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

Question

For $x \in ]0,1[$, we want to prove, without convergence theorem, that : $$\sum_{k=1}^{+ \infty} \dfrac{x^{k+1}}{ k (k+1)}= x + (1-x) \ln (1-x) $$

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My attempt :

$ \begin{align*} \sum_{k=1}^m \dfrac{x^k}{k}&= \sum_{k=1}^m \int_{0}^{x} t^{k-1} dt \\ &= \int_{0}^{x} \sum_{k=1}^m t^{k-1} dt \\ &=\int_{0}^{x} \dfrac{1- t^m}{1-t} dt \\ &= \int_{0}^{x} \dfrac{1}{1-t} dt - \int_{0}^{x} \dfrac{t^m}{1-t} dt \\ &= \ln(1-x) - \int_{0}^{x} \dfrac{t^m}{1-t} dt \\ \int_{0}^{x} \dfrac{t^m}{1-t} dt & \leq x^m \int_{0}^{x} \dfrac{t^m}{1-t} dt \\ & = x^m \ln(1-x) \\ & \xrightarrow[ m \to \infty ]{} 0 \\ \sum_{k=1}^{\infty} \dfrac{x^k}{k}&= - \ln(1-x) \end{align*} $

I integrate term by term , and how to justify it ?

Answer 1

$$S=\sum_{k=1}^{+ \infty} \dfrac{x^{k+1}}{ k (k+1)}$$ $$= x + (1-x) \ln (1-x) $$ Use $$\sum_{k=1}^{\infty} \frac{x^k}{k}=-\ln(1-x)$$ $$S=\sum_{k=1}^{\infty} x^{k+1}\left( \frac{1}{k}-\frac{1}{k+1}\right)=x \sum_{k=1}^{\infty} \frac{x^k}{k}-\sum_{j=2}^{\infty} \frac{x^j}{j}.$$ $$S=-x\ln(1-x)+\ln(1-x)+x=x+(1-x)\ln(1-x)$$

Answer 2

$\int \sum_n f_n(x) dx=\sum_n \int f_n(x)$ if $f_n$ are non-negative continuous (or measuarble ) functions. This is a special case of Tonelli's Theorem. It is also a consequence of Monotone Convergence Theorem.

Answer 3

If $\sum_n f_n'(x) = g'(x)$ with uniform convergence and $\sum_n f_n(x_0) =g(x_0)$ for a particular $x_0$, then $\sum_n f_n(x) = g(x)$ with uniform convergence on compact sets.

The proof is something like $$\Big\lvert\sum_{n=0}^m f_n(x) - g(x)\Big\rvert \leq\Big\lvert\sum_{n=0}^m f_n(x_0) - g(x_0)\Big\rvert+\Big\lvert\int_{x_0}^x\sum_{n=0}^mf_n'(y)-g'(y)dy\Big\lvert \leq \varepsilon(1+\lvert x-x_0\rvert)\leq\varepsilon(1+M).$$

Now, $$\sum_{k=1}^\infty x^{k-1} = \frac1{1-x}$$ uniformly on compact subsets of $[0,1)$. Since $\sum_{k=1}^\infty \frac{0^k}k = 0$, we obtain that $$\sum_{k=1}^\infty \frac{x^k}k = -\log(1-x).$$ We can now apply it again, since $\sum_{k=1}^\infty \frac{0^{k+1}}{k(k+1)} = 0$, and deduce that $$\sum_{k=1}^\infty \frac{x^{k+1}}{k(k+1)} = x + (1-x)\log(1-x).$$