$\infty \sum k = 0 {(- 1)}^{k} (\frac{k + 1}{4^{k + 1}}) =$ $sum_(k=0)^oo(-1)^k ((k+1)/4^(k+1)) =$ ?

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$\infty \sum k = 0 {(- 1)}^{k} (\frac{k + 1}{4^{k + 1}}) =$ $sum_(k=0)^oo(-1)^k ((k+1)/4^(k+1)) =$ ?

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Answer 1 · 2017-03-03T16:03:36

See below.

Explanation:

$\frac{1}{{(x + 1)}^{2}} = - \frac{d}{d x} (\frac{1}{x + 1})$ $1/(x+1)^2= -d/(dx)(1/(x+1))$ but

$\frac{1}{x + 1} = \frac{1}{2 + x - 1} = \frac{1}{2} (\frac{1}{1 + \frac{x - 1}{2}}) = \frac{1}{2} \infty \sum k = 0 {(- 1)}^{k} {(\frac{x - 1}{2})}^{k}$ $1/(x+1)=1/(2+x-1)=1/2(1/(1+(x-1)/2)) = 1/2sum_(k=0)^oo(-1)^k((x-1)/2)^k$ for $∣ ∣ ∣ \frac{x - 1}{2} ∣ ∣ ∣ < 1$ $abs((x-1)/2)<1$

but

$- \frac{d}{d x} (\frac{1}{2} \infty \sum k = 0 {(- 1)}^{k} {(\frac{x - 1}{2})}^{k}) = - \frac{1}{2^{2}} \infty \sum k = 1 {(- 1)}^{k} \frac{k}{2^{k - 1}} {(x - 1)}^{k - 1} = \infty \sum k = 0 {(- 1)}^{k} (\frac{k + 1}{2^{k + 2}}) {(x - 1)}^{k}$ $-d/(dx)(1/2sum_(k=0)^oo(-1)^k((x-1)/2)^k)=-1/2^2sum_(k=1)^oo(-1)^k k/2^(k-1)(x-1)^(k-1) = sum_(k=0)^oo(-1)^k ((k+1)/2^(k+2))(x-1)^k$

Making $x = \frac{1}{2}$ $x = 1/2$ we have

$\infty \sum k = 0 {(- 1)}^{k} (\frac{k + 1}{4^{k + 1}}) = \frac{1}{{(\frac{1}{2} + 1)}^{2}} = {(\frac{2}{3})}^{2}$ $sum_(k=0)^oo(-1)^k ((k+1)/4^(k+1)) = 1/(1/2+1)^2 = (2/3)^2$

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$\infty \sum k = 0 {(- 1)}^{k} (\frac{k + 1}{4^{k + 1}}) =$ $sum_(k=0)^oo(-1)^k ((k+1)/4^(k+1)) =$ ?

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Explanation:

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sum_(k=0)^oo(-1)^k ((k+1)/4^(k+1)) =∞∑k=0(−1)k(k+14k+1)=sum_(k=0)^oo(-1)^k ((k+1)/4^(k+1)) = ?

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$\infty \sum k = 0 {(- 1)}^{k} (\frac{k + 1}{4^{k + 1}}) =$ $sum_(k=0)^oo(-1)^k ((k+1)/4^(k+1)) =$ ?