Question

Let r be the radius of convergence of the power series
\(\frac{1}{3}+\frac{x}{5}+\frac{x^2}{3^2}+\frac{x^3}{5^2}+\frac{x^4}{3^3}+\frac{x^5}{5^3}+\frac{x^6}{3^4}+\frac{x^7}{5^4}+...\)
Then the value of r² is equal to _________. (Rounded off to two decimal places)

Answer 1

Correct Answer: 3

We identify the pattern of the given power series:

\[\frac{1}{3} + \frac{x}{5} + \frac{x^2}{3^2} + \frac{x^3}{5^2} + \frac{x^4}{3^3} + \frac{x^5}{5^3} + \frac{x^6}{3^4} + \frac{x^7}{5^4} + \cdots\]

Step 1: Determine the general term

For even (n=2k):

\[a_{2k} = \frac{1}{3^{k+1}}\]

For odd (n=2k+1):

\[a_{2k+1} = \frac{1}{5^{k+1}}\]

Step 2: Use radius of convergence formula

\[R = \frac{1}{\limsup_{n\to\infty} |a_n|^{1/n}}\]

Compute limsup for each subsequence:

Even terms:

\[|a_{2k}|^{1/(2k)} = 3^{-(k+1)/(2k)} \to 3^{-1/2}\]

Odd terms:

\[|a_{2k+1}|^{1/(2k+1)} = 5^{-(k+1)/(2k+1)} \to 5^{-1/2}\]

We take the larger of the two limits (since limsup):

\[\limsup |a_n|^{1/n} = 3^{-1/2}\]

Thus:

\[R = \frac{1}{3^{-1/2}} = \sqrt{3}\]

Step 3: Find ( r^2 )

\[r^2 = (\sqrt{3})^2 = 3\]

Final Answer (rounded to two decimals):

\[\boxed{3.00}\]