If x_n has a subsequence converging to infinity than the corespondent subsequence of n x_n + 2^x_n will converge to infinity. If x_n has a subsequence converging to - infinity then we get - infinity for the other sequence. If x_n has a subsequence converging to a nonzero finite y then the corresponding subsequence of the other sequence will go to infinity (if y > 0) or - infinity (if y < 0)

From here it follows that x_n must have limit 0 and that n x_n has limit x - 1.

The conclusion follows.

Whose problem is it?

It's my! Proposed problems, RMT No.4/2006. A nice statement! What's your opinion?

Please give me some time to remember my solution!

Somehow, I deduced that n x_n is bounded from the convergence of n x_n + 2^x_n. So it follows that x_n must have limit 0.

Proof ( x(n) is bounded ):

nx(n)+2^x(n) is bounded, being convergent. It results nx(n)

If a subsequence diverges then a subsequence of the subsequence converges to something.

The result you said is not direct is called D J Newman's Lemma, well known in analysis. See D J's problem book.

May I suggest a similar, but slightly different one?

If (1/n)x_n + 2^(x_n) is convergent to x then… same continuation, with 2 replaced by a.

If $x_n$ has a subsequence converigng to $c> 0$ or to infty, then (reducing to the subsequence if necessary) $nx_n+a^x_n>nx_n$ will converge to +infty for all $a$.

The same with $cnx_n$ will converge to -infty by writing $nx_n+a^x_n

First, let us remark that if the conclusion of the exercise is true, we must have (nx_{n}+1^{x_{n}})_{n} converges also to x, what means that (nx_{n})_{n} converges to x-1. And so x_{n}=O((1/n)), what implies, (a^{x_{n}})_{n} converges also to 1, and so, (nx_{n}+a^{x_{n}})_{n} converges also to x, for all real number a>0.

All what we have to prove, is that the result is true for a=1. To do this, we must prove that the sequence (nx_{n})_{n} is bounded. If it is not the case, there exists a subsequence (x_{ϕ(n)})_{n} such that

ϕ(n)x_{ϕ(n)}→-∞.

Since (nx_{n}+2^{x_{n}})_{n} converges, we must have

2^{x_{ϕ(n)}}→+∞

and so x_{ϕ(n)}→+∞, which is in contradiction with ϕ(n)x_{ϕ(n)}→-∞. What permits to conclude.