Question

To find The interval of convergence of the series and explicit formula for f(x)

Short answer

The interval of convergence is (-1,1) and the formula for \(f(x) = \frac{{2x + 1}}{{1 - {x^2}}}{\rm{. }}\)

Step-by-step solution

Observe that if \(1 \le |x|\)the series \{f(x)\)  is divergent 

The series\(1 + 2x + {x^2} + 2{x^3} + {x^4} + \cdots \)is divergent if\(|x| \ge 1\)since the general term of the series does not converges to 0 when\(n \to \infty \).

Consider the power series,

\(\begin{aligned}g(x) &= \sum\limits_{k = 0}^\infty {{x^{2k}}} \\g(x) &= 1 + {x^2} + {x^4} + {x^6} + \cdots \end{aligned}\)

And,

\(h(x) = \sum\limits_{k = 0}^\infty {{x^{2k}}} \)

\(h(x) = x + {x^3} + {x^5} + {x^7} + \cdots \)

Find the interval of convergence of f(x) is (-1,1)

Notice that both series \(g(x)\) and \(h(x)\) are geometric series with the ratio \(r = {x^2}\), for all \(x\).

Therefore, the series converges for all real\(x\)such that\(|x| < 1\)and diverges for all\(|x| \ge 1\)with\(g(x) = \frac{1}{{1 - {x^2}}}\)and\(h(x) = \frac{x}{{1 - {x^2}}}\).

That is,\( - 1 < x < 1\).

Hence, the interval of convergence of \(f(x)\) is \(( - 1,1)\).

Calculate the interval of convergence of f(x)

Obtain the formula for f(x).

\(f(x) = 1 + 2x + {x^2} + 2{x^3} + {x^4} + \cdots \)

\(f(x) = \left( {1 + {x^2} + {x^4} + \cdots } \right) + \left( {2x + 2{x^3} + 2{x^7} + \cdots } \right)\)

\(f(x) = \left( {1 + {x^2} + {x^4} + \cdots } \right) + 2\left( {x + {x^3} + {x^7} + \cdots } \right)\)

\(f(x) = g(x) + 2h(x)\)

\({\rm{Use }}g(x) = \frac{1}{{1 - {x^2}}}{\rm{ and }}h(x)\)

\(f(x) = \frac{x}{{1 - {x^2}}},f(x)\)

\(f(x) = \frac{1}{{1 - {x^2}}} + 2\left( {\frac{x}{{1 - {x^2}}}} \right)\)

\(f(x) = \frac{1}{{1 - {x^2}}} + \frac{{2x}}{{1 - {x^2}}}\)

\(f(x) = \frac{{1 + 2x}}{{1 - {x^2}}}\)

Thus, the formula for \(f(x) = \frac{{1 + 2x}}{{1 - {x^2}}}\), for all real \(x\) such that \(|x| < 1\). Therefore, the interval of convergence of \(f\) is \(( - 1,1)\) and \(f(x) = \frac{{2x + 1}}{{1 - x}}\).