Progressive integer sum

Algebra Level 4

Given that $a+a^2 + a^3 + \cdots$ is a non-zero integer , can $a^2 + a^3 + a^4 + \cdots$ also be an integer?

No Yes

3 solutions

Rishabh Jain
Jul 1, 2016

Relevant wiki: Arithmetic and Geometric Progressions Problem Solving

For both series to converge $|a|<1$ , also since first series is a non zero integer $a\ne 0$ . Hence $a\in(-1,1)-\{0\}$ (i.e $'a'$ is a non integer).

Write second series as:

$(\color{#D61F06}{a}+a^2+a^3+\cdots)-\color{#D61F06}{a}$

$=\text{(Some integer)}-\text{A non integer}$

And we know subtraction of a non integer from an integer cannot result in an integer. Hence second series can never be an integer.

The question is not really clear as it does not specify that series has to converge ? Some of integers is an integer.

Giuseppe Paesano - 4 years, 11 months ago

Even for the first series to converge 'a' has to lie in (-1,1) other wise how could it be a integer... !

Rishabh Jain - 4 years, 11 months ago

What if I write the second series as $a$ x(First series)? Is it correct? If so, it is possible

João Vitor Cordeiro de Brito - 4 years, 11 months ago

Read Calvin's comment on the below solution..... It just makes hard to decide whether second series would be a integer or not!

$\text{Second series }: \underbrace{a}_{\text{A fraction}}\times (\underbrace{\text{First series}}_{\text{An integer}})$

Now the problem is that multiplying a fraction by a integer does not always return a non integer ( Eg. 9*1/3 =3 (A integer)) and hence writing second series as integer times a fraction would be of no help.

Rishabh Jain - 4 years, 11 months ago

Right. We have to be careful about to present this

Thinking of it as "multiply by $a$ ", we have "a = integer / integer" which is just a rational number.
Thinking of it as "subtract these two sequences to get $a$ ", we get "a = integer - integer = integer".

Calvin Lin Staff - 4 years, 11 months ago

Thanks for answering! I agree with you. In fact, I thought the same way.. But the problem is stated as "can ... also be an integer?" And not "will ... also be an integer?" I'm not a native english speaker, but this misunderstood led me to the mistake

João Vitor Cordeiro de Brito - 4 years, 11 months ago

@João Vitor Cordeiro de Brito – Can / Will matters if the statement is sometimes true.

"If $a$ is an integer, can $a/2$ be an integer?" Yes it can, for example when $a = 2$ .
"If $a$ is an integer, will $a/2$ be an integer?" No, not necessarily so.

Calvin Lin Staff - 4 years, 11 months ago

@Calvin Lin – Then, $Yes$ , the sequence can actually $Can$ also be an integer. Isn't it? Please, help me because I'm confused

João Vitor Cordeiro de Brito - 4 years, 11 months ago

@João Vitor Cordeiro de Brito – If there is a value of $a$ such that $a + a^2 + ...$ is a non-zero integer AND $a^2 + a^3 + \ldots$ is an integer, then the answer to this question is YES.
If no such value exists, then the answer is NO.

The claim, proved in the solutions, is that no such value exists.

Calvin Lin Staff - 4 years, 11 months ago

A Former Brilliant Member
Jul 1, 2016

If $|a|>1$ , The series won't converge.

$a+a^2+a^3+\ldots=\dfrac{a}{1-a}$ .

$a^2+a^3+a^4+\ldots=\dfrac{a^2}{1-a}$ . For it to be an integer, $a$ also has to be an integer for $(-1,1)$ but there is no non-zero integer in that interval.

I don't understand your reasoning. Why must $a$ also be an integer? $\frac{5}{2} \times 2$ is an integer even though $\frac{5}{2}$ isn't an integer.

Calvin Lin Staff - 4 years, 11 months ago

Oh right. I've made the same mistake! hahahahaah

@Svatejas Shivakumar , do you know how to fix it?

Pi Han Goh - 4 years, 11 months ago

Great.

Challenge master note : Find the error in the working below.

Suppose the first infinite geometric sum is equal to $-2$ , then solving for $a$ and substituting it into the second infinite sum of $\dfrac{a^2}{1-a}$ , we get another integer $-4$ . So the answer is be "Yes, it is possible".

Pi Han Goh - 4 years, 11 months ago

@Pi Han Goh Well obviously by solving, we get a = -2 and this is a contradiction!! Because the radius of convergence of the GP is when a lies between -1 and 1

Aaghaz Mahajan - 3 years ago

Yuppers! Great!

Pi Han Goh - 3 years ago

The range of the infinite geometric sum should be (-1,1)

A Former Brilliant Member - 4 years, 11 months ago

Hmmm, nope. It is possible to have an infinite geometric sum outside of that range.

For example, (2/3) + (2/3)^2 + (2/3)^3 + ... = 2

Pi Han Goh - 4 years, 11 months ago

Daniel Heiß
Aug 21, 2016

Write x=a+a^2+a^3+... Then we have x=a*(1+x) and as x is non-zero, also (1+x) is non-zero by this equation. So we get a=x/(x+1) which is not integer. This is a alternative to the argument below using convergence. The Rest of the solution is as in the comments below

Progressive integer sum

3 solutions

0 pending reports