Is there any known relationship between $\sum_{i=1}^{n} f(i)$ and $\sum_{i=1}^{n} \dfrac {1}{f(i)}$

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \color{#f00}{\sum_{1 = 1}^{n}\mathrm{f}\pars{i} + \sum_{1 = 1}^{n}{1 \over \mathrm{f}\pars{i}}} & \geq \color{#f00}{2n} \end{align}

When $n=\infty$, the reciprocal sum will either blow up or oscillate forever by the arguments in the comments. For finite $n$, there is rarely much you can do.

Here's a well known example of when you do know the reciprocal:

$$\sum_{d|n}d=n\sum_{d|n}\frac{1}{d},$$

where the sum is over the divisors of $n$. Another example is $f(i)=p^i$ in which case you get:

$$\sum_{i=1}^nf(i)=\frac{1-p^{n+1}}{1-p}=:S$$ $$\sum_{i=1}^n\frac{1}{f(i)}=\frac{S}{p^n}$$

For a more classical example, take $f(i)=i^r$. Then the sum of $f(i)$ will be a Faulhaber polynomial, whereas the reciprocal sum will be a generalized harmonic number. The two aren't really related to each other in an obvious way In general, you have:

$$\sum_{i=1}^n \frac{1}{f(i)}=\frac{\sum_{i=1}^n\prod_{j\neq i}f(j)}{\prod_{i=1}^nf(i)}.$$

The numerator is quite annoying here, and generally intractable.

If you are lucky with the expression of $f$, you can play a bit applying the Mobius inversion formula.

Let $f, g : \mathbb{N} \mapsto G$ two functions from $\mathbb{N}$ to an addictive abelian group $G$.

Then

\begin{equation} f(n) = \sum_{d|n} g(d) \Leftrightarrow g(n) = \sum_{d|n} \mu(\frac{n}{d}) f(d) = \sum_{d|n} \mu(d) f(\frac{n}{d}) \end{equation}

where $\mu$ is the Mobius function defined as

• $\mu(n) = 1$ if $n=1$
• $\mu(n) = (−1)^k$ if $n$ is a square-free positive integer with $k$ different prime factors.
• $\mu(n) = 0$ if $n$ has a squared prime factor.

Then some results could follow from letting $g = \frac{1}{f}$.