And, more importantly, can anyone point me to a proof either way? My barely remembered high-school math is sufficient to demonstrate that for $p=2$ the expression explodes, but not enough to say whether there is some value of $p$ near 1 for which the expression still converges. Invoking the basic principle that I would have heard of it if such a value existed has not sufficiently resolved the question for me.
Asked
Active
Viewed 89 times
2
-
10$\left(p+\frac1n\right)^n > p^n$. – Daniel Fischer Jan 09 '14 at 17:12
-
Presumably, you mean "does (this limit) exist for any $p>1$?" As written, the title is meaningless. – Thomas Andrews Jan 09 '14 at 17:13
-
Yes, I overlook a missing word. Corrected. – Jan 09 '14 at 17:18
2 Answers
6
Note that for any constant $p > 1$ $$ \left( p + \frac{1}{n} \right)^n > p^n > 1$$ and thus $$ \lim_{n \to \infty}\left( p + \frac{1}{n} \right)^n \ge \lim_{n \to \infty} p^n = \infty$$ and $$\lim_{n \to \infty}\left( 1 + \frac{1}{n} \right)^n = e \approx 2.71828... > 1 $$
LinAlgMan
- 2,924
2
Another way of looking at this is
$$ \left( p + \frac{1}{n}\right)^n = \left( p \left(1+ \frac{1/p}{n}\right)\right)^n= \left( p \right)^n \left(1+ \frac{1/p}{n}\right)^n. $$
The second term goes to $e^{1/p}$ as $n \to \infty$ so overall, this limit tends to
$$ \left(\lim_{n\to\infty} p^n\right) e^{1/p} $$
which is going to be infinite for any $p > 1$. For $p = 1$, the limit is $e$, for $|p| < 1$ the limit is $0$, and for $p \leq -1$, the limit does not exist.
BaronVT
- 13,613