# Putnam problem from 1949 - lim sup

1. Aug 13, 2011

### jbunniii

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.

1. The problem statement, all variables and given/known data

Let $(a_n)$ be an arbitrary sequence of real, positive numbers. Show that

$$\limsup_{n \rightarrow \infty} \left(\frac{a_1 + a_{n+1}}{a_n}\right)^n \geq e$$

2. Relevant equations

$$e = \lim_{n \rightarrow \infty}\left(1 + \frac{1}{n}\right)^n$$

3. 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: Aug 13, 2011
2. Aug 13, 2011

### e(ho0n3

It looks right to me.

3. Aug 13, 2011

### upsidedowntop

I'm happy with it.

4. Aug 13, 2011

### jbunniii

Cool, thanks for taking the time to check it out.