Answer clearly wrong; what went wrong?

Question

I was trying to answer this question Not sure which test to use?

Here was my clearly wrong response:

$\ln(1 + \frac{1}{3^k})$ = $\frac{\ln((1+\frac{1}{3k})^k)}{k}$

$\sum_{k=1}^\infty 2^{k-1}\ln((1+3^{-k})^k )$

$\sum_{k=0}^\infty 2^{k}\ln(\sum_{k=0}^\infty \binom{k+1}{k}(3^{-k})^k$)

$\sum_{k=0}^\infty 2^{k}\ln(\sum_{k=0}^\infty (k+1)3^{{-k}^2})$

$\sum_{k=0}^\infty 2^{k}\ln(\sum_{k=0}^\infty \frac{(k+1)}{3^{{k}^2}})$

$\sum_{k=0}^\infty 2^{k}\ln(\sum_{k=0}^\infty \frac{(k)}{3^{{k}^2}} + \frac{(1)}{3^{{k}^2}})$

$\sum_{k=0}^\infty 2^{k}\ln(1.70390704...)$

$ \ln(1.70390704...)\sum_{k=0}^\infty 2^{k}$...and here we see that this clearly diverges when it clearly should converge. What went wrong?

EDIT: Here is a revised answer:

$\ln(1 + \frac{1}{3^k})$ = $\frac{\ln((1+\frac{1}{3k})^k)}{k}$

$\sum_{k=1}^\infty 2^{k-1}\ln((1+3^{-k})^k )$

$\sum_{k=0}^\infty 2^{k}\ln(\sum_{i=0}^\infty \binom{k}{i}(3^{-k})^i$)

$\sum_{k=0}^\infty 2^{k}\ln(\sum_{i=0}^\infty \binom{k}{i}(3^{-ik}$)

But this is no better! Now where do I go from here? More specifically, how would I evaluate

$\sum_{i=0}^\infty \binom{k}{i}(3^{-ik})$?

Answer 1

Don't use the same variable $k$ in both sums!

Answer 2

$$ (a+b)^k = \sum_{i=0}^k \binom{k}{i} a^{k-i} b^{i} $$

Using this for $(1+3^{-k})^k$, we get (here $a=1$ and $b=3^{-k}$) $$ (1+3^{-k})^k = \sum_{i=0}^k \binom{k}{i} 3^{-ki}. $$

What you wrote seems completely different: you use $k$ as the summation index in the new sum, so nobody knows which $k$ is used where, you have a sum from $0$ to $\infty$, and the coefficient in your sum is $\binom{k+1}{k}$ (again, which $k$ are these?).