<p>Which representation is valid? \(\displaystyle\sum_{n=0}^\infty\int_0^1 x^n e^{-x}\,dx=?\) [JEE Advanced 2011]</p>
Step-by-Step Solution
Key Concept: \sum\int_0^1 xⁿe^(-x)dx = \int_0^1 e^(-x) \cdot \sumxⁿ dx = \int_0^1 e^(-x)/(1-x)dx (geometric series, valid for x\in [0,1)).
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<p>By uniform convergence on $[0,1)$:</p>
<p>$$\sum_{n=0}^\infty\int_0^1 x^n e^{-x}dx=\int_0^1 e^{-x}\sum_{n=0}^\infty x^n\,dx=\int_0^1\frac{e^{-x}}{1-x}dx$$</p>
<p>(The geometric series $\sum x^n = 1/(1-x)$ converges for $|x|<1$; at $x=1$ the integrand has a singularity but the integral converges.)</p>
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Correct Answer: A