# 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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Holo, John B, Leucippus, José Carlos Santos, Christopher
If this question can be reworded to fit the rules in the help center, please edit the question.

• 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

## 2 Answers

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