# On the sum of the series $\sum_{n=0}^{\infty} \left( \frac{1}{3}\right)^{n + \frac{k_n}{2^n}}$

I have this series

$$\sum_{n=0}^{\infty} \left( \frac{1}{3}\right)^{n + \frac{k_n}{2^n}}$$

where all the entries in $(k_n)$ are integers and the sequence $\left( \frac{k_n}{2^n} \right)$ is non-decreasing and convergent. I need to prove that for any two different sequences $(k_n)$ this series sums up to different numbers. I thought that if I take two sequences where one is lexicographically smaller, giving a bigger summand, then the sum corresponding to the other sequence will automatically be smaller. However, this is not the case. For sequences

$$k_n = \begin{cases} 0 & n \le 2 \\ 2^{n+2} & \mathrm{else} \end{cases}$$ and $$l_n = \begin{cases} 0 & n \le 1 \\ 2^{n-2} & \mathrm{else} \end{cases}$$

where $(k_n)$ is lexicographically smaller, the sum corresponding to $(l_n)$ is still bigger.

This problem is in an article by A. Lindenbaum http://matwbn.icm.edu.pl/ksiazki/fm/fm23/fm2314.pdf Theorem 9.

I'd appreciate a hint or something to help me prove this statement.

• @Dr.MV well, in this case the first is lexicographically smaller and it's sum is bigger, what do you mean whether they work? – David Nov 27 '16 at 16:55
• @Dr.MV what I meant was that for any(!) two different sequences, their series sum up to different numbers. – David Nov 27 '16 at 22:19