Anyone knows how to represent this as a power series?

x^y = exp(y * ln(x))

and

exp(x) = 1 + x + x²/2! + x³/3! + x^4/4! + ...

with n! = n*(n-1)*(n-2)*...*2*1 (factorials)

and

ln(1+x) = x - x²/2 + x³/3 - x^4/4 + ....

so (1.01)^0.2

= exp(0.2 ln(1.01))

= exp(0.2 (0.01 - 0.01²/2 + 0.01³/3 - ....) )

= exp(0.0019901)

= 1 + 0.0019901 + 0.0019901²/2 + 0.0019901³/6 + ...

= 1.002

Most calculus books include the example of the binomial series:

(1+x)^p = 1 + px + (p(p-1)/2)x^2 + (p(p-1)(p-2)/6)x^3 + ...

Set x=.01 and p=1/5.

It's asking for the binomial expansion

1.01 = 1 + 0.01

(1+0.01)^(1/5) = (1.01)^(1/5)

The expansion is

(1+x)^n = 1 + nx + [n(n-1)/2!]*x² + [n(n-1)(n-2)/3!]x³ + ... + n!/(n-r)!r! * x^n

Start with (1 + x)^n

The first few terms are

1^n + n(1)^(n - 1)x + [n(n - 1)/2](1)^(n - 2)x^2 + . . .

= 1 + nx + [n(n - 1)/2]x^2 + . . .

Now x = 0.01 and n = 1/5 = 0.2, so the series is

1 + (0.01)(0.2) + [(0.2)(-0.8)/2](0.01)^2 + . . .

= 1 + 0.002 - 0.08(0.0001) + . . .

= 1 + 0.002 - 0.000008 + . . .

= 1.001992

Compare this to the exact value, which is 1.00199204766653...

Since you needed three digits, the first two terms would have sufficed.

The first two terms are 1 + nx, so we have

1 + (0.2)(0.01) = 1 + 0.002 = 1.002, which is good to three digits.