- #1
- 3,488
- 257
This is a nice Putnam problem from the 1949 exam. It also appears as challenge problem 2.14.16 in Thomson, Bruckner and Bruckner, Elementary Real Analysis. I think I solved it correctly, but it seemed a little too easy so if anyone would like to check my solution, I would appreciate it.
Let (a_n) be an arbitrary sequence of real, positive numbers. Show that
\limsup_{n \rightarrow \infty} \left(\frac{a_1 + a_{n+1}}{a_n}\right)^n \geq e
e = \lim_{n \rightarrow \infty}\left(1 + \frac{1}{n}\right)^n
Suppose not. Then there exists N \in \mathbb{N} and \alpha \in \mathbb{R} such that
\left(\frac{a_1 + a_{n+1}}{a_n}\right)^n \leq \alpha < e
for all n \geq N. Also, there exists M \in \mathbb{N} such that
\left(1 + \frac{1}{n}\right)^n > \alpha
for all n \geq M. Therefore, for all n \geq \max(N,M), we have
0 < \left(\frac{a_1 + a_{n+1}}{a_n}\right)^n < \left(1 + \frac{1}{n}\right)^n
and therefore
0 < \frac{a_1 + a_{n+1}}{a_n} < 1 + \frac{1}{n} = \frac{n+1}{n}
After rearrangement, this is equivalent to
\frac{a_n}{n} - \frac{a_{n+1}}{n+1} > \frac{a_1}{n+1}
This inequality holds for all n \geq \max(M,N). If I write out the inequality for n through n+k-1, and then sum these inequalities, then the LHS telescopes, and I obtain
\frac{a_n}{n} - \frac{a_{n+k}}{n+k} > \sum_{j = n}^{n+k-1}\frac{a_1}{j+1}
This holds for any positive integer k. If I hold n fixed and let k grow large, the right hand side is unbounded. In particular, there is some k so large that the right-hand side is larger than a_n/n. Thus for that k, we have
\frac{a_n}{n} - \frac{a_{n+k}}{n+k} > \frac{a_n}{n}
or equivalently
\frac{a_{n+k}}{n+k} < 0
and thus
a_{n+k} < 0
This is a contradiction because the sequence contains only positive numbers.
Homework Statement
Let (a_n) be an arbitrary sequence of real, positive numbers. Show that
\limsup_{n \rightarrow \infty} \left(\frac{a_1 + a_{n+1}}{a_n}\right)^n \geq e
Homework Equations
e = \lim_{n \rightarrow \infty}\left(1 + \frac{1}{n}\right)^n
The Attempt at a Solution
Suppose not. Then there exists N \in \mathbb{N} and \alpha \in \mathbb{R} such that
\left(\frac{a_1 + a_{n+1}}{a_n}\right)^n \leq \alpha < e
for all n \geq N. Also, there exists M \in \mathbb{N} such that
\left(1 + \frac{1}{n}\right)^n > \alpha
for all n \geq M. Therefore, for all n \geq \max(N,M), we have
0 < \left(\frac{a_1 + a_{n+1}}{a_n}\right)^n < \left(1 + \frac{1}{n}\right)^n
and therefore
0 < \frac{a_1 + a_{n+1}}{a_n} < 1 + \frac{1}{n} = \frac{n+1}{n}
After rearrangement, this is equivalent to
\frac{a_n}{n} - \frac{a_{n+1}}{n+1} > \frac{a_1}{n+1}
This inequality holds for all n \geq \max(M,N). If I write out the inequality for n through n+k-1, and then sum these inequalities, then the LHS telescopes, and I obtain
\frac{a_n}{n} - \frac{a_{n+k}}{n+k} > \sum_{j = n}^{n+k-1}\frac{a_1}{j+1}
This holds for any positive integer k. If I hold n fixed and let k grow large, the right hand side is unbounded. In particular, there is some k so large that the right-hand side is larger than a_n/n. Thus for that k, we have
\frac{a_n}{n} - \frac{a_{n+k}}{n+k} > \frac{a_n}{n}
or equivalently
\frac{a_{n+k}}{n+k} < 0
and thus
a_{n+k} < 0
This is a contradiction because the sequence contains only positive numbers.
Last edited:
