$Convergence$

Solution:

Summation of $\frac{x^{n}}{x^{2n} + 1}$ for $n ∈ Natural Numbers$ $->$ Summation of $\frac{1}{{x^{n}} + \frac{1}{x^{n}}}$ for $n ∈ Natural Numbers$

Since $n ∈ Natural Numbers$ ,

From $x^{2} - x + 1 = 0$

$x + \frac{1}{x} = 1$

$x^{2} + \frac{1}{x^{2}} = -1$

$x^{3} + \frac{1}{x^{3}} = -1-1 = -2$

$x^{4} + \frac{1}{x^{4}} = (-1)^{2} - 2 = -1$

$x^{5} + \frac{1}{x^{5}} = 2-1 = 1$

$x^{6} + \frac{1}{x^{6}} = (-2)^{2} - 2 = 2$

$...$

$therefore,$

for $n = 1,2,4,5,7,8,...$

$x^{n} + \frac{1}{x^{n}} = 1,-1,-1,1,1,-1,...$ $REMEMBER:$ All values for n has a corresponding values for $x + \frac{1}{x}$

for $n = 3,6,9,12,15,18,...$

$x^{n} + \frac{1}{x^{n}} = -2,2,-2,2,-2,2,...$ $REMEMBER:$ All values for n has a corresponding values for $x + \frac{1}{x}$

Then,

Summation of $\frac{x^{n}}{x^{2n} + 1}$ for $n ∈ Natural Numbers$ $->$ Summation of $\frac{1}{{x^{n}} + \frac{1}{x^{n}}}$ for $n ∈ Natural Numbers$

= $\frac{1}{1} + \frac{1}{-1} + \frac{1}{-2} + \frac{1}{-1} + \frac{1}{1} + \frac{1}{2} + ...$

= $\frac{1}{1} + \frac{1}{-1} + \frac{1}{-1} + \frac{1}{1} + ... + \frac{1}{-2} + \frac{1}{2} + ...$

As noted above, $a+-a+a+-a+a+-a+...$ $converges$ to $0$ and $a+-a+-a+a+a+-a+-a+a+...$ $converges$ to $0$

Therefore,

Summation of $\frac{x^{n}}{x^{2n} + 1}$ for $n ∈ Natural Numbers$ $->$ Summation of $\frac{1}{{x^{n}} + \frac{1}{x^{n}}}$ for $n ∈ Natural Numbers$

= $\frac{1}{1} + \frac{1}{-1} + \frac{1}{-1} + \frac{1}{1} + ... + \frac{1}{-2} + \frac{1}{2} + ...$

= $\frac{1}{1} + \frac{-1}{1} + \frac{-1}{1} + \frac{1}{1} + ... + \frac{-1}{2} + \frac{1}{2} + ...$ $converges$ to $0$ . $Ans.$