Skipping sums

For the original geometric series to be convergent, it has to have a common ratio $|r|=|a|<1$ .

This means that each term in the series is smaller in absolute value than the preceding term.

For the expression

$(a+a^2)+(a^3+a^4)+(a^5+a^6)+\cdots=(a+a^2)+a^2(a+a^2)+a^4(a+a^2)+\cdots =(a+a^2)(1+a^2+a^4+\cdots)$

to be positive, $a$ , which is larger in absolute value than $a^2$ , has to be positive.

The two series in question

$a + a^3 + a^5 + a^7 + \cdots$

$a^2 + a^4 + a^6 + a^8 + \cdots$

can now be compared term by term. All of the terms in both series are positive, and each term of the first series is larger than the corresponding term of the second series.

The sum of the first series is therefore larger than the sum of the second series.

First thing you should conclude is that $0< a< 1$ . Why? Well, first sentence says that the sum of $a + a^{2} + a^{3} + a^{4} + ...$ is a positive number, and that means that this infinite series converges to some positive value. If $a> 1$ , then the sum would go to the infinity. Also, $a$ can't be negative. Here is the math:

$a + a^{2} + a^{3} + a^{4} + a^{5} + ... = \frac{a}{1-a}> 0$

$\Rightarrow (a> 0 \wedge 1-a > 0)$ or $(a < 0 \wedge 1-a < 0)$ - both numerator and denominator must have the same sign.

Solution of the system $a> 0 \wedge 1-a > 0$ is set of numbers $0< a< 1$ , and the system $a < 0 \wedge 1-a < 0$ has no solutions.

Hence $0< a< 1$ .

Now, once we cleared that, we can move on with our problem:

$a + a^{3} + a^{5} + a^{7} + ...= \frac{a}{1-a^{2}}$ and

$a^{2} + a^{4} + a^{6} + ...= \frac{a^{2}}{1-a^{2}}$

We can notice that they have the same denominator, so the one with greater numerator will be greater. Since $0< a< 1$ , then $a> a^{2}$ .

Hence, $a + a^{3} + a^{5} + a^{7} + ...> a^{2} + a^{4} + a^{6} + ...$ .

The original infinite geometric series sums to:

$a + a^2 + a^3 + a^4 + a^5 + ... = \frac{a}{1-a} = N \Rightarrow a = \frac{N}{N+1}$ , where $N \in \mathbb{R^{+}}.$

Let $S_{odd}$ and $S_{even}$ be the sum of the odd and the even exponent terms respectively. These yield:

$S_{odd} = a + a^3 + a^5 + ... = \frac{1}{a} \cdot \frac{a^2}{1 - a^2}$ and $S_{even} = a^2 + a^4 + a^6 + ... = \frac{a^2}{1-a^2}.$

Since $S_{odd} = \frac{1}{a} \cdot S_{even} = \frac{N+1}{N} \cdot S_{even} = (1 + \frac{1}{N}) \cdot S_{even},$ we conclude $\boxed{S_{odd} > S_{even}}.$

If we want $a+a^2+a^3+a^4+...$ to be a positive number but no diverge to infinity, we must assume $0<a<1$

So,

$S_{even}=aS_{odd}$

$\boxed{S_{even}<S_{odd}}$

Since the geometric series' sum is a number, it has to converge and since it is a positive number, $0 < a < 1$ .

Looking at the two sums in question, we can rewrite the second as a product of the first: $a(a+a^3+a^5+\ldots)$ . Since $a$ is smaller than 1, the second sum must be smaller than the first.

Let $A=a^2+a^4+a^6+...=a(a+a^3+a^5+...)$ and $B=a+a^3+a^5+...$

$\frac{A}{B}=a$ . Since $a+a^2+a^3+...>0$ , $0<a<1$ ; therefore, $0<\frac{A}{B}<1 \Rightarrow \boxed{B>A}$

Wrote a bit of python code:

    x = 2

    b = 0

    c = 1


    for i in range(1,10):
        a = x ** b
        print a
        c += 2

    for i in range(1,10):
        a = x ** c
        print a
        c += 2

Plugging in a few values for x, we can see that the system with the odd exponents is always larger.

Use the formula $\sum_{i=0}^{\infty} r^{i} = \frac{1}{1-r}$ , which is valid for $|r| < 1$ .

Since the original series converges, its sum is $S_1 = \frac{a}{1-a}$ . We are also told that it is positive, which implies that $a > 0$ .

We are next asked to compare two sums, each of which is a geometric series with ratio $a^2$ . Since $|a| < 1$ it follows that $|a^2| < 1$ , so we know both of these series must also converge. Their sums are $S_{2} = \frac{a}{1-a^2}$ and $S_3 = \frac{a^2}{1-a^2}$ . Thus we see that $S_3 = a * S_2$ . Since $0 < a < 1$ , this implies that $S_3 < S_2$ .

For the original series to converge to a positive number, 0<a<1. You get the second series by multiplying the first by a, so the second series converges to a smaller number.

let $U_{n}= \sum_{i=0}^{n} a^{2i+1}$ and $V_{n}= \sum_{i=1}^{n} a^{2i}$ with n \in N $U_{n} - V_{n} = \sum_{i=0}^{n} a^{2i+1} - \sum_{i=1}^{n} a^{2i} = \sum_{i=0}^{n} a^{2i+1} - \sum_{i=0}^{n} a^{2i} + 1 = \sum_{i=0}^{n} a^{2i+1} - a^{2i} + 1 = \sum_{i=0}^{n} a^{2i}( a - 1 ) + 1$ Therefore: If $a > 1$ then $U_{n} - V{n}$ is positive. so $U_{n} > V_{n}$

For the value of the infinite series to be a real number (i.e. converge) the $**|r| < 1**$ . Because both series (one of even exponents, one of odd exponents) have real values, it's fair to assume they have a real number sum. Hence $a^2 + a^4 + a^6...$ = $(a + a^3 + a^5...) * a$ . Because we know $|a| < 1$ and that a proper (positive) fraction of a (positive) value has to be less than that same value, we conclude that $(a + a^3 + a^5...) > (a^2 + a^4 + a^6...)$