Prove $$\sum _{n=1}^{\infty }\:\frac{x^n}{n} = -\ln\left(1-x\right),$$ for $|x| < 1$.
Pretty much title I've searched and seen some solutions using Taylor Series but isn't there a more intuative way to do this?
Prove $\sum _{n=1}^{\infty }\:\frac{x^n}{n}$ is equal to $-\ln\left(1-x\right)$, for $|x| < 1$
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$\begingroup$
sequences-and-series
discrete-mathematics
logarithms
taylor-expansion
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1What is your definition of the function $\ln x$? – 2017-01-04
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0Natural logarithm. Is there a more recognizable way to write this? – 2017-01-04
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0How do you *define* the *natural logarithm*? – 2017-01-04
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1By integrating term by term the geometric series $\frac{1}{1-x}=1+x+x^2+...$, for $x$ with $|x|<1$? – 2017-01-04
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0@martini Inverse of $e^x$ – 2017-01-04
2 Answers
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Since $|x|<1$, the series in the LHS is absolutely convergent by comparison with a geometric series.
Additionally, $\frac{x^n}{n}=\int_{0}^{x}y^{n-1}\,dy$, hence:
$$ \sum_{n=1}^{N}\frac{x^n}{n} = \int_{0}^{x}\frac{1-y^N}{1-y}\,dy =-\log(1-x)-\int_{0}^{x}\frac{y^N}{1-y}\,dy.$$
by exchanging $\sum$ and $\int$. Now we just need to consider the limit as $N\to +\infty$, and that is just $-\log(1-x)$ since $|x|<1$ grants:
$$ \left|\int_{0}^{x}\frac{y^N}{1-y}\,dy\right|\leq |x|^N \int_{0}^{x}\frac{dy}{1-y}\to 0.$$
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0Nice proof :) I don't usually see partial sums used for this :D – 2017-01-04
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0Another bound is $\int_{0}^{x}\frac{y^N}{1-y}\,dy\leq\frac1{1-x}\int_{0}^{x}y^N\,dy=\frac{x^{N+1}}{(N+1)(1-x)}\to 0$. – 2017-01-04
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Here is intuitive:
$$-\ln(1-x)=\int_0^x\frac1{1-t}\ dt=\int_0^x\left(1+t+t^2+t^3+\dots\ \right)\ dx\\=x+\frac{x^2}2+\frac{x^3}3+\frac{x^4}4+\dots$$
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0? Is someone serial downvoting me? – 2017-01-04
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0Thanks that is more than enough :) – 2017-01-04
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0@X.Daniels :) No problem, glad to help! – 2017-01-04
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0I do not see the reason for the downvote. Ok, perhaps you need to justify(?) integration term by term of the infinite sum (i.e. interchange of $\int$ and $\sum$), but other than that is ok. So +1 from me. – 2017-01-04
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0I agree with Jimmy, there is no reason to downvote (so +1 back). Maybe, we should justify the exchange of $\int$ and $\sum$ by saying "$1+t+t^2+\ldots=\frac{1}{1-t}$ is an analytic function in the open disk $|t|<1$" or something like that. – 2017-01-04
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0@JackD'Aurizio Nah, you got that all covered. – 2017-01-04