Prove that $(n+\frac{1}{2})\log(1+\frac{1}{n})-1=\frac{1}{3(2n+1)^2}+\frac{1}{5(2n+1)^4}+\cdots$

Question

In From Real to Complex Analysis there is an exercise in which you must prove that, if $d_n=\log n!-(n+\frac{1}{2})\log n+n$, then $$\begin{split}d_n-d_{n+1}&=\left(n+\frac{1}{2}\right)\log\left(1+\frac{1}{n}\right)-1\\ &=\frac{1}{3(2n+1)^2}+\frac{1}{5(2n+1)^4}+\cdots\end{split}$$ I can do the first line, but I am stuck on the second. I tried the Taylor series expansion of $\log x$ centred at 1 to get $$\log\left(1+\frac{1}{n}\right)=\sum\limits_{r=1}^{\infty}\frac{(-1)^{r-1}}{rn^r},$$ but now, after some manipulations, I am just left with proving that $$\sum\limits_{r=1}^{\infty}\frac{(-1)^{r-1}(2n+1)^{2r+1}(2r+1)-2rn^r}{2rn^r(2r+1)(2n+1)^{2r}}=1,$$which isn't helpful. I also had the idea to try to find an infinite series of the form of the expansion of some logarithm that simplifies to to the given sum (this sum does not have alternating sign, so I was thinking of trying to find a logarithm expansion in which pairs of terms simplify to the given sum, but I couldn't do this). Any help would be appreciated.

Answer 1

Hint:

$$\sum_{r=1}^\infty\dfrac1{(2r+1)(2n+1)^{2r}}$$

$$=-1+\sum_{r=0}^\infty\dfrac1{(2r+1)(2n+1)^{2r}}$$

$$=(2n+1)\sum_{r=1}^\infty\dfrac{\left(\frac1{2n+1}\right)^{2r+1}}{2r+1}$$

Now we know $\ln(1-z)=-\dfrac{z^r}r$ for $-1<|z|<1$

$\ln(1+y)-\ln(1-y)=?$

Answer 2

Hint: \begin{align*}\frac{x^2}{3}+\frac{x^4}{5}+\cdots&=\frac{1}{x}(\frac{x^3}{3}+\frac{x^5}{5}+\cdots)\\ &=\frac{1}{2x}(\ln(1+x)-\ln(1-x)-2x)\end{align*}