A sum

Calculus Level 2

$\large \sum_{n=1}^{\infty} \frac{2^{2^{n}}-1}{2^{2^{n+1}}} = \frac ab$

If the equation above holds true for coprime positive integers $a$ and $b$ , find the value of $a+b$ .

The answer is 5.

4 solutions

Chew-Seong Cheong
Jan 31, 2017

$\begin{aligned} S & = \sum_{n=1}^\infty \frac {2^{2^n}-1}{2^{2^{n+1}}} \\ & = \frac {2^2-1}{2^4} + \frac {2^4-1}{2^8} + \frac {2^8-1}{2^{16}} + \frac {2^{16}-1}{2^{32}} + ... \\ & = \frac 1{2^2} - \frac 1{2^4} + \frac 1{2^4} - \frac 1{2^8} + \frac 1{2^8} - \frac 1{2^{16}} + \frac 1{2^{16}} - \frac 1{2^{32}} + ... \\ & = \frac 1{2^2} \\ & = \frac 14 \end{aligned}$

$\implies a+b = 1+4 = \boxed{5}$

Moderator note:

In dealing with sum-notation, it can be extremely helpful to simply expand into individual terms; here the pattern which allows terms to cancel then becomes much more visible.

LaTeX: $S_{\infty} = \displaystyle\sum_{n=1}^{ \infty} \frac{2^{2^n} - 1}{2^{2^{n+1}}} = \displaystyle\sum_{n=1}^{ \infty} [\frac{2^{2^n}}{2^{2^{n+1}}} - \frac{1}{2^{2^{n+1}}}]$

LaTeX: $\displaystyle\sum_{n=1}^{ \infty} [\frac{2^{2^n}}{{2^{2^n}} \times {2^2}} - \frac{1}{{2^{2^n}} \times {2^2}}] = \displaystyle\sum_{n=1}^{ \infty} [\frac{1}{2^2} - \frac{1}{{2^{2^n}} \times {2^2}}] = \frac{1}{4} \times \displaystyle\sum_{n=1}^{ \infty}[1 - \frac{1}{2^{2^n}}]$

LaTeX: $\lim_{n \to \infty} \color{#3D99F6}{\frac{1}{2^{2^n}}}= 0^+$ and LaTeX: $\lim_{n \to \infty} \color{#3D99F6}{1 - \frac{1}{2^{2^n}}} = 1 +0^+ = 1$

Then LaTeX: $\lim_{n \to \infty} S_{\infty} = \frac{1}{4} \times 1 = \color{#3D99F6}{\boxed{\frac{1}{4}}}$

Then we have : LaTeX: $a + b = \color{#D61F06}{\boxed{ 1 + 4 = 5}}$

Frédéric Deleria - 4 years, 1 month ago

There's an error in the calculations here. $2^{2^{n+1}} = 2^{2^{n}} \times 2^{2^{n}}$ . So $\frac{2^{2^{n}}-1}{2^{2^{n+1}}}=\frac{2^{2^{n}}}{2^{2^{n}}\times2^{2^{n}}}-\frac{1}{2^{2^{n}}\times2^{2^{n}}}=\frac{1}{2^{2^{n}}}\times(1-\frac{1}{2^{2^{n}}})$ . Which isn't too helpful in this form.

But can be written as:

$\frac{1}{2^{2^{n}}}-\frac{1}{2^{2^{n+1}}}$ Which we can easily use to telescope to infinity leaving $\frac{1}{2^{2^{1}}}=\frac{1}{4}$ .

Alex Burgess - 2 years, 4 months ago

from fractions import Fraction

sum_ = 0
for i in range(1,8):
    numer = 2**(2**i)-1
    denom = 2**(2**(i+1))
    sum_ += float(numer)/denom

x= Fraction(sum_).limit_denominator(1000)  
print '\n\nSummation = %s'     % x          
print 'a = %d' % x.numerator    
print 'b = %d' % x.denominator 
print 'a + b = %d'  % (x.numerator+ x.denominator)

Summation = 1/4
a = 1
b = 4
a + b = 5

Michael Fitzgerald - 3 years, 2 months ago

But can we do this when we have infinite sum?

Malika Oubilla - 2 years, 9 months ago

Yes. the series converges.

Chew-Seong Cheong - 2 years, 9 months ago

A Former Brilliant Member
Oct 8, 2017

LaTeX: $\large \sum_{n=1}^{\infty} \frac{2^{2^{n}}-1}{2^{2^{n+1}}}$ LaTeX: $\large \sum_{n=1}^{\infty} \frac{2^{2^{n}}-1}{2^{2^{n}*2}}$ LaTeX: $\large \sum_{n=1}^{\infty} \frac{2-1}{2^{2}}$ LaTeX: $\large \sum_{n=1}^{\infty} \frac{1}{4}$ Here, a = 1, b = 4. =>a+b = 5.

Why can we simplify $\sum_{n=1}^{\infty} \frac{2^{2^{n}}-1}{2^{2^{n}*2}}$ to $\sum_{n=1}^{\infty} \frac{2-1}{2^{2}}$ ?

Samuel Leško - 3 years, 5 months ago

Every term after the first one have an additive inverse which makes it yield 0. Try telescoping series!

A Former Brilliant Member - 3 years, 5 months ago

The sum of 1/4 from 1 to n is n/4...

Alex Burgess - 2 years, 4 months ago

@ A Brilliant Member - Those last two lines of yours should not have summation notation in front of them. They are just equal to the equivalent of, or exactly, 1/4. Putting the summation from n = 1 to oo in front of them means you would be adding an infinite number of 1/4 terms. That is not what you mean. You need to edit away those last two summation notations.

Dennis Rodman - 2 years, 2 months ago

Ben Major
Jan 30, 2017

You can rewrite the fraction as a sum of 2 fractions: (1/2)^(2^n)-(1/2)^(2^(n+1)). Now why is this more helpful? Well, if you expand it out, you get 1/4-1/16+1/16-1/128+1/128 etc. It sure seems like all the terms after 1/4 cancel out. Well, you can confirm this intuitive solution by considering simple exponent rules. Whenever you subtract a (1/2)^(2^(n+1)) term, you add it back with the next term because the n in the tem (1/2)^(2^n) just increased by 1 as the counter went up by one, making it (1/2)^(2^(n+1)). This term and the one preceding it are opposites, so adding them will cancel out, leaving you with the original term, 1/4=a/b, so a+b=1+4=5. Note: formally, the partial telescoping sum, that is, the sum but only taken up to a finite n can be rewritten as a(1)-a(2)+a(3)-a(4)+...+a(n)-a(n+1). Using the reasoning above we can see that all but a(n+1) will cancel out in the partial sum. Now, we can take lim(n->infty) of a(1)-a(n+1) (the partial sum). Since the exponent of a(n+1) is negative, as n tends to infty the value of this term will tend to 0, leaving you with a(1) as the value of the sum.

What a fun problem!

Stewart Gordon
Dec 10, 2017

$\large\sum_{n=1}^\infty \frac{2^{2^n}-1}{2^{2^{n+1}}} \\ = \large\sum_{n=1}^\infty \frac{2^{2^n}}{2^{2^{n+1}}} - \frac1{2^{2^{n+1}}} \\ = \large\sum_{n=1}^\infty \left(2^{2^n - 2^{n+1}} - 2^{-2^{n+1}} \right) \\ = \large\sum_{n=1}^\infty \left(2^{-2^n} - 2^{-2^{n+1}} \right) \\ = 2^{-2^1} - 2^{-2^2} + 2^{-2^2} - 2^{-2^3} + 2^{-2^3} - 2^{-2^4} + \ldots \\ = 2^{-2^1} \\ = \boxed{\frac14}$

A sum

The answer is 5.

4 solutions

Moderator note:

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