Weird series!

Number Theory Level pending

Let A \mathbb{A} be the set of all the numbers that an be expressed as a b a^{b} , for a , b > 1 a, b>1 and a , b ∈ Z a, b\in\mathbb{Z} . What is the value of ∑ k ∈ A 1 k − 1 \sum_{k\in\mathbb{A}} \frac{1}{k-1} ?

Enter 0 if you think the series diverges.


The answer is 1.

Eric Hernandez by Eric Hernandez , 20, Mexico

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1 solution

Eric Hernandez
Aug 14, 2014

It is very well known that ∑ k = 1 ∞ 1 x − k = 1 x − 1 \sum_{k=1}^{\infty} \frac{1}{x^{-k}}=\frac{1}{x-1} . We'll use this fact later.

By grouping, we see that the diverging sum ∑ k = 2 ∞ 1 k \sum_{k=2}^{\infty} \frac{1}{k} can be reorganized as ( 1 2 + 1 4 + 1 8 . . . ) + ( 1 3 + 1 9 + 1 27 . . . ) + ( 1 5 + 1 25 + 1 125 . . . ) . . . (\frac{1}{2}+\frac{1}{4}+\frac{1}{8}...)+(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}...)+(\frac{1}{5}+\frac{1}{25}+\frac{1}{125}...)... , skipping all the sums whose first term denominator is in A \mathbb{A} . (Because they would already be in series before them).

Therefore, ( 1 2 + 1 3 + 1 4 . . . ) = ( 1 2 + 1 4 + 1 8 . . . ) + ( 1 3 + 1 9 + 1 27 . . . ) + ( 1 5 + 1 25 + 1 125 . . . ) (\frac{1}{2}+\frac{1}{3}+\frac{1}{4}...)=(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}...)+(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}...)+(\frac{1}{5}+\frac{1}{25}+\frac{1}{125}...) .

Now, the right side can be simplified, using the fact above, to ( 1 + 1 2 + 1 4 . . . ) (1+\frac{1}{2}+\frac{1}{4}...) . Here, all the fractions with ( d e n o m i n a t o r + 1 ) ∈ A (denominator+1)\in\mathbb{A} are being excluded.

Because everything in the right side but 1 is also on the left side (it's an infinite sum!), we can cancel out and get ( 1 3 + 1 7 + 1 8 . . . ) = 1 (\frac{1}{3}+\frac{1}{7}+\frac{1}{8}...)=1 .

But the left side is exactly the sum we wanted!

Therefore, the answer is 1 1 .

0 Helpful 0 Interesting 0 Brilliant 0 Confused

I do not quite understand what you are trying to do. I don't see how you're including 1 7 \frac{1}{7} but excluding (say) 1 6 \frac{1}{6} .

Note that when dealing with infinite sums that diverge, you have to be extremely careful with how you cancel terms. You cannot cancel terms against things that appear "much later on".

Calvin Lin Staff - 6 years, 10 months ago

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On the step when I wrote the equation above: ( 1 2 + 1 4 + 1 8 . . . ) + ( 1 3 + 1 9 + 1 27 . . . ) + ( 1 5 + 1 25 + 1 125 . . . ) . . . (\frac{1}{2}+\frac{1}{4}+\frac{1}{8}...)+(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}...)+(\frac{1}{5}+\frac{1}{25}+\frac{1}{125}...)... , the 1 4 + 1 16 + 1 64 . . . \frac{1}{4}+\frac{1}{16}+\frac{1}{64}... series is being excluded, because it is already counted in the 1 2 + 1 4 + 1 8 . . . \frac{1}{2}+\frac{1}{4}+\frac{1}{8}... series. In fact, the 1 a b + 1 a b 2 + 1 a b 3 . . . \frac{1}{a^{b}}+\frac{1}{a^{b^{2}}}+\frac{1}{a^{b^{3}}}... series (which I'll simply call the a b a^{b} series from now on) is always included in the a a series! This means that we must skip in that step all of the x x series, for x ∈ A x\in\mathbb{A} . (By the definition of A \mathbb{A} ).

Eric Hernandez - 6 years, 10 months ago

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