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There is recurrence relation formula below.

$$\frac{1-c}{(n+1)^c}=(a_{n+1})^{1-c}-(a_{n})^{1-c}$$ $$c>0,\ c\neq1\ \ \ \ \ a_{n+1}>a_{n},\ a_1=1$$

Question. How can I find the limit below? $$\lim_{n\to\infty}a_{n}=\ ?$$

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up vote 6 down vote accepted

Here’s the answer that I was working on when I got side-tracked by Ross’s post.

Let $b_n = a_n^{1-c}$; then the $b_n$ satisfy the recurrence $b_{n+1} = b_n + \frac{1-c}{(n+1)^c}$, with initial condition $b_1 = 1$. This recurrence expresses $b_n$ as a sum: $$b_n = 1 + (1-c) \sum\limits_{k=0}^{n-2}\frac{1}{(n-k)^c} = 1 + (1-c)\sum\limits_{k=2}^n\frac{1}{k^c}. $$

If $0<c<1$, $\sum k^{-c}$ diverges, so $\lim\limits_{n\to\infty}b_n = \infty$. For each $n$, $a_n = b_n^{1/(1-c)}>b_n$, so $\lim\limits_{n\to\infty}a_n = \infty$ as well.

If $c>1$, $\sum_{k=2}^\infty k^{-c} = \zeta(c)-1$, so $\lim\limits_{n\to\infty}b_n = 1+(1-c)(\zeta(c)-1) = (1-c)\zeta(c)+c$, and $\lim\limits_{n\to\infty}a_n = \left((1-c)\zeta(c)+c\right)^{1/(1-c)}$, provided that $(1-c)\zeta(c)+c>0$.

Indeed this is the case for all $c>1$. To prove this, one can note that $\zeta(c)-1=\sum\limits_{n\ge2}n^{-c}$ and compare the series to the integral $I(c)=\int\limits_1^{+\infty}x^{-c}\mathrm{d}x$. Since the function $x\mapsto x^{-c}$ is decreasing on $x\ge1$, $n^{-c}\le\int\limits_{n-1}^nx^{-c}\mathrm{d}x$ for every $n\ge2$ hence $\zeta(c)-1<I(c)=1/(c-1)$ and the proof is complete.

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+1 Sorry for the distraction. –  Ross Millikan Aug 20 '11 at 5:17
    
@Ross: No problem. I’m somewhat prone to carelessness, so a distraction that makes me double-check myself is valuable! –  Brian M. Scott Aug 20 '11 at 5:31
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