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Problem: Investigate the convergence of $\sum a_n$, where $a_n =\int_{0}^{1} \frac{x^n \sin(\pi x)}{1-x} \, \mathrm{d}x$.

I'm thinking about changing $\frac{1}{1-x}$ to $\sum x^k$ and then exchange the integral and the sum but it came back to the original form. Or I just take the sum of $a_n$ switch the sum and integral since the integrand is nonnegative and measurable function. Then the integrand becomes $\frac{\sin(\pi x)}{(1-x)^2}$,so I think the series should diverge.

Spenser
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Topoguy
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2 Answers2

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The series diverges. We have that $$\sin\pi x\geq \pi x(1-x),\quad\forall x\in[0,1]$$ (see here), so $$a_n=\int_0^1\frac{x^n\sin \pi x}{1-x}dx\geq\int_0^1\pi x^{n+1}dx=\frac{\pi}{n+2}.$$ Hence $\sum a_n$ diverges by the comparison test.

Community
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Spenser
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The monotone convergence theorem applies. Let $s_n$ denote the $n^{th}$ partial sum. Then$$ s_n=\int_0^1\dfrac{\sum_{i=0}^n x^i \sin(\pi x)}{1-x}dx=\int_0^1\dfrac{(1-x^{n+1})\sin(\pi x)}{(1-x)^2}dx$$Let's look at the integrand, $f_n(x)=\dfrac{(1-x^{n+1})\sin(\pi x)}{(1-x)^2}$. We know that $$f_n\to \dfrac{\sin(\pi x)}{(1-x)^2}$$ pointwise, so you just need to show that this function is monotonic increasing, which is a simple calculation (that I can do out if you don't see it). Then by the monotone convergence theorem, $$\lim_{n\to\infty} s_n=\sum_{i=0}^{\infty}a_i=\lim_{n\to\infty}\int_0^1 \dfrac{(1-x^{n+1})\sin(\pi x)}{(1-x)^2}dx=\int_0^1\dfrac{\sin(\pi x)}{(1-x)^2}dx=\infty$$

Moya
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