To evaluate the infinite series
[tex]\[
\sum_{n=1}^{\infty}\left[\frac{k^3}{n^4}+\frac{2k}{n^2}\right],
\][/tex]
let's break it down into two separate series which we can analyze individually.
First, consider the series
[tex]\[
\sum_{n=1}^{\infty} \frac{k^3}{n^4}.
\][/tex]
Here, [tex]\( \frac{k^3}{n^4} \)[/tex] can be written as [tex]\( k^3 \cdot \frac{1}{n^4} \)[/tex]. We can factor [tex]\( k^3 \)[/tex] out of the summation since it is constant with respect to [tex]\( n \)[/tex]:
[tex]\[
k^3 \sum_{n=1}^{\infty} \frac{1}{n^4}.
\][/tex]
The series [tex]\(\sum_{n=1}^{\infty} \frac{1}{n^4}\)[/tex] is a well-known p-series with [tex]\( p = 4 \)[/tex], which converges and is equal to [tex]\(\zeta(4)\)[/tex], where [tex]\(\zeta(s)\)[/tex] is the Riemann zeta function. Specifically,
[tex]\[
\zeta(4) = \frac{\pi^4}{90}.
\][/tex]
Thus, we have
[tex]\[
k^3 \sum_{n=1}^{\infty} \frac{1}{n^4} = k^3 \cdot \frac{\pi^4}{90}.
\][/tex]
Next, consider the series
[tex]\[
\sum_{n=1}^{\infty} \frac{2k}{n^2}.
\][/tex]
Here, [tex]\( \frac{2k}{n^2} \)[/tex] can be written as [tex]\( 2k \cdot \frac{1}{n^2} \)[/tex]. We can similarly factor [tex]\( 2k \)[/tex] out of the summation:
[tex]\[
2k \sum_{n=1}^{\infty} \frac{1}{n^2}
\][/tex]
The series [tex]\(\sum_{n=1}^{\infty} \frac{1}{n^2}\)[/tex] is another well-known p-series with [tex]\( p = 2 \)[/tex], which converges and is equal to [tex]\(\zeta(2)\)[/tex]. Specifically,
[tex]\[
\zeta(2) = \frac{\pi^2}{6}.
\][/tex]
Thus, we have
[tex]\[
2k \sum_{n=1}^{\infty} \frac{1}{n^2} = 2k \cdot \frac{\pi^2}{6} = \frac{2k\pi^2}{6} = \frac{k\pi^2}{3}.
\][/tex]
Now, we sum the results of the two separate series:
[tex]\[
k^3 \frac{\pi^4}{90} + \frac{k \pi^2}{3}.
\][/tex]
Therefore, the evaluated value of the given infinite series is:
[tex]\[
\sum_{n=1}^{\infty}\left[\frac{k^3}{n^4}+\frac{2k}{n^2}\right] = \frac{\pi^4 k^3}{90} + \frac{\pi^2 k}{3}.
\][/tex]
