0
 STATEMENT:

$t \in R$, $0 < t < 1$, $$\sum_{i=1}^\infty t^i = \frac{t}{1-t}$$

 EXAMPLE:

let t = $\frac{1}{2}$

then,

$\frac{1}{2}$ + $\frac{1}{4}$ + $\frac{1}{8}$ + ... = 1

let n=$\frac{2}{3}$

then,

$\frac{2}{3}$ + $\frac{4}{9}$ + $\frac{8}{27}$ + ... = 3

ex.1 $\frac{1}{2}$ + $\frac{1}{4}$ + $\frac{1}{8}$ + ...=1

ex.2 $\frac{9}{10}$ + $\frac{9}{100}$ + $\frac{9}{1000}$ + ...=1

 PROOF:

$$\sum_{i=1}^\infty t^i = (1-t) + t(1-t) + t^2(1-t) + t^3(1-t) ... = 1, \; \; (t \in R, 0 < t < 1)$$

$$ = 1 + t + t^2 + t^3 + ... = \frac{1}{1-t} $$

$$ = t + t^2 + t^3 + ... = \frac{t}{1-t} $$

$$ = \sum_{i=1}^\infty t^i = \frac{t}{1-t}$$

I found this by myself and I think this is true. I wonder if I can apply it to something. If you know something about this topic, please let me know. I'm not very good at math and I might be wrong, so please forgive and help me.

Ashutosh Gupta
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Kuta
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  • You seem to have mixed $n$ and $t$. These are examples of geometric series – Henry Apr 06 '16 at 13:48
  • As you are dealing with dots indicating infinity this is perhaps not an absolutely rigorous proof, but the idea is correct. You have actually just derived a well known sum for certain geometric series as expanded here – K.Power Apr 06 '16 at 13:48

1 Answers1

2

This is well known to be a geometric series. It is very useful in analysis.

Also, the series converge for $|t| < 1$, so $t$ can be negative.

Here is a somewhat more rigorous proof: The partial sums are given by $$ \sum_{j=0}^{m-1} t_j = \frac{1-t^m}{1-t} $$ so if $|t| < 1$, $$ \lim_{m \to \infty}\sum_{j=0}^{m-1} t_j = \lim_{m \to \infty} \frac{1}{1-t} - \frac{t^m}{1-t} = \frac{1}{1-t} $$ Since $\frac{t^m}{1-t} \to 0$.

Henricus V.
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