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If $k > 0$ how can the interchange between the summation and the integration for the following expression be formally justified:

$$\int_0^{\frac{\pi}{2}} \sum_{n = 0}^\infty \frac{(-1)^n (2k \cos x)^{2n}}{(2n + 1) \binom{2n}{n}} \, dx$$

My thoughts were to apply the following form of Fullini's theorem. It says that for general $f_n$, if $\int \sum |f_n| < \infty$ or $\sum \int |f_n| < \infty$, then $\int \sum f_n = \sum \int f_n$. Applying this result we have \begin{align} \int_0^{\frac{\pi}{2}} \sum_{n = 0}^\infty \left |\frac{(-1)^n (2k \cos x)^{2n}}{(2n + 1) \binom{2n}{n}} \right | \, dx &< \int_0^{\frac{\pi}{2}} \sum_{n = 0}^\infty \frac{(2k)^{2n}}{(2n + 1)\binom{2n}{n}} \, dx\\ &= \frac{\pi}{2} \sum_{n = 0}^\infty \frac{(2k)^{2n}}{(2n + 1)\binom{2n}{n}}\\ &= \frac{\pi}{2} \frac{\arcsin (k)}{k\sqrt{1 - k^2}} < \infty, \end{align} where the series expansion used can be found here. This result however is only valid for $|k| < 1$ whereas in my case $k > 0$.

omegadot
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    If $k>1$ and $x \simeq 0$ then $\sum_{n = 0}^\infty \frac{(-1)^n (2k \cos x)^{2n}}{(2n + 1) \binom{2n}{n}}$ is divergent by the divergence test. – N. S. May 03 '21 at 04:04
  • Let $j=[0,\pi /2]$....If $|k|<1$ then the partial sums $f_n$ in your series are continuous and converge uniformly, so the full sum $f$ is continuous (so $f$ is integrable), and $|\int_jf -\int_jf_n|=$ $|\int_j(f-f_n)|\le$ $\le \int_j|f-f_n|\le$ $ (\pi /2)\max {|f(x)-f_n(x)|:x\in j}\to 0$ by uniform convergence of $f_n$ to $f.$ – DanielWainfleet May 03 '21 at 05:45

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