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Prove the following equality ($|x|<1$).

$$\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{3x^2}{1+x^3}+\frac{4x^3}{1+x^4}+\cdots\\ =\frac{1}{1-x}+\frac{3x^2}{1-x^3}+\frac{5x^4}{1-x^5}+\frac{7x^6}{1-x^7}+\cdots\\$$

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    $\begingroup$ Are you sure there are no typos? L.H.S seems to diverge.. $\endgroup$
    – rkm0959
    Commented Dec 1, 2015 at 13:18
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    $\begingroup$ @GyuminRoh There was no typo. $\endgroup$
    – Kay K.
    Commented Dec 1, 2015 at 13:23
  • $\begingroup$ @KayK. Gyumin is right; your series on the LHS diverges. Are the numerators correct? $\endgroup$ Commented Dec 1, 2015 at 13:32
  • $\begingroup$ Wait a second. My apology there was a typo. $\endgroup$
    – Kay K.
    Commented Dec 1, 2015 at 13:47
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    $\begingroup$ Hint: expand each term as the sum of a geometric series, then reorder the double summation $\endgroup$ Commented Dec 1, 2015 at 14:06

4 Answers 4

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Integrate both sides. Let L.H.S be $f(x)$ and R.H.S be $g(x)$.

(1). L.H.S

$$\int \frac{ix^{i-1}}{1+x^{i}} = \ln (1+x^i)$$

This gives $$\int f(x) dx= \sum_{i=1}^\infty \ln (1+x^i) = \ln (\prod_{i=1}^\infty (1+x^i))$$

(2). R.H.S

$$\int \frac{(2i+1)x^{2i}}{1-x^{2i+1}} = -\int \frac{-(2i+1)x^{2i}}{1-x^{2i+1}} = -\ln (1-x^{2i+1})$$

This gives $$\int g(x) dx=\sum_{i=0}^\infty -\ln (1-x^{2i+1}) = -\ln (\prod_{i=0}^\infty (1-x^{2i+1}))$$

(3). Finishing the problem.

Since $f(0)=g(0)=0$, it now suffices to show that $$ \prod_{i=1}^\infty (1+x^i) \cdot \prod_{i=0}^\infty (1-x^{2i+1}) = 1$$

Now here is the part that I'm not sure if it is true or not.

Let $$h(x)=\prod_{i=1}^\infty (1+x^i) \cdot \prod_{i=0}^\infty (1-x^{2i+1})$$

Of course, the domain of $h(x)$ is $|x|<1$.

Now we have $$h(x)=\prod_{i=1}^\infty (1+x^i) \cdot \prod_{i=0}^\infty (1-x^{2i+1})=\prod_{i=1}^\infty (1+x^{2i}) \cdot \prod_{i=0}^\infty (1+x^{2i+1}) \cdot \prod_{i=0}^\infty (1-x^{2i+1}) =\prod_{i=1}^\infty (1+x^{2i}) \cdot \prod_{i=0}^\infty (1-x^{4i+2})=h(x^2)$$

Therefore, for all $|x|<1$, we have $$h(x)=h(x^2)=h(x^4)= \cdots \lim_{n \to \infty} h(x^{2^n}) = h(0) = 1$$ as desired.

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  • $\begingroup$ Taking a look at my solution again, I don't think this solution is close to correct T_T. Is there any way to somehow fix it? $\endgroup$
    – rkm0959
    Commented Dec 1, 2015 at 14:06
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    $\begingroup$ (+1) Looks good to me! You can use telescopic cancellation $\displaystyle \prod_{k=1}^{\infty}\frac{1}{1-x^{2k-1}} = \prod_{k=1}^{\infty} \frac{1-x^{2k}}{1-x^k} = \prod_{k=1}^{\infty}(1+x^k)$ to prove your $h(x) = 1$, take logarithm on both sides and differentiate! $\endgroup$
    – r9m
    Commented Dec 1, 2015 at 14:10
  • $\begingroup$ @r9m So is there any problem in my last argument on $h(x)=h(x^2)=h(x^4)= \lim_{n \to \infty} h(x^{2^n}) = h(0)=1$? Something like continuity or etc... $\endgroup$
    – rkm0959
    Commented Dec 1, 2015 at 14:14
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    $\begingroup$ Not really! In principle both are same. I was just suggesting another approach, that's all! :-) $\endgroup$
    – r9m
    Commented Dec 1, 2015 at 14:16
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    $\begingroup$ Thanks. Actually I made this question and had some ugly proof. So I wanted to see if there's a better one. I think you did it beautifully. $\endgroup$
    – Kay K.
    Commented Dec 1, 2015 at 14:28
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Multiplying $x$ to both sides, the identity is equivalent to

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

Expanding and rearranging, each series can be written as

\begin{align*} \sum_{k=1}^{\infty} \frac{kx^k}{1+x^k} &= \sum_{k=1}^{\infty} \sum_{j=1}^{\infty} (-1)^{j-1} k x^{jk} = \sum_{n=1}^{\infty} \Bigg( \sum_{d|n} (-1)^{d-1}\frac{n}{d} \Bigg) x^{n} \\ \sum_{k=1}^{\infty} \frac{(2k-1)x^{2k-1}}{1-x^{2k-1}} &= \sum_{k=1}^{\infty} \sum_{j=1}^{\infty} (2k-1)x^{j(2k-1)} = \sum_{n=1}^{\infty} \Bigg( \sum_{\substack{d | n \\ d \text{ odd}}} d \Bigg) x^{n} \end{align*}

So it is sufficient to prove that

$$ \color{blue}{\sum_{d|n} (-1)^{d-1}\frac{n}{d} = \sum_{\substack{d | n \\ d \text{ odd}}} d} \quad \text{for } n = 1, 2, \cdots. \tag{1}$$

To this end, write $n = 2^e m$ for $e \geq 0$ and $m$ is odd. Then

$$ \sum_{d|n} (-1)^{d-1}\frac{n}{d} = \sum_{d'|m} \underbrace{\sum_{i=0}^{e} (-1)^{2^i d'-1} 2^{e-i}}_{=1} \frac{m}{d'} = \sum_{d'|m} \frac{m}{d'} = \sum_{d|m} d = \sum_{\substack{d | n \\ d \text{ odd}}} d $$

and hence $\text{(1)}$ is proved.

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    $\begingroup$ (+1) Beautiful solution! :) small typo in first line $\dfrac{kx^{\n}}{1+x^k}$ $\endgroup$
    – r9m
    Commented Dec 1, 2015 at 16:31
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Check that for $|x| < r < 1$,

$$\frac{nx^{n-1}}{1+x^n}, \frac{nx^{n-1}}{1-x^n}, \frac{2nx^{2n-1}}{1-x^n}$$ are all $O(nr^n)$, which is summable.

Then, you can switch the order of summation in the following infinite sum :

$$0 = \sum_{n \ge 1} (\frac{ nx^{n-1}}{1+x^n} - \frac {nx^{n-1}}{1-x^n} + \frac {2nx^{2n-1}}{1-x^{2n}}) = (\sum_{n \ge 1} \frac{ nx^{n-1}}{1+x^n}) - \sum_{n \ge 1}(\frac {nx^{n-1}}{1-x^n} - \frac {2nx^{2n-1}}{1-x^{2n}})$$

The second sum is a telescoping sum $\sum (a_n - a_{2n})$ so reordering again, it is $$(\sum a_n) - (\sum a_{2n}) = \sum_{n \text{ odd}} a_n + \sum_{n \text{ even}} (a_n - a_n) = \sum_{n\text{ odd}} a_n$$

Hence $$\sum_{n \ge 1} \frac{ nx^{n-1}}{1+x^n} - \sum_{n \text{ odd}}\frac {nx^{n-1}}{1-x^n}=0$$

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Let me add my own proof that I had in my mind.

$$Let\quad f(x)=\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{3x^2}{1+x^3}+\frac{4x^3}{1+x^4}+\frac{5x^4}{1+x^5}+\frac{6x^5}{1+x^6}+\frac{7x^6}{1+x^7}+\cdots$$ $$f(x)-\frac{1}{1-x}=\color{red}{\frac{-2x}{1-x^2}}+\frac{2x}{1+x^2}+\frac{3x^2}{1+x^3}+\frac{4x^3}{1+x^4}+\frac{5x^4}{1+x^5}+\frac{6x^5}{1+x^6}+\frac{7x^6}{1+x^7}+\cdots\\ =\frac{3x^2}{1+x^3}+\color{red}{\frac{-4x^3}{1-x^4}}+\frac{4x^3}{1+x^4}+\frac{5x^4}{1+x^5}+\frac{6x^5}{1+x^6}+\frac{7x^6}{1+x^7}+\cdots\\ =\cdots=\frac{3x^2}{1+x^3}+\frac{5x^4}{1+x^5}+\frac{6x^5}{1+x^6}+\frac{7x^6}{1+x^7}+\cdots+\lim_{n\to\infty}\frac{-2^nx^{2^n-1}}{1-x^{2^n}} $$ $$f(x)-\frac{1}{1-x}-\frac{3x^2}{1-x^3} =\frac{5x^4}{1+x^5}+\color{red}{\frac{-6x^5}{1-x^6}}+\frac{6x^5}{1+x^6}+\frac{7x^6}{1+x^7}+\cdots+\lim_{n\to\infty}\frac{-2^nx^{2^n-1}}{1-x^{2^n}}\\ =\cdots=\frac{5x^4}{1+x^5}+\frac{7x^6}{1+x^7}+\cdots+\lim_{n\to\infty}\frac{-2^nx^{2^n-1}}{1-x^{2^n}}+\lim_{n\to\infty}\frac{-3\cdot2^nx^{3\cdot2^n-1}}{1-x^{3\cdot2^n}}$$ $$\cdots$$ $$f(x)-\sum_{k=0}^{\infty}\frac{(2k+1)x^{2k}}{1-x^{2k+1}}=\sum_{k=0}^{\infty}\lim_{n\to\infty}\frac{-(2k+1)\cdot2^nx^{(2k+1)\cdot2^n-1}}{1-x^{(2k+1)\cdot2^n}}=0$$ $$\therefore f(x)=\sum_{k=0}^{\infty}\frac{(2k+1)x^{2k}}{1-x^{2k+1}}$$

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