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Hi could anyone help me with this question

Determine the sum of the power series:

$$S=-\sum_{n=1}^{\infty}\frac{(1-x)^n}{n}$$

Where x=1.74

I tried to differentiate this expression, but I do not know how to proceed from here.

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Differentiating gives:

$$\frac{\mathrm{d}S}{\mathrm{d}x}=\sum_{n=1}^{\infty}(1-x)^{n-1}=\sum_{n=0}^{\infty}(1-x)^{n}=\frac{1}{1-(1-x)}=\frac{1}{x}$$

As $|1-x|<1$. We therefore can integrate with respect to $x$ to give:

$$S=\int\frac{\mathrm{d}S}{\mathrm{d}x}\:\mathrm{d}x=\ln(x)+C$$

We have that for $x=1$, $S=0$ and therefore $C=0$:

$$S=\ln(x)=\ln(1.74)$$

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  • $\begingroup$ Hi may i know why do u need to remove the negative sign in front of the summation? Thanks $\endgroup$ – ys wong Apr 8 '14 at 12:25
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Your sum is:

$$ \sum^{\infty}_{n=1} \frac{(1-x)^n}{n}=?$$

First review thees sums:

$$ \sum^{\infty}_{n=1} x^n=\frac{x}{1-x}$$

Also:

$$ \sum^{\infty}_{n=1} (1-x)^n=\frac{1-x}{1-(1-x)}=\frac{1-x}{x}$$

$$ \sum^{\infty}_{n=1} (1-x)^n=\frac{1-x}{x} / (1-x)$$

$$ \sum^{\infty}_{n=1} (1-x)^{n-1}=\frac{1}{x} / \int dx $$

$$ \sum^{\infty}_{n=1} \frac{(1-x)^n}{n}=\int \frac{1}{x}dx=ln(x) $$

So the answer to your question would be $ln 1.74$

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Then you get a sum of a geometric sequence.

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If you define $$f(x) = -\sum_{i=1}^\infty \frac{(1-x)^n}{n},$$ you can see that (some justification here is needed) $$f'(x) = -\sum_{i=1}^\infty (1-x)^{n-1}$$ which you probably know how to calculate. Then, having $f',$ you can calculate $f$ as one of the indefinite integrals of $f'$.

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