Find:
limn→∞∫n0(1+xn)n+1exp(−2x)dx
The sequence (1+xn)n+1exp(−2x) converges pointwise to exp(−x). So if we could apply Lebesgue's monotone convergence theorem, we have:
limn→∞∫n0(1+xn)n+1exp(−2x)dx=limn→∞∫RI(0,n)(x)(1+xn)n+1exp(−2x)dx=∫∞−∞exp(−x)dx=+∞
Can someone help me to prove that the sequence of integrands is monotone?
Answer
Using the indicator function, we can write
∫n0(1+xn)n+1e−2xdx=∫∞0(1+xn)n+1e−2xξ[0,n](x)dx
Then, note that for 0≤x≤n, we have
(1+xn)n+1=(1+xn)(1+xn)n≤2ex
EDIT: The OP requested a source for the preceding inequality.
In THIS ANSWER, I showed using only the limit definition of the exponential function along with Bernoulli's Inequality that the exponential function satisfies the inequality ex≥(1+xn)nfor x>−n.
We can assert, therefore, that
(1+xn)n+1e−2xξ[0,n](x)≤2e−x
Applying the Dominated convergence theorem, we have
limn→∞∫n0(1+xn)n+1e−2xdx=∫∞0limn→∞((1+xn)n+1ξ[0,n](x))e−2xdx=∫∞0ex(1)e−2xdx=1
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