What is the sum from n=0 to infinity of (1+(-1)^n)/n ?

The sum from n=0 to infinity of (1+(-1)^n)/n is 0

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sum from n=0 to infinity of (1+(-1)^n)/n

n = 0 \sum \infty \frac{1 + ( - 1 ) ^{n}}{n}

0

Solution steps

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

Multiply by the conjugate of

1 + (- 1)^{n} : \frac{0}{1 - ( - 1 ) ^{n}}

= n = 1 \sum \infty \frac{\frac{0}{1 - ( - 1 ) ^{n}}}{n}

\frac{\frac{0}{1 - ( - 1 ) ^{n}}}{n} = 0

= n = 1 \sum \infty 0

\sum 0 = 0

= 0

n = 0 \sum \infty 8^{\frac{1}{n}}

n = 1 \sum \infty \frac{4}{e ^{n}}

n = 1 \sum \infty \frac{1}{( 7 n + 4 ) ^{2}}

n = 0 \sum \infty n^{5}

n = 1 \sum \infty 3 n^{- 2}

What is the sum from n=0 to infinity of (1+(-1)^n)/n ?
The sum from n=0 to infinity of (1+(-1)^n)/n is 0