Σ_(k=−∞) ^(+∞) ((k^2 +k+1)/(k^4 +1))=?


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Question Number 214997 by MrGaster last updated on 25/Dec/24

$\underset{k = - \infty}{\sum^{+ \infty}} \frac{k^{2} + k + 1}{k^{4} + 1} = ?$
	Answered by mr W last updated on 26/Dec/24

	$\underset{k = - \infty}{\sum^{+ \infty}} \frac{k^{2} + k + 1}{k^{4} + 1}$ $= \underset{k = - \infty}{\sum^{+ \infty}} \frac{k}{k^{4} + 1} + \underset{k = - \infty}{\sum^{+ \infty}} \frac{k^{2} + 1}{k^{4} + 1}$ $= 0 + \underset{k = - \infty}{\sum^{+ \infty}} \frac{k^{2} + 1}{k^{4} + 1}$ $= 2 \underset{k = 1}{\sum^{+ \infty}} \frac{k^{2} + 1}{k^{4} + 1} + 1$ $= 2 \underset{k = 1}{\sum^{+ \infty}} \frac{k^{2} + 1}{(k^{2} + i) (k^{2} - i)} + 1$ $= \underset{k = 1}{\sum^{+ \infty}} (\frac{1 - i}{k^{2} - i} + \frac{1 + i}{k^{2} + i}) + 1$ $= \underset{k = 1}{\sum^{+ \infty}} \frac{1 - i}{(k - \frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}) (k + \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}})} + \underset{k = 1}{\sum^{+ \infty}} \frac{1 + i}{(k - \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}) (k + \frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}})} + 1$ $= - \frac{i}{\sqrt{2}} \underset{k = 1}{\sum^{+ \infty}} (\frac{1}{k - \frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}} - \frac{1}{k + \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}}) + \frac{i}{\sqrt{2}} \underset{k = 1}{\sum^{+ \infty}} (\frac{1}{k - \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}} - \frac{1}{k + \frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}}) + 1$ $= - \frac{i}{\sqrt{2}} [\underset{k = 1}{\sum^{\infty}} \frac{1}{k} - γ - ψ (1 - \frac{1 + i}{\sqrt{2}}) - \underset{k = 1}{\sum^{\infty}} \frac{1}{k} + γ + ψ (1 + \frac{1 + i}{\sqrt{2}})] + \frac{i}{\sqrt{2}} [\underset{k = 1}{\sum^{\infty}} \frac{1}{k} - γ - ψ (1 - \frac{1 - i}{\sqrt{2}}) - \underset{k = 1}{\sum^{\infty}} \frac{1}{k} + γ + ψ (1 + \frac{1 - i}{\sqrt{2}})] + 1$ $= \frac{i}{\sqrt{2}} [ψ (1 - \frac{1 + i}{\sqrt{2}}) - ψ (1 + \frac{1 + i}{\sqrt{2}}) - ψ (1 - \frac{1 - i}{\sqrt{2}}) + ψ (1 + \frac{1 - i}{\sqrt{2}})] + 1$ $= \frac{i}{\sqrt{2}} [ψ (\frac{1 + i}{\sqrt{2}}) + π \cot \frac{(1 + i) π}{\sqrt{2}} - ψ (\frac{1 + i}{\sqrt{2}}) - \frac{\sqrt{2}}{1 + i} - ψ (\frac{1 - i}{\sqrt{2}}) - π \cot \frac{(1 - i) π}{\sqrt{2}} + ψ (\frac{1 - i}{\sqrt{2}}) + \frac{\sqrt{2}}{1 - i}] + 1$ $= \frac{i}{\sqrt{2}} [\sqrt{2} i + π \cot \frac{(1 + i) π}{\sqrt{2}} - π \cot \frac{(1 - i) π}{\sqrt{2}}] + 1$ $= \frac{π i}{\sqrt{2}} [\cot \frac{(1 + i) π}{\sqrt{2}} - \cot \frac{(1 - i) π}{\sqrt{2}}]$ $= \frac{π i}{\sqrt{2}} [\frac{\cot \frac{π}{\sqrt{2}} \cot \frac{i π}{\sqrt{2}} - 1}{\cot \frac{π}{\sqrt{2}} + \cot \frac{i π}{\sqrt{2}}} + \frac{\cot \frac{π}{\sqrt{2}} \cot \frac{i π}{\sqrt{2}} + 1}{\cot \frac{π}{\sqrt{2}} - \cot \frac{i π}{\sqrt{2}}}]$ $= \frac{\sqrt{2} π i (\cot^{2} \frac{π}{\sqrt{2}} + 1) \cot \frac{i π}{\sqrt{2}}}{\cot^{2} \frac{π}{\sqrt{2}} - \cot^{2} \frac{i π}{\sqrt{2}}}$ $= \frac{\sqrt{2} π (\cot^{2} \frac{π}{\sqrt{2}} + 1) \coth \frac{π}{\sqrt{2}}}{\cot^{2} \frac{π}{\sqrt{2}} + \coth^{2} \frac{π}{\sqrt{2}}} ✓$ $\approx 4.414011009$
	Commented by mr W last updated on 26/Dec/24

	$i s t h i s c o r r e c t ?$ $d o y o u h a v e a n o t h e r e a s i e r p a t h ?$
	Commented by MathematicalUser2357 last updated on 28/Dec/24

	$4.41401 10087 82893$
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