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Consider a function $f(k)$ defined for positive integers $k=1,2,\cdots \infty$ the function is satisfies the condition $f(1)+f(2)+\cdots = p/(p-1)$, where $0<p<1$, how to determine f(k) from this information?

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How about $f(1) = p/(p-1)$ and $f(k) = 0$ for all $k > 1$? Is there some information missing? –  Alex Nov 12 '11 at 6:02
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Another function, $f(n):=-p^n$, certainly satisfies the conditions. This is because we have the following geometric series for $0<p<1$: $$ \sum_{n=1}^\infty f(n) = - \sum_{n=1}^\infty p^n = -\frac{p}{1-p} = \frac{p}{p-1}$$ –  matt Nov 12 '11 at 6:07
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There are infinitely many (indeed uncountably many) functions $f$ such that $\sum_{k=1}^\infty f(k)=\frac{p}{p-1}$. –  André Nicolas Nov 12 '11 at 6:33

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