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I cannot compute

$$\lim_{n\to\infty}\int_1^2\left(1+\frac{\ln x}{n}\right)^n \,\mathrm{d}x.$$

The reason is that I cannot find an antiderivative for $(1+\frac{\ln x}{n})^n $, or, I cannot see if the sequence $\{(1+\frac{\ln x}{n})^n\}_{n=1}^\infty $ is uniformly convergent.

What should we do to find the limit in this case?

SvanN
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serenus
  • 851

1 Answers1

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First, note that for every $x \in [1,2]$,

$$ \lim_{n\to\infty} \left(1+\frac{\ln x}{n}\right)^n = e^{\ln x} =x.$$ This limit is increasing, so that all the functions $f_n(x) := \left(1+\frac{\ln x}{n}\right)^n$ are bounded from above by $f(x) := x$ on $[1,2]$. We also have pointwise convergence $f_n \to f$ on $[1,2]$, so that by Lebesgue's dominated convergence theorem,

$$ \lim_{n\to\infty} \int_1^2 f_n(x)\, \mathrm{d}x = \int_1^2 f(x)\, \mathrm{d}x = \int_1^2 x\, \mathrm{d}x = \tfrac{3}{2}.$$

SvanN
  • 2,307
  • I think we do not need even to know that the sequence ${(1+\frac{lnx}{n})^n }{n=1}^\infty$ is increasing, because note that on $[1,2]$, $lnx\leq lne=1$, so $(1+\frac{lnx}{n})^n\leq (1+\frac{1}{n})^n \leq e$ and we can take the function $f$ as $f(x):=e$ on $[1,2]$. But, of course we should know that the sequence ${(1+\frac{1}{n})^n }{n=1}^\infty$ is increasing. – serenus Jan 18 '19 at 10:40
  • @serenus You're still using that $(1+1/n)^n$ is increasing in $n$ if you want to bound it by $e$ for all $n$. Though indeed it is not strictly necessary that it be increasing; the functions $f_n$ just need to be bounded by some integrable function on $[1,2]$. The fact that the limit is increasing just made the argument even simpler: we can bound them by their (pointwise) limit. – SvanN Jan 18 '19 at 10:42