/sci/ - Science & Math

No. I don't remember the proof though.

1 = 1/3 + 1/9 + 1/27 + 1/81 + ...

The number of fractions would have to be odd.

1/odd + 1/odd = (odd + odd)/odd = even/odd = even != 1

>>4289210
I'm sorry. I neglected to clarify that the series should be finite.

>>4289210

that doesn't equal 1...

>>4289194
1/3 + 1/3 + 1/3 = 1

now you know that 1/3 = 3/9 and for every part of the sum multiply by 3 the numerator and denominator enough times..
and you'll get unique value looking fractions which are basically all 1/3 and their sum would still be 1
+ all the numbers will be primes

i don't know who has proved that <span class="math">3^x =[/spoiler]prime but it just happens to be so

1/3 + 1/5 + 1/7.... nope this isn't going to work

1/1 + 0

Well, consider this. Let your first term be 1/3. How can you sum 2/3rds? Only by picking fractions with a denominator in the form 3*x. Pick one of those fractions. So how do you sum up 2*x/(3*x)- 1/(3*x)? By picking a fraction with a denominator in the form of (3*x)y. Now, can we reduce this series to 0? Why or why not?

Finite series? Sure.
1/1 = 1

It's highly unlikely such a series of fractions exist because the smallest fraction in that series would be the reciprocal of an odd perfect number, which are believed not to exist.

An odd perfect number's divisors including 1 would sum to itself.

<div class="math">d_0 + d_1 + d_2 + d_3 + ... + d_{n - 1} = d_n ~with \; d_k\cdot d_{n-k} = d_n</div>
<div class="math">d_0 + d_1 + d_2 + d_3 + ... + d_{n} = 2d_n</div>
<div class="math">\frac{d_0 + d_1 + d_2 + d_3 + ... + d_{n}}{d_n} = 2</div>
<div class="math">\frac{1}{d_n} + \frac{1}{d_{n-1}} + \frac{1}{d_{n-2}} + ... + \frac{1}{d_2} + \frac{1}{d_1} + 1 = 2</div>
<div class="math">\frac{1}{d_n} + \frac{1}{d_{n-1}} + \frac{1}{d_{n-2}} + ... + \frac{1}{d_2} + \frac{1}{d_1} = 1</div>

It's not possible, since no one else managed to prove it I guess I will

odd numbers are represented by 2k+1 where k is an integer, the equation becomes:

1/(2k'1+1) + 1/(k'2+1) + ... + 1/(k'n+1) ?= 1

my keyboard doesn't do subscripts and I don't feel like downloading mathtype so view the k'1 as k subscript 1, and k'n as k subscript n, and so forth

n = the number of fractions we're summing & I'm also going to denote P'n as: k'1 * k'2 * ... * k'n

if you combine the fractions in the first equation you get:

{2^n-1[P'n/k'1 + P'n/k'2 + ... + P'n/k'n] + n} / {2^n(P'n) + 2^n-1[P'n/k'1 + P'n/k'2 + ... + P'n/k'n] + 1} ?= 1

get rid of the fraction

2^n(P'n) + 2^n-1[P'n/k'1 + P'n/k'2 + ... + P'n/k'n] + 1 ?= 2^n-1[P'n/k'1 + P'n/k'2 + ... + P'n/k'n] + n

get rid of the bulky fucking middle term:

2^n(P'n) + 1 ?= n

isolate P'n

P'n ?= (n - 1)/2^n

the first equation now depends on the product of a set of integers being equal to a term that is zero when n=1 and between zero and one when n>1

this is possible in only one scenario:

n=1, P'n=0, 1/2(0)+1 = 1

1/1 = 1

Q.E.D.

>>4289773
You do realise that you just proved it to be possible right? Its been said earlier
1/1 = 1

OP didn't ask for non-integer terms.

>>4289811
>You do realise that you just proved it to be possible right? Its been said earlier
1/1 = 1

yes, but only for that one case which OP probably did not think of or he would have disallowed the option of 1 as a denominator in his post. Everyone can tell that 1=1.

>OP didn't ask for non-integer terms.

Not sure what you mean by that, I never said that he did.

>>4289893
Yes, series of distinct fractions with an s implies it's possible that there is only one fraction?

Hurrrrrr