Infinite Sum Practice (2)

Calculus Level 4

$S=\dfrac{1^5}{5^1}+\dfrac{2^5}{5^2}+\dfrac{3^5}{5^3}+\dfrac{4^5}{5^4} +\dots$

The value of $S$ is of the form $\dfrac{A}{B}$ where $A$ and $B$ are co-prime positive integers. Find the value of $A-B$

The answer is 3023.

1 solution

Brian Charlesworth
Oct 25, 2014

We can rewrite $S$ as $\sum_{k=1}^{\infty} k^{5} * (\frac{1}{5})^{k}$ .

Now in general, for $|x| \lt 1$ , we have that $\sum_{k=0}^{\infty} x^{k} = \dfrac{1}{1 - x}$ .

Differentiate both sides, (the LHS term by term), and then multiply both sides by $x$ to find that

$\sum_{k=1}^{\infty} k * x^{k} = \dfrac{x}{(1 - x)^{2}}$ .

Differentiate both sides again and multiply both sides by $x$ to find that

$\sum_{k=1}^{\infty} k^{2} * x^{k} = \dfrac{x(x + 1)}{(1 - x)^{3}}$ .

Continuing with this process, we obtain the following results:

$\sum_{k=1}^{\infty} k^{3} * x^{k} = \dfrac{x(x^{2} + 4x + 1)}{(1 - x)^{4}}$ ,

$\sum_{k=1}^{\infty} k^{4} * x^{k} = \dfrac{x(x^{3} + 11x^{2} + 11x + 1)}{(1 - x)^{5}}$ , and

$\sum_{k=1}^{\infty} k^{5} * x^{k} = \dfrac{x(x^{4} + 26x^{3} + 66x^{2} + 26x + 1)}{(1 - x)^{6}}$ .

So to find $S$ we just need to plug $x = \frac{1}{5}$ into this last formula. This is straightforward, (but somewhat tedious), and yields

$S = \dfrac{3535}{512}$ .

Thus $A = 3535, B = 512$ and $A - B = \boxed{3023}$ .

Is there a solution not requiring calc? And where is a good website to learn differentiation and integration of infinite sums?

Trevor Arashiro - 6 years, 7 months ago

There is a non-calculus approach given here for dealing with $\sum_{k=1}^{\infty} k*x^{k}$ ,and a good discussion here about how to deal with $\sum_{k=1}^{\infty} k^{2} * x^{k}$ (for $x = \frac{1}{2}$ ) without resorting to calculus. However, for a fifth power sum as is the case here I think that calculus is the best tool for the job.

As for a good website, try this one . There are useful sections accessed by the "Back" and "Next" options at the bottom of this page as well.

Brian Charlesworth - 6 years, 7 months ago

The easier way to extend it algebraically, is to define

$S_i = \sum_{k = 1 } ^ \infty k ( k+1) \ldots (k+i) \times x ^ k$

We get that $S_ 0 = \frac{ 1}{ 1-x}$ ,
$S_1 ( 1-x) = S_0$ and so $S_1 = \frac{1}{ (1-x)^2 }$ ,
$S_2 (1-x) = 2 S_1$ and so $S_2 = \frac{ 2} { (1-x)^3 }$ ,
$S_3 ( 1-x) = 3 S_2$ and so $S_3 = \frac{ 6} { (1-x)^n }$ .
More generally, $S_n = \frac{ n!} { (1-x)^n }$ .

Now, you just need to figure out how to write $k^n$ in terms of $k, k (k+1), k(k+1)(k+2)$ , etc.

The (underlying) reason motivating the choice of coefficients as $\prod (k + i )$ is due to accounting for the constants we get from integration / differentiation.

Calvin Lin Staff - 6 years, 7 months ago

I was very lucky cause I recently learnt this method :D

Julian Poon - 6 years, 7 months ago

Infinite Sum Practice (2)

The answer is 3023.

1 solution

0 pending reports