What is the sum from n=0 to infinity of-1/n ?

The sum from n=0 to infinity of-1/n is diverges

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sum from n=0 to infinity of-1/n

n = 0 \sum \infty - \frac{1}{n}

diverges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Cauchy's Convergence Condition:

diverges

= - diverges

= diverges

n = 0 \sum \infty cos (\frac{nπ}{2})

n = 0 \sum \infty [7 (0.1)^{n}]

n = 1 \sum \infty \frac{3}{n} - \frac{3}{n + 2}

n = 2 \sum \infty n e^{- 3 n^{2}}

n = 0 \sum \infty \frac{1}{5 + n}

What is the sum from n=0 to infinity of-1/n ?
The sum from n=0 to infinity of-1/n is diverges