Infinite

Let the sum of infinite series be represented by S.

So, S= $\frac { 1 }{ 5 } +\frac { 16 }{ 25 } +\frac { 81 }{ 125 } +\frac { 256 }{ 625 } +.....\\ =\sum _{ n=0 }^{ \infty }{ \frac { { n }^{ 4 } }{ { 5 }^{ n } } }$

Now, we know that $\sum _{ n=0 }^{ \infty }{ { x }^{ n } } =\frac { 1 }{ 1-x }$ for all 0<x<1.

So, differentiating on both sides, we get

$\sum _{ n=0 }^{ \infty }{ { nx }^{ n-1 } } =\frac { 1 }{ { (1-x) }^{ 2 } }$

Multiplying both sides by x and differentiating again , we get

$\sum _{ n=0 }^{ \infty }{ { { n }^{ 2 }x }^{ n-1 } } =\frac { { x }^{ 2 }+x }{ { (1-x) }^{ 3 } }$

Again multiplying both sides by x and differentiating, we get

$\sum _{ n=0 }^{ \infty }{ { { n }^{ 3 }x }^{ n-1 } } =\frac { { x }^{ 2 }+4x+1 }{ { (1-x) }^{ 4 } }$

Once again multiplying by x and differentiating on both sides, we get

$\sum _{ n=0 }^{ \infty }{ { { n }^{ 4 }x }^{ n-1 } } =\frac { { x }^{ 3 }+11x+{ 11x }^{ 2 }+1 }{ { (1-x) }^{ 5 } }$

For the last time multiplying by x on both sides, we get

$\sum _{ n=0 }^{ \infty }{ { { n }^{ 4 }x }^{ n } } =\frac { { x }^{ 4 }+11{ x }^{ 3 }+{ 11x }^{ 2 }+x }{ { (1-x) }^{ 5 } }$

Now , we know that the above summation is valid for all x less than 1 and greater than 0. So, we just need to put the value of x = $\frac { 1 }{ 5 }$ to get the value of S.

Hence. by substituting the value of x = $\frac { 1 }{ 5 }$ , we get

$\sum _{ n=0 }^{ \infty }{ \frac { { n }^{ 4 } }{ { 5 }^{ n } } } =\frac { { \frac { 1 }{ { 5 }^{ 4 } } }+{ \frac { 11 }{ { 5 }^{ 3 } } }+\frac { 11 }{ { 5 }^{ 2 } } +\frac { 1 }{ 5 } }{ { (1-\frac { 1 }{ 5 } ) }^{ 5 } } =\quad \frac { 285 }{ 128 } \\ \\$

$\sum_{x=1}^{\infty}{\frac{x^4}{5^x}} = \frac{1}{128} 5^{-x} (-32x^4-160x^3-360x^2-460x+57 5^{x+1}-285)$

For $x\rightarrow\infty$ $\frac{1}{128} 5^{-x} (-32x^4-160x^3-360x^2-460x+57 5^(x+1)-285)=\boxed{\frac{285}{128}}$

Hint: Putting x = 1/5, the generating function of this summation S can be expressed as 5*S = { (1 + 11x + 11 x^2 + x^3)/(1 - x)^5}.