The radius of convergence of the power series $\sum_{n=1}^\infty\left(\frac{2^n}{n}+\frac{3^n}{n^2}\right)z^n$ is ____.

Question

Match List-I with List-II and choose the correct option:

LIST-I (Function)	LIST-II (Value)
(A) $ \int_{\gamma} \frac{1}{z-a} \, dz $, where $ \gamma: \|z-a\|=r, r > 0 $	(III) $ 2i\pi $
(B) $ \int_{\gamma} \frac{z+2}{z} \, dz $, where $ \gamma: z = 2e^{it}, 0 \le t \le \pi $	(IV) $ i\pi $
(C) $ \int_{\gamma} \frac{e^{2z}}{(z-1)(z-2)} \, dz $, where $ \gamma: \|z\|=3 $	(II) $ 2i\pi(e^4 - e^2) $
(D) $ \int_{\gamma} \frac{z^2 - z + 1}{2(z-1)} \, dz $, where $ \gamma: \|z\|=2 $	(I) $ -4 + 2i\pi $

Choose the correct answer from the options given below:

Answer 1

Step 1: Split the series into two separate power series: $\sum \dfrac{2^n}{n}z^n$ and $\sum \dfrac{3^n}{n^2}z^n$, and find the radius of convergence of each using the ratio test.
Step 2: For $b_n=\dfrac{2^n}{n}$, $\left|\dfrac{b_{n+1}}{b_n}\right|=\dfrac{2^{n+1}/(n+1)}{2^n/n}=\dfrac{2n}{n+1}\to 2$. So this series has radius of convergence $R_1=\dfrac{1}{2}$.
Step 3: For $c_n=\dfrac{3^n}{n^2}$, $\left|\dfrac{c_{n+1}}{c_n}\right|=\dfrac{3^{n+1}/(n+1)^2}{3^n/n^2}=3\left(\dfrac{n}{n+1}\right)^2\to 3$. So this series has radius of convergence $R_2=\dfrac{1}{3}$.
Step 4: When two power series with different radii of convergence $R_1\neq R_2$ are added termwise, the resulting series has radius of convergence equal to $\min(R_1,R_2)$: inside $|z|<\min(R_1,R_2)$ both converge absolutely so the sum converges, while for $\min(R_1,R_2)<|z|<\max(R_1,R_2)$ the series with the smaller radius has terms that do not tend to $0$, forcing the sum's terms to also not tend to $0$, so the sum diverges there.
Step 5: Since $R_1=\dfrac{1}{2}$ and $R_2=\dfrac{1}{3}$ are different, the radius of convergence of the given series is $\min\left(\dfrac{1}{2},\dfrac{1}{3}\right)=\dfrac{1}{3}$.\[\boxed{R=\dfrac{1}{3}}\]

LIST-I (Function)	LIST-II (Value)
(A) \( \int_{\gamma} \frac{1}{z-a} \, dz \), where \( \gamma: \|z-a\|=r, r > 0 \)	(III) \( 2i\pi \)
(B) \( \int_{\gamma} \frac{z+2}{z} \, dz \), where \( \gamma: z = 2e^{it}, 0 \le t \le \pi \)	(IV) \( i\pi \)
(C) \( \int_{\gamma} \frac{e^{2z}}{(z-1)(z-2)} \, dz \), where \( \gamma: \|z\|=3 \)	(II) \( 2i\pi(e^4 - e^2) \)
(D) \( \int_{\gamma} \frac{z^2 - z + 1}{2(z-1)} \, dz \), where \( \gamma: \|z\|=2 \)	(I) \( -4 + 2i\pi \)

The radius of convergence of the power series \(\sum_{n=1}^\infty\left(\frac{2^n}{n}+\frac{3^n}{n^2}\right)z^n\) is ____.

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The Correct Option is D

Solution and Explanation

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