Do a convergence test for the following series?

Question

$$ 1 - \frac{1}{3} + \frac{1}{3^2} - \frac{1}{3^3} + \frac{1}{3^4} - ....... ∞ $$

I'm able to construct its general term which is :-

$$ u_n = \left( \frac{-1}{3} \right)^{n-1} $$

But I'm not sure what to do next.

Any help would be highly appreciated. Thanks in advance.

@NickLiu Messy notation. Wrong too. – Enrico M. Sep 17 '16 at 10:27 — Enrico M., Sep 17 '16 at 10:27
One has to ask: which series can you sum? – Did Sep 18 '16 at 15:55 — Did, Sep 18 '16 at 15:55

E.H.E · Answer 1 · 2016-09-17T10:27:07.203

0

Hint: $${\frac {1}{1+x}}=\sum _{n=0}^{\infty }(-x)^{n}\quad {\text{ for }}|x|<1$$

edited Sep 17 '16 at 10:27

answered Sep 17 '16 at 10:21

E.H.E

score 0 · Answer 2 · answered Sep 17 '16 at 10:23

This is nothing but

$$\sum_{k = 0}^{+\infty} \left(-\frac{1}{3}\right)^k$$

Which is nothing but the geometric series.

The result is

$$\frac{1}{1 + \frac{1}{3}} = \frac{1}{\frac{4}{3}} = \frac{3}{4}$$

Hence the series converges.

Geometric Series

$$\sum_{k = 0}^{+\infty}\ x^k = \frac{1}{1 - x} ~~~~~~~~~~~ \text{for}\ |x| < 1$$

In your case it's simply $x = -1/3$

So you have to change signs by consequence getting:

$$\sum_{k = 0}^{+\infty} \left(-\frac{1}{3}\right)^k = \frac{1}{1 - \left(-\frac{1}{3}\right)}$$

From which the result above.

2 Answers2