Take the following sum. $$f(k,p)=\frac{\sum_{i=1}^k2^{p_{i-1}}3^{k-i}}{2^{p_k}-3^k}$$ where the set $p$ is an arbitrarily increasing set of positive integers. Explain why $f(k,p)$ is only natural when $f(k,p)=1$.
So far I've come up with using long division to get that $$f(k,p)=\sum_{n=1}^{\infty}\sum_{i=1}^k\frac{3^{nk-i}}{2^{np_k-p_{i-1}}}$$
And that turns out to just be an infinite sum of incrementing powers of 3 over arbitrarily increasing powers of 2. So, effectively, $$f(k,p)=\sum_{n=1}^{\infty}\frac{3^{n-1}}{2^{p_n}}=\frac{1}{\sqrt[n]{2^{p_n}}-3}$$ $$\therefore p_n=n\log_2\Big(\frac{1}{f(k,p)}+3\Big)$$
But the above just shows that every $p_n$ could be that and the algebra would work. But, we need $p$ to be a sequence of positive integers, so in reality, it can't be defined by the $p_n$ formula above.
Again, the question is why can't $f(k,p)$ be a natural number greater than 1 when we have the restriction on $p$?
The statement is false. The sum can be any positive integer desired. Even more, we can start with any desired finite sequence $p_n$ and extend it to form any sum that we have not already exceeded.
If our target is $N$, just take $p_n$ the smallest number greater than $p_{n-1}$ that leaves the sum below $N$. The sum is bounded above and monotonic, so it converges. If $p_n=p_{n-1}+1$ the $n^{th}$ term is $\frac 32$ times the $n-1^{st}$ term so we can make the terms get larger and larger until we get close.
For example, if we want to represent $4$ the $p_n$ start $$1,2,3,5,7,11,15,19,21,27,30,33,35,43,48,51,54$$ by which point we are at $3.999999998$. If $s_{n}$ is the sum of the first $n$ terms we have $$p_n=\max\left(p_{n-1}+1,\left\lceil \log_2 \left(\frac {3^{n-1}}{N-s_{n-1}}\right)\right\rceil\right)$$