What is the sum from n=1 to infinity of (-5)^n ?

The sum from n=1 to infinity of (-5)^n is diverges

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sum from n=1 to infinity of (-5)^n

n = 1 \sum \infty (- 5)^{n}

diverges

Solution steps

n = 1 \sum \infty (- 5)^{n}

n = k \sum \infty (a_{n}) = n = 0 \sum \infty (a_{n}) - a_{0} - a_{1} - \dots a_{k - 1}

= n = 0 \sum \infty (- 5)^{n} - (- 5)^{0}

n = 0 \sum \infty (- 5)^{n} =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int cosh (0)

\int (3 cos (2 θ)) d θ

\int \frac{x}{( y ^{2} + x ^{2} ) ^{\frac{3}{2}}} d x

x^{2}, - x + 5, 0

\frac{8 x ^{3} - 2 x + 9}{x}

What is the sum from n=1 to infinity of (-5)^n ?
The sum from n=1 to infinity of (-5)^n is diverges