Discussion
The radius of convergence of the power series \(\displaystyle\sum_{n=0}^{\infty} \frac{z^{n}}{(n^{2} + 1)2^{n}}\) is :
Answer: D
The radius of convergence is 2.
With \(a_{n} = \dfrac{1}{(n^{2} + 1)2^{n}}\), the root test gives
\[ |a_{n}|^{1/n} = \frac{1}{2(n^{2} + 1)^{1/n}} \to \frac{1}{2} \]
since \((n^{2} + 1)^{1/n} \to 1\). So \(R = 1/(1/2) = 2\). The polynomial factor \(n^{2} + 1\) helps convergence on the boundary circle \(|z| = 2\), where the series converges absolutely, but it does not change the radius.
1/2 is the reciprocal mistake (taking the limit itself as the radius), 1 ignores the \(2^{n}\), and \(\infty\) would need the coefficients to shrink faster than any geometric sequence.
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