What is the sum from n=-infinity to infinity of 1/n ?

The sum from n=-infinity to infinity of 1/n is diverges

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sum from n=-infinity to infinity of 1/n

n = - \infty \sum \infty \frac{1}{n}

diverges

Solution steps

n = 1 \sum \infty \frac{1}{n}

Apply Cauchy's Convergence Condition:

diverges

= diverges

\frac{d}{d x} (e^{- x} x^{2} - 4 e^{- x} x + 2 e^{- x})

f (x) = x^{2} - 5

y^{^{'''}} + 2 y^{^{''}} + 5 y^{^{'}} - 26 y = 0

y = \frac{x ^{2}}{2} + 4 x + 2, (- 3, - \frac{11}{2})

y^{^{'}} + \frac{1}{2} y = y^{3}

What is the sum from n=-infinity to infinity of 1/n ?
The sum from n=-infinity to infinity of 1/n is diverges