Let's evaluate the infinite series [tex]\(\sum_{n=1}^{\infty}\left[\frac{k^3}{n^4}+\frac{2 k}{n^2}\right]\)[/tex] step by step, focusing on the properties and behavior of the series components.
### Step 1: Break Down the Series
The given series can be split into two separate series:
[tex]\[ \sum_{n=1}^{\infty}\left[\frac{k^3}{n^4}+\frac{2 k}{n^2}\right] = \sum_{n=1}^{\infty} \frac{k^3}{n^4} + \sum_{n=1}^{\infty} \frac{2 k}{n^2} \][/tex]
### Step 2: Evaluate Each Series Separately
We will evaluate [tex]\(\sum_{n=1}^{\infty} \frac{k^3}{n^4}\)[/tex] and [tex]\(\sum_{n=1}^{\infty} \frac{2 k}{n^2}\)[/tex] individually.
#### Series 1: [tex]\(\sum_{n=1}^{\infty} \frac{k^3}{n^4}\)[/tex]
This is a p-series with [tex]\(p = 4\)[/tex] (since it is of the form [tex]\(\sum_{n=1}^{\infty} \frac{1}{n^p}\)[/tex] where [tex]\(p > 1\)[/tex]):
[tex]\[ \sum_{n=1}^{\infty} \frac{k^3}{n^4} = k^3 \sum_{n=1}^{\infty} \frac{1}{n^4} \][/tex]
The sum of the p-series [tex]\(\sum_{n=1}^{\infty} \frac{1}{n^4}\)[/tex] is known to be [tex]\(\frac{\pi^4}{90}\)[/tex]:
[tex]\[ \sum_{n=1}^{\infty} \frac{k^3}{n^4} = k^3 \cdot \frac{\pi^4}{90} = \frac{\pi^4 k^3}{90} \][/tex]
#### Series 2: [tex]\(\sum_{n=1}^{\infty} \frac{2 k}{n^2}\)[/tex]
This is also a p-series but with [tex]\(p = 2\)[/tex]:
[tex]\[ \sum_{n=1}^{\infty} \frac{2 k}{n^2} = 2 k \sum_{n=1}^{\infty} \frac{1}{n^2} \][/tex]
The sum of the p-series [tex]\(\sum_{n=1}^{\infty} \frac{1}{n^2}\)[/tex] is known to be [tex]\(\frac{\pi^2}{6}\)[/tex]:
[tex]\[ \sum_{n=1}^{\infty} \frac{2 k}{n^2} = 2 k \cdot \frac{\pi^2}{6} = \frac{2\pi^2 k}{6} = \frac{\pi^2 k}{3} \][/tex]
### Step 3: Combine the Results
Now, add the results of the two series:
[tex]\[ \sum_{n=1}^{\infty}\left(\frac{k^3}{n^4} + \frac{2 k}{n^2}\right) = \frac{\pi^4 k^3}{90} + \frac{\pi^2 k}{3} \][/tex]
Therefore, the sum of the series [tex]\(\sum_{n=1}^{\infty}\left[\frac{k^3}{n^4}+\frac{2 k}{n^2}\right]\)[/tex] is:
[tex]\[ \boxed{\frac{\pi^4 k^3}{90} + \frac{\pi^2 k}{3}} \][/tex]
