I could use some pointers in how I might go about simplifying a complicated limit. Here it is (pre-simplification):

$\displaystyle \displaymode \lim_{n\to\infty}\left(\lim_{k\to\infty}\left( \frac{\sum_{i=1}^{\lfloor \frac{n+\log_2 {k}}{2\cdot 3^n} \rfloor} \left( \sum_{j=1}^{n+\lfloor \log_2 {k}\rfloor - 2\cdot 3^n i} \binom{n+j-1}{j} \right)}{k} \right) \right)$

Edit:

I just thought of something... Can I take the limit as $\displaystyle k\to 2^{2\cdot 3^n}$? That should simplify it significantly, but I am not sure if I am allowed to do that.

Then the limit would be:

$\displaystyle \displaymode \lim_{n\to\infty} \left( \frac{ \sum_{i=1}^n \binom{n+i-1}{i} }{4^{3^n}} \right)$

Edit 2:

Never mind. I figured it out. Thanks anyway.