Possible Values of x for Convergence of Power Series

[Fernando Rios]

Homework Statement

The following series are not power series, but you can transform each one into a power
series by a change of variable and so find out where it converges.

Relevant Equations

Σ((√(x^2+1))^n 2^n/(3^n+n^3))

We transform the series into a power series by a change of variable:
y = √(x²+1)

We have the following after substituting:
∑(2ⁿyⁿ/(3ⁿ+n³))

We use the ratio test:
ρ_n = |(2ⁿ⁺¹yⁿ⁺¹/(3ⁿ⁺¹+(n+1)³)/(2ⁿyⁿ/(3ⁿ+n³)| = |(3ⁿ+n³)2y/(3ⁿ⁺¹+(n+1)³)|

ρ = |(3^∞+∞³)2y/(3^∞+1+(∞+1)³)| = |2y|

|2y| < 1

|y| = 1/2

-1/2 < y < 1/2

We find the possible values of "x":
-1/2 < √(x²+1) < 1/2

This inequality has no solution, so I conclude the series doesn't converge for any "x", but the book says the answer is |x| < √(5)/2. Could you please tell me what am I doing wrong?

 

[Mark44]

Homework Statement:: The following series are not power series, but you can transform each one into a power
series by a change of variable and so find out where it converges.
Relevant Equations:: Σ((√(x^2+1))^n 2^n/(3^n+n^3))

We transform the series into a power series by a change of variable:
y = √(x²+1)

We have the following after substituting:
∑(2ⁿyⁿ/(3ⁿ+n³))

We use the ratio test:
ρ_n = |(2ⁿ⁺¹yⁿ⁺¹/(3ⁿ⁺¹+(n+1)³)/(2ⁿyⁿ/(3ⁿ+n³)| = |(3ⁿ+n³)2y/(3ⁿ⁺¹+(n+1)³)|

ρ = |(3^∞+∞³)2y/(3^∞+1+(∞+1)³)| = |2y|

No. You should never substitute $\infty$ into an arithmetic expression.

BTW, I get a limit of $\frac 2 3 |y|$.

|2y| < 1

|y| = 1/2
-1/2 < y < 1/2

Aside from the error I mentioned above, in your substitution, y will always be nonnegative, since $y = \sqrt{x^2 + 1}$.

We find the possible values of "x":
-1/2 < √(x²+1) < 1/2

This inequality has no solution, so I conclude the series doesn't converge for any "x", but the book says the answer is |x| < √(5)/2. Could you please tell me what am I doing wrong?

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