Finding the sum $\sum\limits_{n=2}^{\infty}5^{-(n+1)}\ln\big(\frac{n^5}{n+1}\big)$ 
I want to find the sum of this series: 
  $$\sum_{n=2}^{\infty}5^{-(n+1)}\ln\bigg(\frac{n^5}{n+1}\bigg)$$


I have went through theses steps:
$$\sum_{n=2}^{\infty}\frac{1}{5^{n+1}}\ln\bigg(\frac{n^5}{n+1}\bigg)=\sum_{n=2}^{\infty}\frac{1}{5^{n+1}}[\ln(n^5)-\ln(n+1)]=\sum_{n=2}^{\infty}\frac{1}{5^{n+1}}[5\ln(n)-\ln(n+1)]$$
But I can't get past this.
 A: An eloboration on the comment and something just so this question will not remain unanswered:
You can distribute the $\frac{1}{5^{n+1}}$ to get:
$$\sum_{n=2}^{\infty}\frac{\ln(n)}{5^n}−\frac{\ln(n+1)}{5^{(n+1)}}$$
Luckily the second term $−\frac{\ln(n+1)}{5^{(n+1)}}$ cancels with the $\frac{\ln(n)}{5^n}$ when $n$ becomes $n+1$. The terms get smaller and smaller because $\ln x$ grows super slowly compared to $5^n$, so the terms go to zero as $n$ goes to infinity. So all that's left is 
$$\frac{\ln(2)}{25}\approx0.0277259$$
A: First, note that the series converges.
$$\sum_{n=2}^{\infty}5^{-(n+1)}\ln\bigg(\frac{n^5}{n+1}\bigg)$$
To see the convergence, observe that $5^{(n+1)}$ grows much faster than $\ln\bigg(\frac{n^5}{n+1}\bigg)$.
$5^{(n+1)}$ is in the denominator and $\ln\bigg(\frac{n^5}{n+1}\bigg)$ in the numerator.
$$ 
\begin{array}{rcl|l}
  \sum_{n=2}^{\infty}5^{-(n+1)}\ln\bigg(\frac{n^5}{n+1}\bigg)  & = &  \sum_{n=2}^{\infty}\frac{1}{5^{n+1}}[5\ln(n)-\ln(n+1)  & \text{basic properties of logorithms}  \\
  & = &  \sum_{n=2}^{\infty}\frac{1}{5^{n+1}}*5\ln(n)-\sum_{n=2}^{\infty}\frac{1}{5^{n+1}}*\ln(n+1)  & \text{distribute}  \\
  & = &  \sum_{n=2}^{\infty}\frac{1}{5^{n+1}}*5\ln(n)-\sum_{k=3}^{\infty}\frac{1}{5^{k}}*\ln(k) & \text{substitute k = n + 1 }  \\
&  &    & \text{for index of summation in rightmost sum}  \\
  & = &  \sum_{n=2}^{\infty}\frac{1}{5^{n}}*\ln(n)-\sum_{k=3}^{\infty}\frac{1}{5^{k}}*\ln(k)  & \frac{1}{5^{n+1}} = \frac{1}{5}*\frac{1}{5^{n}}  \\
  & = &  \frac{1}{5^{2}}*\ln(2) + \sum_{n=3}^{\infty}\frac{1}{5^{n}}*\ln(n)-\sum_{k=3}^{\infty}\frac{1}{5^{k}}*\ln(k)  & \text{pull $n=2$ term out of the leftmost sum}  \\
  & = &  \frac{1}{5^{2}}*\ln(2) + 0  & \text{the two series are identical}  \\
  & = &  \frac{\ln(2)}{25}  &   \\
\end{array}
 $$
