Let the number sequence $a_n$ be recursively defined by $a_0 = 1$ and
$a_n= a_{n-1}.(\frac{n}{n+1})^n$ if $n ≥ 0$
Does the series $\sum_{n=0}^{\infty}a_n$ diverge or converge?
(Please note that it's the convergence of the series that I am trying to determine, not the sequence)
How should I handle the recursive formula of this problem? I did tried using the common convergence tests but none seemed to work. Can someone please give me a hint? Thank you all!
Best Answer
We have $(1+x)^n = 1 + xn + \ldots$, thus $(\frac{n+1}n)^n = (1+\frac1n)^n = 1 + n \frac1n + ... \ge 1 + n \frac1n = 2$ and $a_{n+1} = \frac{a_n}{(1+n^{-1})^n} \le \frac{a_n}2$. Thus $a_{n+1} \le \frac{a_n}2 \le \frac{a_{n-1}}4 \le \ldots \le \frac{a_1}{2^n}$. Hence $\sum_{n \ge 1} a_n < \infty$.