Question: "Find the sum:∑k=0∞(−2)kk+2k+1xk"

erycletrefeebr

Answered question

2022-02-25

Find the sum:

\sum_{k = 0}^{\infty} {(- 2)}^{k} \frac{k + 2}{k + 1} x^{k}

bedevijuo3e

Beginner2022-02-26Added 6 answers

So we want to find the value of

\sum_{k = 0}^{\infty} \frac{(k + 2) {(- 2 x)}^{k}}{k + 1}

As you already determined, we know

\ln (x + 1) = \sum_{k = 0}^{\infty} \frac{{(- 1)}^{k} x^{k + 1}}{k + 1}

If you multiply both sides by x you get

x \ln (x + 1) = \sum_{k = 0}^{\infty} \frac{{(- 1)}^{k} x^{k + 2}}{k + 1}

Now, the interesting step is to take the derivative of both sides.

\ln (1 + x) + \frac{x}{x + 1} = \sum_{k = 0}^{\infty} \frac{k + 2}{k + 1} {(- 1)}^{k} x^{k + 1}

Now divide both sides by x to get

\frac{1}{x} \ln (x + 1) + \frac{1}{x + 1} = \sum_{k = 0}^{\infty} \frac{k + 2}{k + 1} {(- x)}^{k}

Substitute 2x into x to get,

\ln (2 x + 1) + \frac{1}{2 x + 1} = \sum_{k = 0}^{\infty} \frac{k + 2}{k + 1} {(- 2 x)}^{k}

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