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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.

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    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

1 Answers 1

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Hint. For $n>2$, we have \begin{eqnarray*} 0<\frac{n-2}{n!} &=&\frac{n}{n!}-\frac{2}{n!}=\frac{1}{(n-1)!}-\frac{2}{n!}<\frac{1}{(n-1)!}. \end{eqnarray*}

Added. Since $\frac{1}{(n-1)!}\to 0$, as $n\to \infty$, applying limits, you get

$\lim_{n\to \infty}\frac{n-2}{n!}=0.$

Thus

$\lim_{n\to \infty}2\cdot\frac{n-2}{n!}=0.$

Remark. See comments bellows. I misunderstood the question. OP wants to solve the recurrence

$u(n+1)=1+\dfrac{1}{n}u(n),\qquad u(1)=2.$

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    @Ig$n$ace: fixed.2012-04-22