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

If we knew $\sum_{i=1}^{n} f(i)=S$ ($n$ can be $\infty$), is there anything we could say about $\sum_{i=1}^{n} \dfrac {1}{f(i)}$ in terms of $S$? The only thing I was able to find on the web is link to a dead question. If there is no known relationship, is it possible for there to exist one that we just haven't discovered?

• @William If the infinite sum (series) is convergent, then its general term $f(i)$ goes to zero, so $1/f(i)$ doesn't go to zero and the other sum is divergent. – Clement C. Jul 12 '16 at 19:34
• You need to know $$g(x) = \sum_{k=1}^{n}\exp\left[xf(k)\right]$$ – Count Iblis Jul 12 '16 at 19:36

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.

• Has it been proved impossible to establish a general relationship between $\sum_{i=1}^{n} f(i)$ and $\sum_{i=1}^{n} \dfrac {1}{f(i)}$, or is it possible that we just haven't found one yet? – Ovi Jul 12 '16 at 20:44
• @Ovi It's impossible in the general sense. The two expressions are completely independent real numbers except for that the product of the two must be at most $n^2$, assuming $f(i)$ are all positive. However, if you are okay with a relationship involving other symmetric expressions (my answer), or you make additional assumptions on $f$ (this answer, other answers), then you can find a relationship. – 6005 Jul 12 '16 at 21:39


• Are you serious?? – Dragonemperor42 Jul 13 '16 at 8:18
• @Roby5 $$0 \leq \sum_{i = 1}^{n} \left[\sqrt{\mathrm{f}\left(i\right)} - {1 \over \sqrt{\mathrm{f}\left(i\right)}} \right]^{\, 2}$$ – Felix Marin Jul 13 '16 at 8:21
• Well, I don't think the OP was looking for this. Anyways, $(+1)$ for the rather amusing effort. – Dragonemperor42 Jul 13 '16 at 8:23
• @Roby5: on the opposite, that's a non-trivial result and possibly the best for $f(i)>0$, as the bound is tight. – Yves Daoust Jul 13 '16 at 8:48
• @YvesDaoust What happens when $n \rightarrow \infty$? – Dragonemperor42 Jul 13 '16 at 8:54

Let $a_i = f(i)$, for $i = 1, 2, \ldots, n$. Then you are asking for a relationship between $$e_1 = p_1 = \sum_{i=1}^n a_i$$ and $$p_{-1} = \sum_{i=1}^n \frac{1}{a_i}.$$ Where did this terminology come from? $e_i$ is the $i$th elementary symmetric polynomial in variables $a_1, \ldots, a_n$, and is defined for $i = 0$ to $i = n$. We have \begin{align*} e_0 &= 1 \\ e_1 &= a_1 + a_2 + \ldots + a_n \\ e_2 &= a_1a_2 + a_1a_3 + \ldots + a_{n-1}a_n \\ e_3 &= a_1a_2a_3 + a_1a_2a_4 + \ldots + a_{n-2}a_{n-1}a_n \\ &\cdots \\ e_n &= a_1a_2a_3\ldots a_n \end{align*} And $p_i$ (for any integer $i$) is the sum of the $i$th powers of the $a_i$: \begin{align*} &\cdots \\ p_{-1} &= a_1^{-1} + a_2^{-1} + \cdots + a_n^{-1} = \frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n} \\ p_0 &= 1 + 1 + \cdots + 1 = n \\ p_1 &= a_1 + a_2 + \cdots + a_n \\ p_2 &= a_1^2 + a_2^2 + \cdots + a_n^2 \\ &\cdots \end{align*}

## So, is there any known relationship?

If you're looking for an answer involving only these two quantities and simple expresions, the answer is no. There cannot be any relationship, because the two expressions $p_1$ and $p_{-1}$ are independent functions of the variables. By this I mean you can find two sequences $a_1, a_2, \ldots, a_n$ and $a_1', a_2', \ldots, a_n'$ that give the same value for $p_1$, but different values for $p_{-1}$, and vice versa, so there is no simple formula for one in terms of the other, nor any general dependency.

However, if you're willing to allow other quantities into the mix and just want to know how the two quantities are related, symmetric polynomials are where to look. There is an entire lass of identities known as Newton's Identities relating $p_i$ to each other and to $e_i$. (Unfortunately, the wikipedia case only defines $p_k$ for $k \ge 1$, but there's no need to be that restrictive.) So you're looking for a relationship between $e_1$ and $p_{-1}$. What such relationships exist? Well, for example, we have $$p_{-1} = \frac{e_{n-1}}{e_n}$$ at which point an equation relating $p_{-1}$ and $e_1$ is the same an equation relating $e_1, e_{n-1},$ and $e_n$, such as $$p_0 e_n - e_{n-1} p_1 + e_{n-2} p_2 - e_{n-3} p_3 + \cdots + (-1)^n p_n e_0 = 0.$$ In general, the relationship between these symmetric polynomials $p_i$ and $e_i$ is such that if you fix $n$ symmetric polynomials, you can compute all the rest in terms of those $n$ (often, you have to recursively compute them) but until you have fixed $n$ you are interested in, you won't be able to get full relationships between any. So you want a relationship between $e_1$ and $p_{-1}$, but you can only express a relationship that depends on some $n$ symmetric polynomial inputs (and you haven't said which $n$).

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

$$f(n) = \sum_{d|n} g(d) \Leftrightarrow g(n) = \sum_{d|n} \mu(\frac{n}{d}) f(d) = \sum_{d|n} \mu(d) f(\frac{n}{d})$$

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 $g = \frac{1}{f}$.