We have an answer to "Use the Integral Test to determine whether the..."

Cem Hayes

Answered question

2020-10-23

Use the Integral Test to determine whether the infinite series is convergent.

\sum_{n = 1}^{\infty} \frac{5}{4^{\ln n}}

Answer & Explanation

timbalemX

Skilled2020-10-24Added 108 answers

We have given a series
$\sum_{n = 1}^{\infty} \frac{5}{4^{\ln n}}$
We need to check if the series converges or diverges using integral test
$L = lim_{n \to \infty} \int_{1}^{n} \frac{5}{4^{\ln (x)}} d x$
Substitute $\ln (x) = u, x = e^{u}$
$\frac{d x}{x} = d u$
$d x = x d u, d x = e^{u} d u$
So, $\int \frac{5}{4^{\ln (x)}} d x = \int \frac{5}{4^{u}} e^{u} d u = 5 \int (\frac{e}{4})^{u} d u$
Since, $\int a^{x} d x = \frac{a^{x}}{\ln (a)} + c$
Therefore, $5 \int (\frac{e}{4})^{u} d u = 5 \frac{(\frac{e}{4})^{u}}{\ln (\frac{e}{4})}$
Next we plug back $u = \ln (x)$ , we get:
$lim_{n \to \infty} \int_{1}^{n} \frac{5}{4^{\ln (x)}} d x = lim_{n \to \infty} 5 [\frac{(\frac{e}{4})^{\ln (n)}}{\ln (\frac{e}{4})} - \frac{(\frac{e}{4})^{\ln (1)}}{\ln (\frac{e}{4})}]$
$lim_{n \to \infty} 5 [\frac{(\frac{e}{4})^{\ln (n)} - 1}{\ln (\frac{e}{4})}] = \frac{\infty}{\infty}$
So we use L-Hospitals rule
$lim_{n \to \infty} 5 [\frac{\frac{d}{d n} (\frac{e}{4})^{\ln (n)} - 1}{\frac{d}{d n} \ln (\frac{e}{4})}] = 0$
Since we are getting finite value for the integral
Hence the given series is convergent

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