I'm trying to compute the following limit:

$\displaystyle \lim_{n \to \infty} \int_{0}^{\infty} (1+\frac{x}{n})^{(-n)} \sin(x/n) dx$

Clearly the integrand converges to 0 pointwise so the trick is to find the dominating function. So for simplicity let ${f_{n}}$ denote the sequence of functions of the integrand: it is not hard to see that $|f_{n}| \leq (1+x/3)^{-3}x$ for all $n\geq 3$ and the latter function is integrable on $(0,\infty)$. But doesn't Lebesgue dominated convergence theorem requires to bound it for all $n\in\mathbb{N}$? What happens if we can't obtain a dominating function for $n=1$ or $n=2$ ? Can you please clarify this? I have read in a book that you can drop some terms, say $n=1,2,3$ and the bound the rest , but why is this possible? doesn't this violates the hypothesis of Lebesgue's theorem?


The limit of a sequence $\{ a_{n} : n \geq 1\}$ is the same as that of $\{ a_{n+2} : n \geq 1 \}$.


There are a couple of ways to think about it; as acarchau says, since you are concerned only about the limit, and the "tail" has the same limit, it is enough to consider the tail. Equivalently: the limit of a sequence depends only on what it is doing "eventually". Any amount of "weird" stuff that happens up to a finite bound $N$ is irrelevant (that is, if the sequence is "nice" for all $n\geq N$, then that is enough, because as far as the limit is concerned, we don't care what happens at $n=1$, $n=2,\ldots,n=N-1$).

Or, consider the sequence of functions $\{g_n\}$, where $g_n(x) = f_{N+n}(x)$. Then the hypothesis of Lebesgue's Theorem hold exactly as written for $g_n(x)$, so we know that we can get the conclusion for the $g_n$. Now rewrite $g_n$ as $f_{N+n}$, and do a change of variable $m=N+n$; then $m\to\infty$ as $n\to\infty$, so the limit of $f_{N+n}$ as $n\to\infty$ is the same as the limit of $f_m$ as $m\to\infty$, giving the result.

But all in all, it's always the same moral: what happens at finitely many terms "at the beginning" does not affect the end behavior of a sequence. Only the "eventual" behavior matters.

  • $\begingroup$ Got it now! Thank you, I was making things harder, thanks again,very clear explanation. $\endgroup$ – student Nov 5 '10 at 4:23

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