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$$2\cdot(n-2)/n!\; ?$$

Convergence to $?$. It's part of a larger problem but the rest are all null sequences. The serie is : $$u(n+1) = 1 + 1/n\cdot u(n),\qquad (u(n) \text{ is the }n\text{-th} \text{ term}), \qquad u(1) = 2.$$

I managed to redefine the serie to a formula in $n$, being

$$1/n + 1/n\cdot(n-1) + 2\cdot(n-2)/n! + 1/n\cdot(n-1)\cdot(n-2).$$ I assume apart from the problem with $n!$ that all others are null-sequences.

By induction I proved $u(n) \le 3$.

So, if it converges to $2$ or $3$, some of the terms in the formula must be $\ge 1$. I think thus, that $2\cdot(n-2)/n!$ cannot be a null-sequence.

  • 0
    How does $\frac{2(n-1)}{n!}$ compare to $\frac{1}{n}$? Which one is larger? Can you use this to deduce the limit?2012-04-21
  • 0
    Please see if my edit is correct, namely the denominators.2012-04-21
  • 0
    If (in the second displayed line) you are giving a formula for $u(n)$, it cannot be correct. For note from the recurrence that $u(n)>1$ for all $n$.2012-04-21

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