Question

For what values\({\rm{x}}\) does the series \(\sum {_{{\rm{n = 1}}}^{\rm{x}}{{{\rm{(lnx)}}}^{\rm{n}}}} \)converge?

Short answer

Therefore, When the series converges is \(\frac{{\rm{1}}}{{\rm{e}}}{\rm{ < x < e}}\).

Step-by-step solution

Geometric series.

Keep that a geometric series with a common ratio of \({\rm{r}}\)only converges when\({\rm{|r| < 1}}\).

Step 2: The series converges.

Note

\(\sum\limits_{{\rm{n = 1}}}^\infty {{{{\rm{(lnx)}}}^{\rm{n}}}} \)

a typical ratio geometric series\({\rm{r = lnx}}\)

The series will converge when\({\rm{|r| = |lnx| < 1}}\)

Therefore

\(\begin{array}{c}{\rm{ - 1 < lnx < 1}}\\{{\rm{e}}^{{\rm{ - 1}}}}{\rm{ < }}{{\rm{e}}^{{\rm{lnx}}}}{\rm{ < }}{{\rm{e}}^{\rm{1}}}\\\frac{{\rm{1}}}{{\rm{e}}}{\rm{ < x < e}}\end{array}\)

When the series converges,

\(\frac{{\rm{1}}}{{\rm{e}}}{\rm{ < x < e}}\).