Prove that $\lim\limits_{x\to 1^{-}}f(x)=-1+\ln 4$ where $f(x)=\sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^{n+1}}{n(n+1)}$

Question

Prove that $\lim\limits_{x\to 1^{-}}f(x)=-1+\ln 4$ where $$f(x)=\sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^{n+1}}{n(n+1)}.$$

My trial:

Since the series converges absolutely on $[-1,1]$, then it is uniformly convergent on $[-1,1].$ So,

$$f(x)=\sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^{n+1}}{n(n+1)}$$ $$=\sum_{n=1}^{\infty}(-1)^{n-1}x^{n+1}\Big[\frac{1}{n}-\frac{1}{n+1}\Big]$$ $$=\sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^{n+1}}{n}-\sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^{n+1}}{n+1}$$ $$=-\frac{1}{x}\sum_{n=1}^{\infty}(-1)^{n}\frac{x^{n}}{n}-\sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^{n+1}}{n+1}$$ $$=-\frac{1}{x}\sum_{n=0}^{\infty}(-1)^{n+1}\frac{x^{n+1}}{n+1}-\sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^{n+1}}{n+1}$$ $$=-\frac{1}{x}\Big[\sum_{n=0}^{\infty}(-1)^{n+1}\frac{x^{n+1}}{n+1}-x+x\Big]-\sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^{n+1}}{n+1}$$ $$=-\frac{1}{x}\Big[\sum_{n=1}^{\infty}(-1)^{n+1}\frac{x^{n+1}}{n+1}-x\Big]-\sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^{n+1}}{n+1}$$ $$=-\frac{1}{x}\Big[\sum_{n=1}^{\infty}(-1)^{n+1}\frac{x^{n+1}}{n+1}-x\Big]-\sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^{n+1}}{n+1}$$ $$=1-\Big(\frac{1}{x}+1\Big)\ln(1+x)$$ So, $$\lim\limits_{x\to 1^{-}}f(x)=1-\ln 4.$$ Please, where am I missing it? Better solutions will be accepted.

Answer 1

I rewrite part of yours: \begin{align} f(x) &=\sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^{n+1}}{n(n+1)}\\ &=\sum_{n=1}^{\infty}(-1)^{n-1}x^{n+1}\Big[\frac{1}{n}-\frac{1}{n+1}\Big]\\ &=\sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^{n+1}}{n}-\sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^{n+1}}{n+1}\\ &=x\sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^n}{n}+\sum_{n=1}^{\infty}(-1)^{n}\frac{x^{n+1}}{n+1}+x-x\\ &=x\left(\sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^n}{n}\right)+\left(\sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^n}{n}\right)-x\\ &=(1+x)\ln(1+x)-x \end{align}

Answer 2

$ln (1+x)=x-x^{2}/2+... $ but you are writing $ln (1+x)=x^{2}/2-x^{3}/3... $. All your calculations are fine except for this mistake at the end. BTW you can take the limit inside the sum right in the beginning (and get rid of $x$) because of uniform convergence).

Answer 3

A Different Approach $$ \begin{align} \lim_{x\to1^-}\sum_{n=1}^\infty(-1)^{n-1}\frac{x^{n+1}}{n(n+1)} &=\lim_{x\to1^-}\sum_{n=1}^\infty\frac{x^{2n}}{(2n-1)2n} -\lim_{x\to1^-}\sum_{n=1}^\infty\frac{x^{2n+1}}{2n(2n+1)}\tag1\\ &=\sum_{n=1}^\infty\frac1{(2n-1)2n} -\sum_{n=1}^\infty\frac1{2n(2n+1)}\tag2\\ &=\sum_{n=1}^\infty\left(\frac1{2n-1}-\frac2{2n}+\frac1{2n+1}\right)\tag3\\ &=\sum_{n=1}^\infty\left(\frac2{2n-1}-\frac2{2n}\right)-\sum_{n=1}^\infty\left(\frac1{2n-1}-\frac1{2n+1}\right)\tag4\\[9pt] &=2\log(2)-1\tag5 \end{align} $$ Explanation:
$(1)$: separate the series into the difference of the even and the odd terms
$(2)$: apply monotone or dominated convergence to both sums (either works)
$(3)$: partial fractions
$(4)$: separate the series into the difference of two convergent series
$(5)$: the series on the left is twice the alternating harmonic series
$\phantom{(5)\text{:}}$ the series on the right telescopes

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A Modification of the Approach in the Question $$ \begin{align} \lim_{x\to1^-}\sum_{n=1}^\infty(-1)^{n-1}\frac{x^{n+1}}{n(n+1)} &=\lim_{x\to1^-}\sum_{n=1}^\infty(-1)^{n-1}\frac{x^{n+1}}{n}-\lim_{x\to1^-}\sum_{n=1}^\infty(-1)^{n-1}\frac{x^{n+1}}{n+1}\tag6\\ &=\lim_{x\to1^-}x\log(1+x)-\lim_{x\to1^-}(1-\log(1+x))\tag7\\[6pt] &=2\log(2)-1\tag8 \end{align} $$ Explanation:
$(6)$: partial fractions, then separate into two convergent series
$(7)$: the series converge absolutely to $\log(1+x)$ for $x\lt1$
$(8)$: evaluate at $x=1$

Answer 4

It's not hard to see that $f$ is infinitely many times differentiable in $(-1,1)$ (use ratio test) therefore $$f''(x)=\sum_{n=1}^{\infty}(-x)^{n-1}=\dfrac{1}{1+x}$$therefore $$f'(x)=\ln(1+x)+C_1$$and $$f(x)=(1+x)\ln(1+x)-1-x+C_1x+C_2$$and we know that $$C_1=C_2=0$$therefore $$\lim_{x\to1^-}f(x)=2\ln2-1$$