I'm trying to solve the following problem: Find the positive values of $b$ for which the series $\sum_{n=1}^\infty b^{\ln(n)} $ converges.
I started by doing the integral test with the function $b^{\ln(x)}$.
To integrate this: $\int b^{\ln(x)} \, dx $ I used integration by parts:
I picked
- $u=b^{\ln(x)}$
- $du= b^{\ln(x)} \ln(b)(1/x)dx$
- $v=\int dx = x$
Doing the substitution I get:
$$\int b^{\ln(x)} \, d x= b^{\ln(x)}x-\int x b^{\ln(x)}\ln(b)(1/x)\,dx$$
$$\int b^{\ln(x)} dx= b^{\ln(x)}x-\int b^{\ln(x)}\ln(b)(x/x)dx$$
$$\int b^{\ln(x)} dx= b^{\ln(x)}x-\int b^{\ln(x)}\ln(b)dx$$
$$\int b^{ln(x)} dx= b^{ln(x)}x-ln(b)\int b^{ln(x)}dx$$
$$\int b^{\ln(x)} dx + \ln(b)\int b^{\ln(x)}dx= b^{\ln(x)}x$$
$$[1 + \ln(b)]\int b^{\ln(x)} dx = b^{\ln(x)}x$$
$$\bbox[5px,border:2px solid red]{\int b^{\ln(x)} dx = \frac{b^{\ln(x)}x}{[1 + \ln(b)]}}\qquad$$
Then I solve evaluate the improper integral between $1$ and $\infty$
$$\int_1^\infty b^{\ln(x)} dx =\lim_{t\rightarrow \infty} \int_1^t b^{\ln(x)} dx =\lim_{t\rightarrow \infty} \left| \begin{array}{c} \frac{ b^{\ln(x)}x }{[1 + \ln(b)]} \end{array} \right|_1^t$$
$$\lim_{t\rightarrow \infty} \left| \begin{array}{c} \frac{b^{ln(x)}x}{[1 + \ln(b)]} \end{array} \right|_1^t =\lim_{t\rightarrow \infty} \frac{b^{\ln(t)}t}{[1 + \ln(b)]} -\frac{b^{\ln(1)}1}{[1 + \ln(b)]}$$
$$=\lim\limits_{t\rightarrow \infty} \frac{b^{\ln(t)}t}{[1 + \ln(b)]} -\frac{b^{0}1}{[1 + \ln(b)]}$$
$$=\lim\limits_{t\rightarrow \infty} \frac{b^{\ln(t)}t}{[1 + \ln(b)]} -\frac{1(1)}{[1 + \ln(b)]}$$
$$=\lim\limits_{t\rightarrow \infty} \frac{b^{\ln(t)}t}{[1 + \ln(b)]} -\frac{1}{[1 + \ln(b)]}$$
$$=[\frac{1}{[1 + \ln(b)]}]\lim\limits_{t\rightarrow \infty} b^{\ln(t)}t - 1$$
Here is where I get a little confused: I assume that $0<b < 1$ because that way I have an indeterminate form $0\cdots\infty$ which allows me to apply l'Hôpital's rule as following:
$$=\left[\frac{1}{[1 + \ln(b)]}\right] \lim_{t\rightarrow \infty} \frac {b^{\ln(t)}}{t^{-1}} - 1$$
I differentiate numerator and denominator
$$=\left[\frac{1}{[1 + \ln(b)]}\right] \lim_{t\rightarrow \infty} \frac {b^{\ln(t)}\ln(b)t^{-1}}{-t^{-2}} - 1$$
And I end up getting the same indeterminate form again ($0\cdot\infty$)
$$=\left[\frac{1}{[1 + \ln(b)]} \right] \lim_{t\rightarrow \infty} -b^{\ln(t)} \ln(b)t - 1$$
What should I do? The book gives $b<(1/e)$ as an answer, which makes sense because if $b=1/e$ then $\ln(b) = -1$ and the denominator of $[\frac{1}{[1 + \ln(b)]}]$ would be zero. But I'm stuck here. How do I solve evaluate the limit? I'm assuming that my mistake is not in the indefinite integral because Mathematica and other CAS give the same answer. So the problem is likely to be in the improper integral and the limit.
I'd try a shorter path. Note that $$b^{\ln n}=e^{\ln n\ln b}=n^{\ln b}$$
Now, for what values of $s$ does $\sum n^s$ converge?