Let the walk on the positive integer axis $\{0,1,2,...\}$ with the following step probabilities. $p_i:=p_{i,i+1}=1-(1/2)^{i+1}$, $q_i:=p_{i,0}=(1/2)^{i+1}$. I know that the chain is transient.
I want to prove that the only existing invariant measure is $\mu=0$.
For this $\mu$ has to satisfy
$\mu_o=\sum_{i\geq0}{\mu_iq_i}$ and $\mu_i=\mu_{i-1}p_{i-1}$ for $i\neq0$.
Then the hint is that since $p_i$ goes to $1$ very fast it follows that $p:=\prod_{i\geq0}{p_i}>0$ stating that this is equivalent to $\sum_{i=0}^\infty{q_i}=\infty$, which does not make sense since the sum is convergent...
The "hint" is a bit off. Recall that if $0<a_n<1$ then $$\prod_{n=0}^\infty (1-a_n)>0 \iff \sum_{n=0}^\infty a_n<\infty. $$ Since \begin{align} \prod_{n=0}^m (1-p_n) &= \prod_{n=0}^m 2^{-(n+1)}\\ &= 2^{\displaystyle\left(\lg\prod_{n=0}^m 2^{-(n+1)}\right)}\\ &= 2^{\displaystyle-\left(\sum_{n=0}^m (n+1) \right)}\\ &= 2^{\left(-\dfrac{(m+1)(m+2)}{2}\right)}\stackrel{m\to\infty}\longrightarrow 0, \end{align} it follows that $$\sum_{n=0}^\infty\left(1-2^{-(n+1)} \right)=\infty. $$