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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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How does $\frac{2(n-1)}{n!}$ compare to $\frac{1}{n}$? Which one is larger? Can you use this to deduce the limit? – Ayman Hourieh Apr 21 '12 at 10:38
Please see if my edit is correct, namely the denominators. – Américo Tavares Apr 21 '12 at 10:41
If this is homework, please add the [homework] tag; people will still help, so don't worry. – Américo Tavares Apr 21 '12 at 11:09
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 $n$. – André Nicolas Apr 21 '12 at 14:28

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

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


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

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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So it converges to 0. As do the other terms. Not convincing anyway. But appreciate your hint. I use this forum for self-study. – Ignace Apr 21 '12 at 12:20
@Ignace Yes, it converges to $0$. Why not convincing? You can apply the sqweeze theorem. – Américo Tavares Apr 21 '12 at 13:01
Américo, thanks. But the original problem being the series u(n+1) = 1+1/n*(u(n)) begin with 2 suggests a convergence to at least 2. We tackled correctly the part of the formula which I described in the question. So, not convincing in the sense that I don't have a limit >= 2, as I assume, for my problem. – Ignace Apr 21 '12 at 13:17
@Ignace Do you want to solve the recurrence $$u(n+1) = 1 + \frac{1}{n\cdot u(n)},$$ with $u(1) = 2?$ I've thought you had wanted to evaluate the limit $$\lim_{n\to \infty}\frac{n-2}{n!}=0.$$ – Américo Tavares Apr 21 '12 at 13:38
Yes Americo, the goal is to solve 1+ (1/n)*u(n) (not 1/(n*u(n)) as you put it). I just reworked this problem as a sum of formulas in n as a variable. See original question. And 2(n-2)/n! was one of these in the sum. This is something I have troubles with, can you solve it working with the original form u(n+1) = ... or should it be reworked as a formula, as I did, before we can attempt a convergence strategy ? – Ignace Apr 21 '12 at 14:50

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