# Find $A=\sum\limits_{n=1}^{\infty}\frac{\log 2^n}{3^n}$

I can't find

$$A=\sum_{n=1}^{\infty}\frac{\log 2^n}{3^n}$$

-
Is a term of that summation supposed to be $$\frac{\log(2^n)}{3^n}$$ or $$\quad\;\frac{(\log 2)^n}{3^n}\quad?$$ –  Zev Chonoles Jul 14 '13 at 4:12
Suggestion: Use a law of logarithms and factor out $\log 2$ from the infinite sum. –  hardmath Jul 14 '13 at 4:12
I agree with @hardmath. What will be left is an "arithmetic-geometric sum", which you can evaluate. –  Eric Auld Jul 14 '13 at 4:14

Here's what you would do depending on where the power is supposed to be:

$1$. If the general term is $\log(2^n)/3^n$, you have

$$\sum_{n=1}^\infty\frac{\log(2^n)}{3^n}=\log(2)\sum_{n=1}^\infty\frac{n}{3^n}$$ Because $1/3<1$, this series can be evaluated using the following formula:

$$\sum_{n=1}^\infty nz^n=\frac{z}{(1-z)^2}$$This formula is derived by differentiating the geometric series (and multiplying by $z$), which brings us to:

$2$. If the general term is $(\log(2))^n/3^n$, what you have is exactly a geometric series:

$$\sum_{n=1}^\infty\frac{(\log(2))^n}{3^n}=\sum_{n=1}^\infty\left(\frac{\log(2)}{3}\right)^n$$ since $\log(2)<3$, we know that $\log(2)/3<1$, and so this series can be evaulated using the formula

$$\sum_{n=1}^\infty z^k=\frac{z}{1-z}$$

-
+1 Very nice answer. –  DonAntonio Jul 14 '13 at 9:00