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I wanted to examine the convergence of the series $$\displaystyle x_{n+1}= \frac{n}{n+1} x_n$$ and try to find its limit, but I'm having difficulty doing so. The only test I thought would be useful (ratio test) was inconclusive and I'm having trouble proving $0 \leq x_n \leq 1$ that I think could help end up with a geometric sequence. Any suggestions are welcome.

$x_1 = 1$ (This is $\forall n \geq 1$)

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Try finding the first few values $x_n$ takes, and try to spot the pattern – Amr Dec 15 '12 at 22:51
up vote 1 down vote accepted

After a little effort in filling the dots, you will see the following:

$$x_{n+1} = \frac{n}{n+1} x_n = \frac{n}{n+1}\frac{n-1}{n}x_{n-1} = \cdots = \frac{x_1}{n+1} = \frac1{n+1}$$

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oh I got $\frac {2}{n+1}$ a little while ago, but I was convinced that was useless. Thanks – Casquibaldo Dec 15 '12 at 22:55
I suppose its limit is 0. – Casquibaldo Dec 15 '12 at 22:57
Yes sir. It is $0$. – Isomorphism Dec 15 '12 at 23:01
It doesn't converge though... – Casquibaldo Dec 15 '12 at 23:33
@Casquibaldo: $\frac{2}{n+1}$ converges to 0. (Unless you are talking about the series $\sum \frac{2}{n+1}$, which is completely different.) – sdcvvc Dec 15 '12 at 23:37

You're given a sequence such that


$$x_{n+1}=\frac n {n+1} x_n$$

Define $$u_n=nx_n$$

Then the sequence becomes

$$u_1=1\cdot x_1=1$$


It follows that $u_n=1$ for all $n$, so that

$$x_n=\frac 1 n $$

for each $n$. The limit of this series is, in turn, $0$.

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