convergence or divergence of series $\displaystyle \sum^{\infty}_{k=1}3^{-\ln(k)}$ using integral test

Question

Finding convergence or divergence of series $\displaystyle \sum^{\infty}_{k=1}3^{-\ln(k)}$ using integral test

What i try: let $f(x)=3^{-\ln(x)}$

Then $\ln(f(x))=-\ln(x)\cdot \ln(3).$

Then $\displaystyle \frac{f'(x)}{f(x)}=-\frac{\ln(3)}{x}$

$\displaystyle \Longrightarrow f'(x)=-\frac{\ln(3)}{3^{\ln x}x}<0$ for $x\geq 1$

So function $f(x)$ is decreasing function.

Also $$\int^{\infty}_{1}3^{-\ln(x)}dx$$

Put $\ln(x)=t$ and $x=e^t$ and $dx=3^tdt$

Then $$I=\int^{\infty}_{0}\bigg(\frac{e}{3}\bigg)^tdt=-\ln\bigg(\frac{e}{3}\bigg)$$

Is my process is right.if not how do i solve it Help me please

Answer 1

I think you may have made some computational mistake. Regardless, you can simply observe the following: $$ 3^{-\ln(x)} = e^{-\ln(3)\ln{x}} = x^{-\ln(3)} $$ So: $$ \int_1^\infty x^{-\ln(3)} \; \mathrm{d}x = \left[\frac{x^{1 - \ln(3)}}{1 -\ln(3)}\right]_{x = 1}^{x \to \infty} = -\frac{1}{1 - \ln(3)} $$

Answer 2

Your process is correct but you evaluated the integral incorrectly.

$$I=\int {\left(\frac{e}{3}\right)}^t \; dt= \frac{{\left(\frac{e}{3}\right)}^t}{\ln{\left(\frac{e}{3}\right)}}+C$$

Therefore, evaluating the integral yields: $$\frac{-1}{1-\ln{\left(3\right)}} \approx 10.140724$$

You showed that $a_n$ is decreasing across its domain but you also have to show $a_n$ is positive across its domain, which is easy to do. After you prove so, you'll see that the series converges.

Note: you incorrectly said $dx=3^t dt$ instead of $dx=e^t dt$, but correctly substituted it into the integral so I assume that's just a typo or $e=3$.

Answer 3

If you use the definition of natural logarithm ($a=3$): $$ \log_{a}x = \frac{\ln x}{\ln a} $$ an note that $\ln 3 = 1+ \varepsilon, \varepsilon>0$ you sum becomes $$ \sum_{k \geq 1}\frac{1}{x^{1+ \varepsilon}} \leq \int_{1}^{\infty} x^{-1- \varepsilon} dx = \frac{1}{\varepsilon} < \infty $$ So the sum converges.