The Easy Peasy Sum of Series Part 3

Let us consider the sum:

$1+x+x^{2}+x^{3}+x^{4}+...\text{uptill infinite terms}$ where $|x|<1$

By the formula for the sum of an infinite Geometric Progression the above sum can be written as:

$\displaystyle\sum_{n=0}^{\infty} x^{n}=\displaystyle\frac{1}{1-x}$

Now differentiating w.r.t. $x$ we get:

$\displaystyle\sum_{n=0}^{\infty} nx^{n-1}=\displaystyle\frac{1}{\left( 1-x\right)^{2}}$

Multiplying both sides by $x$ :

$\displaystyle\sum_{n=0}^{\infty} nx^{n}=\displaystyle\frac{x}{\left( 1-x\right)^{2}}$

Now again differentiating w.r.t. $x$ :

$\displaystyle\sum_{n=0}^{\infty} n^{2}x^{n-1}=\displaystyle\frac{1+x}{\left( 1-x\right)^{3}}$

Now repeating the above process once more we get:

$\displaystyle\sum_{n=0}^{\infty} n^{3}x^{n-1}=\displaystyle\frac{(1-x)(1+2x)+3(x+x^{2})}{\left( 1-x\right)^{4}}$

Now substitute $x=\displaystyle\frac{1}{3}$ to obtain:

$\displaystyle\sum_{n=0}^{\infty} \frac{n^{3}}{3^{n-1}}=\displaystyle\frac{99}{8}$

Thus the required answer is $\displaystyle\frac{1}{3}\displaystyle\sum_{n=0}^{\infty} \frac{n^{3}}{3^{n-1}}=\boxed{\displaystyle\frac{33}{8}}$

I wrote this bit of Python to converge on the answer:

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from fractions import Fraction as frac
infinity = 200
total = 0
i = 0
while i <= infinity:
    total += frac(i**3,3**i)
    i += 1
print total.numerator/float(total.denominator)

The program spits out 4.125, and 4.125 as a fraction is 33/8, thus the answer is 33+8 = 41.