What is the sum from n=1 to infinity of ((-1+2)^n)/n ?

The sum from n=1 to infinity of ((-1+2)^n)/n is diverges

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sum from n=1 to infinity of ((-1+2)^n)/n

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

diverges

Solution steps

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

Simplify

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

Apply Cauchy's Convergence Condition:

diverges

= diverges

n = 1 \sum \infty \frac{19}{1 + 3 ^{n}}

n = 1 \sum \infty \frac{ln ( n )}{7 n}

n = 4 \sum \infty \frac{5}{3 ^{n}}

n = 0 \sum \infty 7 - n^{4}

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

What is the sum from n=1 to infinity of ((-1+2)^n)/n ?
The sum from n=1 to infinity of ((-1+2)^n)/n is diverges