}{pi}right) \
&=-text{arctan}left(frac{sinh(pi/4)-cosh(pi/4)}{sinh(pi/4)+cosh(pi/4)}right) \
&= text{arccot}left(e^{pi/2}right).
end{align*}$$
answered Jan 6 at 18:40
dxdydzdxdydz
4971411
$endgroup$
add a comment |
$begingroup$
We can compute this sum by considering it as the argument of an infinite product. Here the argument of the nth factor of the product is $-text{arctan}(1/8n^2)$.
$$begin{align*}
sum_{n=1}^inftytext{arctan}left(frac{1}{8n^2}right) &=-text{arg}prod_{n=1}^inftyleft(1-frac{i}{8n^2}right) \
&= -text{arg}prod_{n=1}^inftyleft(1-frac{left(sqrt{i/8}right)^2}{n^2}right)\
&=-text{arg}frac{sinleft(pisqrt{i/8}right)}{pisqrt{i/8}},qquadtext{by Euler's product for the sine} \
&=-text{arg}left(frac{(1-i)sqrt{2}cosh(pi/4)}{pi}+frac{(1+i)sqrt{2}sinh(pi/4)}{pi}right) \
&=-text{arctan}left(frac{sinh(pi/4)-cosh(pi/4)}{sinh(pi/4)+cosh(pi/4)}right) \
&= text{arccot}left(e^{pi/2}right).
end{align*}$$
answered Jan 6 at 18:40
dxdydzdxdydz
4971411
$endgroup$
add a comment |
$begingroup$
We can compute this sum by considering it as the argument of an infinite product. Here the argument of the nth factor of the product is $-text{arctan}(1/8n^2)$.
$$begin{align*}
sum_{n=1}^inftytext{arctan}left(frac{1}{8n^2}right) &=-text{arg}prod_{n=1}^inftyleft(1-frac{i}{8n^2}right) \
&= -text{arg}prod_{n=1}^inftyleft(1-frac{left(sqrt{i/8}right)^2}{n^2}right)\
&=-text{arg}frac{sinleft(pisqrt{i/8}right)}{pisqrt{i/8}},qquadtext{by Euler's product for the sine} \
&=-text{arg}left(frac{(1-i)sqrt{2}cosh(pi/4)}{pi}+frac{(1+i)sqrt{2}sinh(pi/4)}{pi}right) \
&=-text{arctan}left(frac{sinh(pi/4)-cosh(pi/4)}{sinh(pi/4)+cosh(pi/4)}right) \
&= text{arccot}left(e^{pi/2}right).
end{align*}$$
answered Jan 6 at 18:40
dxdydzdxdydz
4971411
$endgroup$
We can compute this sum by considering it as the argument of an infinite product. Here the argument of the nth factor of the product is $-text{arctan}(1/8n^2)$.
$$begin{align*}
sum_{n=1}^inftytext{arctan}left(frac{1}{8n^2}right) &=-text{arg}prod_{n=1}^inftyleft(1-frac{i}{8n^2}right) \
&= -text{arg}prod_{n=1}^inftyleft(1-frac{left(sqrt{i/8}right)^2}{n^2}right)\
&=-text{arg}frac{sinleft(pisqrt{i/8}right)}{pisqrt{i/8}},qquadtext{by Euler's product for the sine} \
&=-text{arg}left(frac{(1-i)sqrt{2}cosh(pi/4)}{pi}+frac{(1+i)sqrt{2}sinh(pi/4)}{pi}right) \
&=-text{arctan}left(frac{sinh(pi/4)-cosh(pi/4)}{sinh(pi/4)+cosh(pi/4)}right) \
&= text{arccot}left(e^{pi/2}right).
end{align*}$$
answered Jan 6 at 18:40
dxdydzdxdydz
4971411
edited Jan 6 at 20:07
answered Jan 6 at 18:40
dxdydzdxdydz
4971411
answered Jan 6 at 18:40
dxdydzdxdydz
4971411
answered Jan 6 at 18:40
dxdydzdxdydz
4971411
4971411
add a comment |
add a comment |
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1
$begingroup$
That last sum doesn't converge since the absolute value of the terms tend to $pi/2$ not $0$. The first two sums do converge though.
$endgroup$
– JimmyK4542
Jan 6 at 18:24
$begingroup$
The last equality is not correct...
$endgroup$
– Fabian
Jan 6 at 18:25
$begingroup$
Try this approach for a sutable $x$.
$endgroup$
– A.Γ.
Jan 6 at 18:31
$begingroup$
I see you guy's point. I put that there since Desmos seemed to say it's equal to the original sum, but it definitely diverges.
$endgroup$
– Tom Himler
Jan 6 at 18:45