Evaluate this infinite sum

Question

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

We can't simplify anything any more. Except.

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

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

Write out the terms of the sum in its original form and you will see how they cancel. You can write down the expression for the (finite) partial sums, and take the limit, if it exists. — Mark Bennet, Nov 02 '14 at 17:43

score 5 · Answer 1 · edited Apr 13 '17 at 12:20

5

$\displaystyle S=\sum_{n=1}^{\infty} x^{3n-2} - x^{3n+1}=\dfrac{1-x^3}{x^2}\sum_{n=1}^\infty(x^3)^n$

$\displaystyle S=\frac{1-x^3}{x^2}\cdot\dfrac{x^3}{1-x^3}=x$

edited Apr 13 '17 at 12:20

Community

answered Nov 02 '14 at 17:40

lab bhattacharjee

Hi! Thanks. If $x^3 < 1$, then how does the sum converge to $\frac{x^3}{1-x^3}$? – Amad27 Nov 02 '14 at 17:42
Wait.. Did you use the ratio formula? the $S = 1/(1-r)$? – Amad27 Nov 02 '14 at 17:42
@Amad27, It's actually $$\frac a{1-r}$$ See link in the edited answer – lab bhattacharjee Nov 02 '14 at 17:44
@Amad27, Background : http://math.stackexchange.com/questions/1002440/interchanging-summation-and-integral ? – lab bhattacharjee Nov 02 '14 at 17:46

score 3 · Accepted Answer · answered Nov 02 '14 at 17:50

3

Just to expand my comment a little, the sum begins $$(x-x^4)+(x^4-x^7)+(x^7-x^{10}) \dots$$

The sum of the first $n$ terms is clearly $x-x^{3n+1}$ (e.g. prove by induction).

Take the limit as $n\to \infty$

answered Nov 02 '14 at 17:50

Mark Bennet

2 Answers2