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Yes, there is such a bound. Below I list a few variants. To avoid trivial cases we need to assume that $f(x)$ is not of the form $h(x)^p-h(x)+r$ for any polynomial $h(x)$ and any constant $r$. I also include sums where the multiplicative also has a polynomial argument. Then non-triviality condition takes the form that if $\psi$ is of order $d$ the polynomial ...
Let $f(x)=1-e^{-x}$. Since $f(x)$ is a concave function on $\mathbb{R}^+$, for any $z>0$ $$\sum_{k=0}^{+\infty}f\left(\frac{z}{2^k}\right)\leq f(z/2)=1-e^{-z/2}$$ holds in virtue of Jensen's inequality, hence the series is convergent. In order to get a "closed form" for the series, just consider the Taylor series of $f(x)$: $$f(x) = ... 1 The trick is to play around with the Taylor Series terms of e^x.$$\cosh(x) = \frac{e^{x}+e^{-x}}{2} \\ = \frac{\sum_{n=0}^\infty\frac{x^n}{n!}+\sum_{n=0}^\infty\frac{(-x)^n}{n!}}{2}$$Now break the sums apart further into their even and odd terms, cancel the odd ones, group the even ones and see what is left. For example, you can write the sum ... 0 If it's just a finite sum over consecutive integer powers of 4 or 8, then there is a closed form and wolfram can definitely help. Sorry to comment in the answer, my account is not rich enough to comment yet. 0 This is only true if you factor out e^{-j\Omega 3}. Then you obtain$$\sum_{n=0}^6 e^{-j\Omega n} = e^{-j\Omega 3} \left(e^{j\Omega 3} + e^{j\Omega 2} + e^{j\Omega} + 1 + e^{-j\Omega} + e^{-j\Omega 2}+e^{-j\Omega 3} \right).$$To evaluate this, you can use that$$2 \cos(x) = e^{x} + e^{-x}.$$1 Making it more compact, it seems to me that the equation you consider can write$$F(k)=\sum_{j=1}^m a_j^k-b=0 where all terms are known except $k$ (which I suppose to be a real number). I really do not know if any analytical method could apply to the problem. If I am right, then only numerical methods could be used (such as Newton as the simplest) ...