What is the equivalent of $f$ when $x$ approaches $1$?

Question

$f(x) = \sum_{n=0}^{+ \infty} \ln(n).x^n$

$g(x) = \sum_{n=1}^{+ \infty} a_n.x^n $, such that: $ a_1=-1$ and $a_n=-\ln \big(1 - \frac{1}{n} \big) - \frac{1}{n} $ for $n \ge 2$.

1. Prove that: $g(x) = (1 -x)f(x) + \ln(1-x) $.

2. We admit that $u_n=\sum_{k=1}^{n} \frac{1}{k} - \ln n$ converges toward $\alpha$. (Euler constant). Compute $g(1)$ and find en equivalent of $f$ near $1$.

3. Prove that $\sum_{k=1}^{+ \infty} \frac{(-1)^{n-1}}{n}$ converge and $\sum_{k=1}^{+ \infty} \frac{(-1)^{n-1}}{n} = \ln2$

4. Using the formula: $n! \sim (\frac{n}{e})^n \sqrt{2 \pi n}$, compute $g(-1)$.

5. Deduce $\lim_{x\to -1} f(x)$.

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1. I have proved that: $ g(x) = (1-x)f(x) + \ln (1 -x) $

2. $$ \begin{align} g(1) & = \sum_{k=1}^{n} a_k \\ & = a_1 + \sum_{k=2}^{n} a_n \\ & = -1 + \sum_{k=2}^n -\ln \big(1 - \frac{1}{k} \big) - \frac{1}{k} \\ & = -1 + \ln(n) - \sum_{k=2}^{n} \frac{1}{k} \\ & = - \sum_{k=1}^{n} \frac{1}{k} - \ln n\\ & = - \alpha \end{align} $$

For the equivalent, I need to find a function $h(x) $such that $\lim \frac{f(x)}{h(x)} = 1$ when $x$ approaches $1$ which I don't know how to do.

3. I am stuck in the second part of proving the serie converges to $\ln 2$, How can I do it?

4. I could not see the link between the given formula and the value of $g(-1)$. How can I compute it?

5. I think I will find it from the equation $ g(x) = (1-x)f(x) + \ln (1 -x) $ once I will know the value of $g(-1)$ which I don't know how to do.

Thank you for your help.

Answer 1

1. First $g(1)$ does not equal to what you wrote. Still your result is good.

2. Just use the previous result :

$$ f\left(x\right)=\frac{g\left(x\right)}{1-x}-\frac{\ln\left(1-x\right)}{1-x} $$ Hence $$ f\left(x\right)\frac{x-1}{\ln(1-x)}=\frac{g(x)}{\ln(1-x)}+1 \underset{x \rightarrow 1}{\rightarrow}1 $$ Hence $$ f\left(x\right) \underset{(1)}{\sim}\frac{\ln(1-x)}{x-1} $$ 3. A beautiful trick is to consider the partial sum and remark that

$$ \left(-1\right)^{n-1}x^n=\left(-1\right)^{n-1}\int_{0}^{1}x^{n-1}\text{d}x=\int_{0}^{1}\left(-x\right)^{n-1}\text{d}x $$ Hence $$ \sum_{k=1}^{N}\frac{\left(-1\right)^{n-1}}{n}=\sum_{k=1}^{N}\int_{0}^{1}\left(-x\right)^{n-1}\text{d}x=\int_{0}^{1}\sum_{k=1}^{N}\left(-x\right)^{n-1}\text{d}x $$ So $$ \sum_{n=1}^{N}\frac{\left(-1\right)^{n-1}}{n}=\int_{0}^{1}\frac{1-\left(-x\right)^{N-1}}{1+x}=\int_{0}^{1}\frac{\text{d}x}{1+x}-\int_{0}^{1}\frac{\left(-x\right)^{n-1}}{1+x}\text{d}x$$

The second integral satisfies $$ \left|\int_{0}^{1}\frac{\left(-x\right)^{n-1}}{1+x}\text{d}x\right| \leq \int_{0}^{1}x^{n-1}=\frac{1}{n}\underset{n \rightarrow +\infty}{\rightarrow}0 $$ Letting $n \rightarrow +\infty$ $$ \sum_{n=1}^{+\infty}\frac{\left(-1\right)^{n-1}}{n}=\int_{0}^{1}\frac{\text{d}x}{1+x}=\ln(2)$$

4. You have $$ g\left(-1\right)=-\sum_{k=2}^{+\infty}\left(-1\right)^k\ln\left(1-\frac{1}{k}\right)+\ln(2) $$

Let $\displaystyle S_N=\sum_{k=2}^{N}\left(-1\right)^k\ln\left(1-\frac{1}{k}\right)$ You have $$ S_{2N}=\sum_{k=2}^{2N}\left(-1\right)^k\ln\left(1-\frac{1}{k}\right)=\sum_{k=1}^{N}\ln\left(1-\frac{1}{2k}\right)-\sum_{k=1}^{N-1}\ln\left(1-\frac{1}{2k+1}\right) $$ hence $$ S_{2N}=\sum_{k=1}^{N}\ln(2k-1)-\sum_{k=1}^{N}\ln(2k)-\sum_{k=1}^{N-1}\ln(2k)+\sum_{k=1}^{N-1}\ln(2k+1) $$ Regrouping we have $$ S_{2N}=2\sum_{k=1}^{N-1}\ln(2k+1)-2\sum_{k=1}^{N-1}\ln(2k)+\ln(2N)$$

Then you have $$ e^{S_{2N}}=\frac{1}{2N}\left(\frac{\prod_{k=1}^{N-1}2k+1}{\prod_{k=1}^{N-1}2k}\right)^2=2N\frac{\left(2N\right)!^2}{\left(2^NN!\right)^4} $$

Now apply Stirling formula as it is written we got $$2N\frac{\left(2N\right)!^2}{\left(2^NN!\right)^4} \underset{N\rightarrow +\infty}{\rightarrow}\frac{2}{\pi}$$ And

$$ g\left(-1\right)=-\ln(\frac{2}{\pi})+\ln(2)=\ln\left(\pi\right) $$

If i havent made too much mistakes lol

5. $$ f\left(-1\right)\underset{(-1)}{\sim}\frac{g(-1)}{2}-\frac{\ln(2)}{2}=\ln\left(\sqrt{\frac{\pi}{2}}\right) $$