Given: A power series: \(\sum_{k=1}^\infty\frac{x^{2k}}{k}\)

It is known that the power series of:

\(e^x=\sum_{n=0}^\infty\frac{x^n}{n!}\)

\(\ln(1-x)=-\sum_{n=0}^\infty\frac{x^k}{k}\)

As the power series of \(\ln(1-y)\) is:

\(\ln(1-y)=-\sum_{n=0}^\infty\frac{y^k}{k}\)

So when \(y=x^2\)

\(\sum_{k=1}^\infty\frac{x^{2k}}{k}=-\ln(1-x^2)\)

Thus, the function for the power series \(\sum_{k=1}^\infty\frac{x^{2k}}{k}=-\ln(1-x^2)\)

It is known that the power series of:

\(e^x=\sum_{n=0}^\infty\frac{x^n}{n!}\)

\(\ln(1-x)=-\sum_{n=0}^\infty\frac{x^k}{k}\)

As the power series of \(\ln(1-y)\) is:

\(\ln(1-y)=-\sum_{n=0}^\infty\frac{y^k}{k}\)

So when \(y=x^2\)

\(\sum_{k=1}^\infty\frac{x^{2k}}{k}=-\ln(1-x^2)\)

Thus, the function for the power series \(\sum_{k=1}^\infty\frac{x^{2k}}{k}=-\ln(1-x^2)\)