Finding the sum to infinity

Question

Question: Find the sum to infinity for the following series $$1, -\frac{1}{2}, \frac{1}{2^2}, -\frac{1}{2^3},\cdots$$

What would be the technique used to find such a sum?

@GaneshTadi That I already realized. Not sure how it would help though. — Gummy bears, Aug 18 '14 at 06:23
oh nice :) try gauss approach : $S = 1 - (\frac{1}{2})(1 - \frac{1}{2} + \frac{1}{2^2} - ...) \implies S = 1-(\frac{1}{2})S$ ; solve $S$ — AgentS, Aug 18 '14 at 06:30
@GaneshTadi First you have to show that the series is convergent. Otherwise your approach would be wrong. specially the $|common ratio|≥1.$ For example $$S=1-1+1-1+1=1+...=1-(1-1+1-...)=1-S$$ — Bumblebee, Aug 18 '14 at 08:27

score 12 · Accepted Answer · answered Aug 18 '14 at 06:30

12

Perhaps a bit simpler:

The $n$'th term of your series (assuming we start at $n=0$) can be written as

$$\frac{(-1)^n}{2^n}$$

which is just $(-1/2)^n$. Since $|-1/2| < 1$, it's a convergent geometric series of the form

$$\sum_{n=0}^{\infty} x^n$$

where $x = -1/2$. The series converges to

$$\frac{1}{1-x} = \frac{1}{1 + \frac{1}{2}} = \frac{2}{3}$$

answered Aug 18 '14 at 06:30

So I'm guessing that the last thing is actually a formula to be memorized? – Gummy bears Aug 18 '14 at 06:31
Yes, you should know the formula for the sum of a geometric series. Specifically, if $|r| < 1$ we have $\sum_{n = 0}^{\infty}ar^n = \frac{a}{1-r}$. – JimmyK4542 Aug 18 '14 at 06:32
Ohh okay. Well that makes it much easier. Thank you! – Gummy bears Aug 18 '14 at 06:33
3

Well, it's used very very often, so it's good to memorize it. :-) But we can also prove it: let $S = 1 + x + x^2 + x^3 + \ldots$. Then $xS = x + x^2 + x^3 + \ldots$. Therefore, $S - xS = 1$, so solving for $S$, we get $S = 1/(1-x)$. Careful: this only works if $|x| < 1$. Do you see why? – Aug 18 '14 at 06:33
Yep I see that now. What if $|x|$ is greater than 1? – Gummy bears Aug 18 '14 at 06:36
Then the series does not converge and that maneuver with $S$ is no longer valid – user139388 Aug 18 '14 at 06:38
3

If $|x|$ is greater than 1, then the magnitudes of $x, x^2, x^3, \ldots$ get bigger and bigger. They don't converge to zero. So the series doesn't converge. Also note that if $|x| = 1$, there are two possibilities assuming $x$ is real. Either $x = 1$, in which case you are summing infinitely many 1's, and the series diverges to $\infty$; or $x = -1$, in which case the series is $1 - 1 + 1 - 1 + 1 - 1$ which doesn't converge either - the partial sums oscillate between $1$ (if you sum an odd number of terms) and $0$ (if you sum an even number). – Aug 18 '14 at 06:38

score 2 · Answer 2 · answered Aug 18 '14 at 08:14

2

Let $\text {eq.}{(1)}$ be,$$S=1 -\frac{1}{2}+ \frac{1}{2^2}-\frac{1}{2^3}+\cdots\infty$$

Then, $\text {eq.}{(2)}$ be

$$\frac{-S}2=-\frac{1}{2}+ \frac{1}{2^2}-\frac{1}{2^3}+\cdots\infty$$

Subtract 2 from 1

$$\frac{3S}2=1$$

Or, $$S=\frac{2}3$$

answered Aug 18 '14 at 08:14

MonK

1,794

The infinity symbol is unnecessary; the three dots are all you need. – zerosofthezeta Aug 18 '14 at 17:01

user139388 · Answer 3 · 2014-08-18T06:27:48.870

1

Write the sum of the $n$ and $(n+1)^{th}$ terms as $$ \frac{1}{2^n} - \frac{1}{2^{n+1}} = \frac{1}{2^{n+1}} $$ then sum them up to get $$ \sum_{n \in 2 \mathbb{N} }^\infty \frac{1}{2^{n+1}} = \sum_{k=0}^\infty \frac{1}{2^{2k+1}} = \frac{1}{2} \sum_{k=0}^\infty \frac{1}{4^k} = 2/3. $$

edited Aug 18 '14 at 06:27

answered Aug 18 '14 at 06:21

user139388

3,449

Ummm what did you do on the second last step? How did you get a four in the denominator? – Gummy bears Aug 18 '14 at 06:25
1

I added an extra step to help make it more clear. The reason is that since $n \in 2 \mathbb{N}$, you can write $n=2k$ for $k \in \mathbb{N}$, and $2^{2k} = 4^k$. – user139388 Aug 18 '14 at 06:28
$2^{2k}=(2^2)^k=4^k$. – Ian Aug 18 '14 at 06:29
I see what you did there now. Thank you. But I would like to know what should be the general approach when solving such problems? – Gummy bears Aug 18 '14 at 06:30

score 1 · Answer 4 · answered Aug 18 '14 at 06:40

$(1+\frac{1}{2^2}+\frac{1}{2^4}+...)-(\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+...)$

Both the brackets above contain GP of $n$ terms with $r=\frac{1}{2^2}=\frac{1}{4}$

$\Large(\frac{(\frac{1}{4})^n-1}{-3/4})-(\frac{1}{2}\frac{(\frac{1}{4})^n-1}{-3/4})$

$\Large\frac{\frac{1}{2}(\frac{1}{4})^n-\frac{1}{2}}{\frac{-3}{4}}=\frac{(\frac{1}{4})^n-1}{\frac{-3}{2}}=\frac{0-1}{\frac{-3}{2}}=\frac{2}{3}$

copper.hat · Answer 5 · 2014-08-18T14:30:20.107

0

Here is another way, given that the series converges:

Let $S$ be the sum, then note that $S = 1 - {1 \over 2} S$, solving gives the desired result.

Note:

To see where the above comes from, note that $S=1-{1 \over 2} + {1 \over 2^2}\cdots$, and so $-{1 \over 2} S = -{1 \over 2} + {1 \over 2^2}\cdots$. Hence we have the equation $S = 1 - {1 \over 2} S$.

(This is contingent on the series being convergent, otherwise one can end up with nonsense.)

edited Aug 18 '14 at 14:30

answered Aug 18 '14 at 07:05

copper.hat

172,524

Is that a general approach or only for this series? – Gummy bears Aug 18 '14 at 08:41
It is specific to this series, but is a useful trick in general. – copper.hat Aug 18 '14 at 14:24

score 0 · Answer 6 · answered Aug 18 '14 at 16:46

0

consider the sum as x. Then 4x = (4 - 2) + 1 - 1/2 ...

So 4x = (2) + x

Then x = 2/3

answered Aug 18 '14 at 16:46

Sadegh

111

Finding the sum to infinity

6 Answers6