# Find the limit of the sequence given by recurrence relation

Find the limit of the sequence given by recurrence relation $c_{n+1} = (1-\frac{1}{n})c_n + \beta_n$ where $\beta_n$ is any sequence with the property $n^2\beta_n \to 0$ as $n \to \infty$

I've proved that $c_{n+1} = \frac{1}{n}\sum\limits_{i=1}^n i\beta_{i}$ but then i got stuck.

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I get:
$c_2 = \beta_1$
$c_3 = (1/2)c_2 + \beta_2 = (1/2)\beta_1 + (2/2)\beta_2$
$c_4 = (2/3)c_3 + \beta_3 = (1/3)\beta_1 + (2/3)\beta_2 + (3/3)\beta_3$
$c_5 = (3/4)c_4 + \beta_4 = (1/4)\beta_1 + (2/4)\beta_2 + (3/4)\beta_3 + (4/4)\beta_4$

Which seems to be a different solution than yours.

Also, this will give a similar limit for $c_n$ under the weaker condition $n\beta_n\rightarrow 0$.

Hour later: Yes, I see you have edited your question to match the solution suggested above. So this is an average of $i\beta_i$ values.

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From the above pattern you can get an explicit formula for $c_n$ that you can verify works. From that you can take its limit under the weaker condition $n\beta_n\rightarrow 0$. – Michael Jun 9 '14 at 14:13
@Michael sorry. copied from a page in a notebook with mistake. edited my question – antimatr0id Jun 9 '14 at 14:21
@Diliuskh: Yes, that is the solution I get. It is the average of $i\beta_i$ values. – Michael Jun 9 '14 at 14:25

We have $n c_{n+1} = (n-1) c_n + \gamma_n$ where $\gamma_n = n \beta_n$, and so $n \gamma_n \to 0$

Let $c_n = \frac{1}{n-1}\sum_{j=1}^{n-1} a_j$. Then

$$\sum_{j=1}^{n} a_j = \sum_{j=1}^{n-1} a_j +\gamma_n \implies a_n=\gamma_n$$

Hence $c_n$ is the Cesaro mean of $\gamma_n$. Because $\gamma_n \to 0$, so does $c_n$ ( unless I made some mistake, it seems that the weaker condition $n\beta_n \to 0$ was enough).

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