<s>[INST] Find formula for ${-\frac{2\ln3}{3} , \frac{4\ln4}{9} ,\frac{6\ln5}{27},\frac{8\ln6}{81} ,....}$ The question is to find a general formula for the nth term, $a_n$, of the sequence:
$${-\frac{2\ln3}{3} , \frac{4\ln4}{9} ,\frac{6\ln5}{27}, \frac{8\ln6}{81} ,\frac{10\ln7}{243},...}$$
Here is what I got, but when I plugged in a few terms, it did not work out right.
I got the formula to be: $$a_n = (-1)^n \frac{2\ln(2+1)\cdot 2^{(n-1)}} {3^n}$$
Please help. Thanks
 [/INST] Each term has a form like $\pm \frac{x\ln y}{z}$.
The sign alternates so it is $(-1)^n$. 
$x$ is $2, 4, 6, 8, \dots$, so $x$ is $2n$. 
$y$ is $3,4,5,\dots$ so $y$ is $n+2$. 
$z$ is $3,9,27, 81, \dots$ so $z$ is $3^n$.
Combining all contributions, we get $a_n = (-1)^n \frac{2n \ln(n+2)}{3^n}$.
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