Weird recurrence [closed]

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 $$N > 1$$.

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

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• It looks like this recurrence defines exactly one function. – Kevin Long Nov 8 '18 at 22:34
• I rephrased the question slightly. – user132290 Nov 8 '18 at 22:45

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).

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