sum from k=1 to infinity of (pi/e)^{-k}

解答

k = 1 \sum \infty (\frac{π}{e})^{- k}

收敛

求解步骤

k = 1 \sum \infty (\frac{π}{e})^{- k}

化简

(\frac{π}{e})^{- k} : \frac{1}{( \frac{π}{e} ) ^{k}}

= k = 1 \sum \infty \frac{1}{( \frac{π}{e} ) ^{k}}

化简

\frac{1}{( \frac{π}{e} ) ^{k}} : (\frac{e}{π})^{k}

= k = 1 \sum \infty (\frac{e}{π})^{k}

n = k \sum \infty (a_{n}) = n = 0 \sum \infty (a_{n}) - a_{0} - a_{1} - \dots a_{k - 1}

= k = 0 \sum \infty (\frac{e}{π})^{k} - (\frac{e}{π})^{0}

k = 0 \sum \infty (\frac{e}{π})^{k} =

收敛

= 收敛 - (\frac{e}{π})^{0}

= 收敛

n = 1 \sum \infty \frac{n}{cos ( n )}

n = 1 \sum \infty \frac{1}{n ^{2} - 4 n + 5}

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

n = 44 \sum \infty \frac{5}{n ^{2} + 2 n}

n = 1 \sum \infty 9 (\frac{4}{5})^{n}