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Find an explicit formula of the function E defined on natural numbers such that

$E(1) = 1$ and $E(N) = 1 + \frac{E(1) + ... + E(N-1)}{N}$ for $N > 1$.

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closed as off-topic by Holo, John B, Leucippus, José Carlos Santos, Christopher Nov 9 '18 at 16:05

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  • $\begingroup$ It looks like this recurrence defines exactly one function. $\endgroup$ – Kevin Long Nov 8 '18 at 22:34
  • $\begingroup$ I rephrased the question slightly. $\endgroup$ – user132290 Nov 8 '18 at 22:45
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Let $N > 2$. Now,

$$ \begin{align} E(N) & = 1 + \frac{E(1) + \dots + E(N-1)}{N} = \frac{\color{red}{[E(1) + \dots + E(N-2)]} + E(N-1)}{N} =\\ & = 1 + \frac{\color{red}{(N-1)(E(N-1)-1)} + E(N-1)}{N} = \color{blue}{1} + \frac{NE(N-1) - (N-1)}{N} = \\ & = \frac{NE(N-1) + \color{blue}{1}}{N} = E(N-1) + \frac{1}{N}. \end{align} $$

We also have that $E(2) = E(1) + \frac{1}{2}$ by a direct calculation, so we get

$$ E(N) = \sum_{j=1}^N\frac{1}{j}. $$

It is likely that we cannot do much better than this, without getting a more complicated expression (of course, that depends on what you intend to do with this).

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Possible approach 1:

  • You might calculate the first few terms, then take first differences i.e. $E(n)-E(n-1)$, and then see if you can spot a pattern

Possible approach 2:

  • $E(N) = 1 + \frac{E(1) + \cdots + E(N-1)}{N}$
  • so $N E(N) = N + {E(1) + \cdots + E(N-1)}$
  • so $(N+1) E(N) + 1 = (N+1) + {E(1) + \cdots + E(N-1)+E(N)}$
  • so $E(N+1) = 1 + \frac{E(1) + \cdots + E(N-1)+E(N)}{N+1} = E(N) + \frac{1}{N+1}$
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