Question

Find all positive value of b for which the series \(\sum\limits_{n = 1}^\infty {{b^{\ln n}}} \) converges.

Short answer

The given series \(\sum\limits_{n = 1}^\infty {{b^{\ln n}}} \)converges for \(b < \frac{1}{c}\).

Step-by-step solution

Using exponents properties in given series:-

We have \(\sum\limits_{n = 1}^\infty {{b^{\ln n}}} \)

Since, \(\begin{aligned}{b^{\ln n}} = {\left( {{e^{\ln b}}} \right)^{\ln n}} = {\left( {{e^{\ln n}}} \right)^{\ln b}} = {n^{\ln b}}\\\end{aligned}\)

\({b^{\ln n}} = \frac{1}{{{n^{ - \ln b}}}}\)

Using p-test:-

So series is \(\sum\limits_{n = 1}^\infty {{b^{\ln n}}} \)=\(\sum\limits_{n = 1}^\infty {\frac{1}{{{n^{ - \ln b}}}}} \)

Which is ap-series with \(p = - \ln b\)

So, this series is converges when \(p > 1\)

For\(p > 1\):-

\(\begin{aligned}&\Rightarrow - \ln b > 1\\&\Rightarrow \ln b < - 1\\&\Rightarrow b < {e^{ - 1}}\\\end{aligned}\)

\( \Rightarrow b < \frac{1}{e}\)

Hence, for\(b < \frac{1}{e}\)the given series converges.