# How to find $f(1)+2f(2)+...+nf(n)+...$ if $f(1)+f(2)+...+f(n)+... =0$

Suppose that $\sum_{n=1}^∞{f(n)} = 0$.

Is there any way to calculate the sum $\sum_{n=1}^∞{nf(n)}$?

• No, in general. Commented May 10, 2017 at 8:24
• Surely if $f(n)=1/n^2$, then the initial assumption does not hold (since the sum is equal to one sixth pi squared, not 0), so this function does not fit the criteria. Commented May 10, 2017 at 8:53
• My bad. Let's use $f(n) = 1/n^2 - (\pi^2/12)(1/2)^n$ then. Commented May 10, 2017 at 9:04

Let us define (for $n \geq 1,\ m\geq 1$) $$f_m(n) = \begin{cases} (-1)^{n-1} & \text{if } n \leq m \\ 0 & \text{else} \end{cases}$$ Then $$\sum_{n=1}^\infty f_2(n) = 0\hspace{1cm} \sum_{n=1}^\infty nf_2(n) = -1$$ but $$\sum_{n=1}^\infty f_4(n) = 0\hspace{1cm} \sum_{n=1}^\infty nf_4(n) = -2$$ So you cannot uniquely determine the value of your sum based on the assumption.
If all you know about the sequence $f(1),f(2),f(3),\ldots$ is that $\sum_{n=1}^\infty f(n)=0$, then it's not possible to say anything about $\sum_{n=1}^\infty nf(n)$. For example, if $S$ is any real number, you can have $\sum_{n=1}^\infty nf(n)=S$ from the sequence $-S,S,0,0,0,0,\ldots$.
In somewhat greater generality, if $f(2k-1)=-a_k$ and $f(2k)=a_k$ where $a_k\to0$ as $k\to\infty$, then $\sum_{n=1}^\infty f(n)=-a_1+a_1-a_2+a_2-a_3+\cdots=0$ while
$$\sum_{n=1}^\infty nf(n)=-1_1+2a_1-3a_2+4a_2-5a_3+6a_3-7a_4+\cdots=a_1+a_2+a_3+\cdots=\sum_{k=1}^\infty a_k$$
and it's well known that a series can do pretty much anything it likes, even if its terms are tending to $0$.