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 \frac{- 5}{n}

diverges

Solution steps

n = 1 \sum \infty \frac{- 5}{n}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= - 5 \cdot n = 1 \sum \infty \frac{1}{n}

Apply Cauchy's Convergence Condition:

diverges

= - 5 diverges

= diverges

n = 1 \sum \infty \frac{n}{( 2 n + 1 ) ^{2}}

n = 1 \sum \infty (\frac{5}{7})^{n - 1}

n = 1 \sum \infty (- \frac{3}{7})^{n}

n = 1 \sum \infty \frac{3 ^{n}}{8 ^{n}}

n = 1 \sum \infty \frac{n + 1}{n 3 ^{n}}

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