Limit of a series

skipjack

Forum Staff
Dec 2006
21,181
2,332
If this were true then the following would be a geometric series
\[ \lim_{n \to \infty} \sum_{k = 0}^{n-1} n^k \]
since for fixed $n$ it has $n$ as a common ratio. But it isn't geometric and it clearly doesn't converge.
For $n \geqslant 2$, \(\displaystyle \sum_{k = 0}^{n-1} n^k = \frac{n^n - 1}{n - 1}\) by the usual formula for the sum of a finite geometric series.

In this case, \(\displaystyle \frac{n^n - 1}{n - 1}\) is unbounded as $n\to\infty$.

For the original problem, to show that \(\displaystyle \lim_{n\to\infty} \frac{1 - (x + 1/n)^n}{1 - x - 1/n} = \frac{1}{1- x}\) for $x \in (0,\,1)$,
start by showing that \(\displaystyle \lim_{n\to\infty} (x + 1/n)^n= 0\).