Geometrically sum?

Let $S = x - x^2 + x^3 - ...$

Then $(-x)S = -x^2 + x^3 - ...$

Subtracting the two, we see that $S(x+1) = x$ (for $abs(x)$ strictly less than 1)

Thus, $S = \frac{x}{x+1}$ = 1 - $\frac{1}{x+1}$

For $abs(x) < 1, \frac{1}{x+1}$ is strictly greater than 0.5.

Hence, 1 - $\frac{1}{x+1}$ is strictly less than 0.5. None of the options are possible.

Relevant wiki: Geometric Progression Sum

The sum to infinity of a geometric series is: $\frac{first term}{1-common ratio}$

The first term of this series is x and the common ratio is -x.

So assuming the series converges, S= $\frac{x}{1+x}$ . Rearranging, we get:

S+Sx=x

S=x-Sx

S=x(1-S)

x= $\frac{S}{1-S}$

In order for the series to converge, x must be less than 1 and greater than -1. x= $\frac{S}{1-S}$ does not satisfy this criterion for S=1,2,3,4. Therefore, none of the given answers are possible.

Given expression is x-x^2+x^3-x^4... = x(1-x)+x^3(1-x).... =(1-x)(x+x^3+x^5....) Clearly for any values of x, the result is either positive or negative and the second factor tells us that it is a big one certainly not among 1, 2, 3 or 4 YOU may use integers like 2 or 3 to test it! or even 1! The result is a big negative or positive integer (Possibly to infinity) The result when we use 1 is 0 which is also not present here in the options!

If $|x| \geq 1,$ the series does not converge.

If $0 < x < 1$ , because it is an alternating series with terms decreasing in magnitude, each partial sum is at most x, thus the sum of the series is also at most x, and thus less than 1.

If $-1 < x < 0$ , each partial sum is negative, and thus the sum is also negative.

If we factor out the lower-order term of each pair of x-terms that are being subtracted from one another, then the original equation:

$x-x^{2}+x^{3}-x^{4}+x^{5}-x^{6}+... = ?$

Thought of like this:

$(x-x^{2})+(x^{3}-x^{4})+(x^{5}-x^{6})+... = ?$

Can be rewritten as:

$x(1-x)+x^{3}(1-x)+x^{5}(1-x)+...$

And a $(1-x)$ can be further factored out, giving:

$(1-x)(x+x^{3}+x^{5}+...)$

From here, we can easily try a few different options for $x$ and rule out some possibilities:

Let $A = (1-x)$ , $B = (x+x^{3}+x^{5}+...)$

if $x < 0$ : $A > 0$ , $B < 0$ , thus $AB < 0$

if $x = 0$ : $A = 1$ , $B = 0$ , thus $AB = 0$

if $0 < x < 1$ : $0 < A < 1$ , $0 < B < 1$ , thus $0<AB < 1$

if $x > 1$ : $A < 0$ , $B > 0$ , thus $AB < 0$

All of these conditions cover every possible real-number value for $x$ , yet we can see that none of the offered solutions (1, 2, 3, or 4) are possible.

Because all terms with an even power are negative, and all terms with an odd power are positive, any negative number will result in a negative value, so we can eliminate those as an option. For all numbers greater than 1, the value will continually fluctuate between increasingly larger positive and negative values as each term is added. For 1, it will continually fluctuate between 1 and 0. Neither of these options result in a defined limit. So if a solution exists, we are limited to values of x between 0 and 1.

For any number between 0 and 1, the absolute value of each term will be less than the last. In other words, the value of the sequence will change by less and less as each term is added, and thus the value of the sequence has a defined limit. But, looking at the first two terms, the second term will subtract from the given value of x, and since the sequence will change by less and less, we know that, for this range of values, the value of the sequence can never exceed the value of x. Therefore, it will never be one of the positive integers listed.

Let y = x - x^2 + x^3 - x^4 + ...

Factor out an x

y = x (1 - x + x^2 - x^3 + x^4 - ...)

Note that the part in parentheses, from the second term on, is just - y. Therefore,

y = x (1 - y)

x = y / (1 - y)

This does, however, introduce some false solutions, so you have to check them.

if y=1, this is undefined

if y=2, x = -1, but plugged back into the original formula gives y = -1 -1 -1 -1 -1 ... so that doesn't work.

Similarly, if y=3 and y=4, x= -3/2 and -4/3, which also race to negative infinity.

Curiously, y = -1, x = -1/2 works just fine.

Relevant wiki: Geometric Progression Sum

Let S= $x-x^2+x^3.....$

We know that by Geometric Progression for infinite addititon,

$S=\frac{a}{1-r}$ where a is 1st term & r is common ratio(r<1).

$x-x^2+x^3-......=\frac{x}{1-x}$ , which is never greater than 1.

Thus, None of these are possible.

Providing that $|x|<1$ , the sum of the power series $x-x^2+x^3-x^4+\cdots$ is $x(1-x+x^2-x^3+\cdots)=\dfrac x{1+x}$ . Let this number be denoted by $c$ . Then $x=\dfrac c{1-c}$ . Since $|x|<1$ , we find $c<\dfrac12$ . So, none of the offered numbers couldn't satisfy the condition $c<\dfrac12$ , and is not possible.

This is an infinite geometric sequence whose first term is $t_1 = x$ and whose ratio is $r = -x$ . The sum of an infinite geometric sequence converges to $S = \frac{t_1}{1-r}$ when $-1 < r < 1$ or in this case $S = \frac{x}{1+x}$ when $-1 < x < 1$ .

When $-1 < x < 0$ , $\frac{x}{1+x} < 0$ (because a negative divided by a positive is a negative) so $S < 0$ , which isn't any of the options.

When $x = 0$ , $\frac{x}{1+x} = 0$ so $S = 0$ , which was given but still isn't any of the options.

When $0 < x < 1$ , $0 < \frac{x}{1+x} < 1$ (because a positive divided by a greater positive must be a value between 0 and 1) so $0 < S < 1$ , which isn't any of the options, either.

Therefore, none of the options are possible.

Let $S = x - x^2 + x^3 - ...$

cases:

$x>=1$ this series does not converge

$x\in(-1,1)$ this series converges as shown below

$x<= -1$ this series diverges to $-\infty$

when S converges the same terms can be rewritten as equal to $x(1-S)$ , so

$S = \frac{x}{1+x} < 1$ for $x\in(-1,1)$