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Problem:

Show that $$\ln2 = \sum_\limits{n=1}^\infty\frac1{n2^n}.$$

My progress:

The problem before this one had me find the Taylor series for $\ln(1-x)$ which was $$-\sum\limits_{n=1}^\infty \frac{x^n}n$$

so I figured I'd use $x=-1$ and plug that into the Taylor series. However, there was a side note stating that the Taylor series I found is only valid for $x\in(-1, 1)$. And in any case, my calculation isn't going anywhere, since I end up with the series $1-\frac12+\frac13-\frac14\cdots$

Question 1: How can I use this to solve the problem stated initially, and in the title?

Question 2: I can see why $x$ is restricted to be less than 1, to prevent taking the log of zero or a negative. Why is it not valid for x greater than 1?

Alec
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    Put $x=1/2$ to get $\ln(1/2)=-\ln(2)$. – Tom Feb 17 '15 at 19:42
  • Question 2: If $x\gt 1$, then the terms $\frac{x^n}{n}$ blow up as $n\to\infty$, so the series does not converge. – André Nicolas Feb 17 '15 at 19:47
  • @Tom - You are a wizard. I don't think I ever would have thought to use that. – Alec Feb 17 '15 at 19:55
  • @AndréNicolas - Very good point! How about $x = -1$? As $n\to\infty$ would the series not converge? – Alec Feb 17 '15 at 19:56
  • @AndréNicolas - Did you mean to alternate addition/subtraction in that sum? – Alec Feb 17 '15 at 20:04
  • Oops, I thought you were expanding $\ln(1+x)$. For $\ln(1-x)$, at $x=-1$ we get convergence, the alternating series $1-\frac{1}{2}+\cdots$. – André Nicolas Feb 17 '15 at 20:08
  • @AndréNicolas - Ah, hehe. But is there a reason why I shouldn't use $x=-1$ for this purpose? That harmonic series does not seem to converge to $\ln2$. – Alec Feb 17 '15 at 20:11
  • The harmonic series $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots$ diverges. It corresponds to $x=1$. The series $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots$ converges. It happens to converge very slowly, so to get any kind of accuracy in using it to approximate $\ln 2$, we would have to use a ridiculously large number of terms. By way of contrast, the series $\sum \frac{1}{n2^n}$ converges rapidly. – André Nicolas Feb 17 '15 at 20:22
  • @AndréNicolas - So in essence, since we are evaluating $n\to\infty$, using $x=-1$ isn't techincally wrong, it's just inefficient? – Alec Feb 17 '15 at 20:28
  • That's right. Call a series alternating if the signs of the terms alternate. Suppose $a_1+a_2+\cdots$ is alternating, and the $|a_n|$ are decreasing, with limit $0$. Then $a_1+a_2+\cdots$ converges. This is called the Alternating Series Test, or sometimes the Leibniz Test. – André Nicolas Feb 17 '15 at 20:34

4 Answers4

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Hint: $\ln 2 = - \ln (1/2)$. Can you use that and your result on the series for $\ln (1-x)$ to solve the problem?

user2566092
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A different perspective: Your series is the Euler transform of $$ 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots$$ and so it has an equal sum.

WimC
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  • Yes, I reached this harmonic series when attempting to use $x=-1$, but it does not seem to converge to $\ln2$. What exactly is happening at $x=-1$? I can't see why this destroyed my attempt. – Alec Feb 17 '15 at 20:12
  • Btw, I meant alternating harmonic. – Alec Feb 17 '15 at 20:27
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Hint:

$$\frac1{1-x}=\sum_{n=0}^\infty x^n\;,\;\;|x|<1\implies-\log(1-x)=\sum_{n=0}^\infty\frac{x^{n+1}}{n+1}=x+\sum_{n=2}^\infty\frac{x^n}n$$

Timbuc
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For similar cases, where you cannot simply evaluate inside the radius of convergence... There is Abel's theorem with conditions that let you evaluate exactly on the radius of convergence.

GEdgar
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