Tighter upper bound on $x$ where $2^x \leq \sum_{i=0}^m{{x \choose i}\lambda^i}$ We have the following inequality:
$$2^x \leq \sum_{i=0}^m{{x \choose i}\lambda^i}$$
All the variables are in $\mathbb{N}_{>0}$
I need to find a tight upper bound for $x$ using $m,\lambda$.
In the case of $\lambda = 1$ we can use the binomial theorem to show $x \leq m$. However for $\lambda>1$ I have no idea how to find a tight upper bound for this.
It can be shown that:
$$2^x \leq \sum_{i=0}^m{{x \choose i}\lambda^i} \leq \left(\frac{\lambda e x}{m}\right)^m$$
And then we can use the solution from here: Upper bound $2^x \leq (ax)^c$
But I need a tighter bound than this. Is there any way to bound $x$ directly from this partial binomial theorem sum?
I thought of maybe doing something like this:
$$2^x = (1 + \lambda)^{x\log_{1 + \lambda}(2)}=(1 + \lambda)^{\frac{x}{\log_2(1 + \lambda)}}=\\ \sum_{i=0}^{{\frac{x}{\log_2(1 + \lambda)}}}{{{\frac{x}{\log_2(1 + \lambda)}} \choose i}\lambda^i} \leq \sum_{i=0}^m{{x \choose i}\lambda^i}$$
But I'm not sure how to continue from here (or if it even helps).
 A: This is more of a long comment than an answer, but I don't get the same upper bound that you get in the $\lambda \leq 1$ case.
Assuming that $\lambda$ (and hence everything) is positive, it seems to me that:
$$\sum_{i=0}^m{{x \choose i}\lambda^i} \leq \sum_{i=0}^x{{x \choose i}\lambda^i} $$
with equality if and only if $m \geq x$.
But the right hand side of this new inequality equals $(1 + \lambda)^x$, by the binomial theorem.
So substituting this back into the original inequality we obtain:
$$2^x \leq (1 + \lambda)^x$$
When $\lambda > 1$ we get this inequality for free and so we don't learn anything new about $x$, which is similar to the problem you experienced.
When $\lambda = 1$ we have equality in the last inequality I typed, which means that we also need equality in the first equality I typed which impies $x \leq m$ as you also found. 
But if $\lambda < 1$ then this inequality puts a rather strong restriction on $x$, namely:
$$x = 0$$
For any $x > 0$ the above inequality with $\lambda < 1$ is violated.
