Show that$$ln(1+\frac{1}{n})^n=1-\frac{1}{2(n+1)}-\frac{1}{2.3(n+1)^2}-\frac{1}{3.4(n+1)^3}-....$$ using logarithmic series.
I try to bring the RHS in the format of logarithmic series but I can't able to do it.
So I try to use binomial theorem on LHS then expand it but I am again fail.
Please someone give hint to solve it .thanks for your help in advance.
On
HINT:
$$\ln\left(1+\dfrac1n\right)^n=n\ln\left(1+\dfrac1n\right)=(n+1)\ln\left(1+\dfrac1n\right)-\ln\left(1+\dfrac1n\right)$$
and $$\ln\left(1+\dfrac1n\right)=-\ln\left(1-\dfrac1{n+1}\right)$$
Now use $\ln(1-y)=-\sum_{r=1}^\infty\dfrac{y^r}r$
On
Letting $x=\frac{1}{n+1}$ then $1+\frac 1n= \frac{1}{1-x}$.
So $$\begin{align}\log(1+1/n)^n &= n\log(1+1/n)\\ &=-\left(\frac{1}{x}-1\right)\log (1-x)\end{align}$$
Then use that $$-\log(1-x)=\sum_{k=1}^{\infty} \frac{x^k}{k}$$
Hint for the last step: If $f(x)=\sum_{k=1}^{\infty} a_kx^k$ then show $$\left(\frac 1x-1\right)f(x)=a_1 + \sum_{k=1}^{\infty}(a_{k+1}-a_k)x^k$$
$$\dfrac1{r(r+1)(n+1)^r}=\dfrac{r+1-r}{r(r+1)(n+1)^r}$$ $$=\dfrac{1/(n+1)^r}r-(n+1)\cdot\dfrac{1/(n+1)^{r+1}}{r+1}$$
$$\implies\sum_{r=1}^\infty\dfrac1{r(r+1)(n+1)^r}=\sum_{r=1}^\infty\dfrac{1/(n+1)^r}r-(n+1)\sum_{r=1}^\infty\dfrac{1/(n+1)^{r+1}}{r+1}$$
$$=-\ln\left(1-\dfrac1{n+1}\right)-(n+1)\left\{-\ln\left(1-\dfrac1{n+1}\right)+\dfrac1{n+1}\right\}$$
$$=?$$
as $\ln(1-y)=-\sum_{r=1}^\infty\dfrac{y^r}r$