sum from n=1 to infinity of (-0.005)^n

解答

n = 1 \sum \infty (- 0.005)^{n}

- 0.00497 \dots

求解步骤

n = 1 \sum \infty (- 0.005)^{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 (- 0.005)^{n} - (- 0.005)^{0}

n = 0 \sum \infty (- 0.005)^{n} = 0.99502 \dots

= 0.99502 \dots - (- 0.005)^{0}

化简

= - 0.00497 \dots

n = 1 \sum \infty \frac{2}{n ln ( 9 n )}

n = 1 \sum \infty \frac{n !}{11 6 ^{n}}

n = 1 \sum \infty \frac{1}{5 ^{n} - 2}

n = 1 \sum \infty 5 (\frac{1}{2})^{n + 1}

n = 0 \sum \infty \frac{1}{1 + n ^{4}}