Prove that lim n → ∞ n 2 ∫ 0 1 n x x + 1 d x = 1 2

Question

Prove that      lim          n      →      ∞            n    2        ∫          0                      1        n                  x          x      +      1        d  x  =            1      2

Dobricap · Accepted Answer

My favorite technique, start with the Taylor expansion of

x^{x + 1}

around x=0. I am only carrying the first two terms because doing this by hand can get very tedious. And the steps can all be rigorously justified.

\begin{array}{rcl} x^{x + 1} & = & x + \ln (x) x^{2} + \dots \\ \int_{0}^{1 / n} x^{x + 1} d x & = & \frac{1}{2 n^{2}} - \frac{1 + 3 \ln (n)}{9 n^{3}} + \dots \\ n^{2} \int_{0}^{1 / n} x^{x + 1} d x & = & \frac{1}{2} - \frac{1 + 3 \ln (n)}{9 n} + \dots \\ lim_{n \to \infty} n^{2} \int_{0}^{1 / n} x^{x + 1} d x & = & \frac{1}{2} . \end{array}

Prove that lim_(n-> infty) n^2 int_0^(1/n) x^(x+1)dx=1/2

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New Questions in Calculus 1