A Painful GP

Algebra Level 4

Evaluate

∑ n = 0 ∞ ∑ k = 0 n 1 2 k 3 n − k \large \sum_{n=0}^\infty \sum_{k=0}^n \frac{ 1 } { 2^k 3^{n-k} }


The answer is 3.00.

View wiki
Calvin Lin by Calvin Lin

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

3 solutions

Kenny Lau
Oct 12, 2015

It is certain that:

a n − b n a − b = ∑ k = 0 n a k b n − k \displaystyle\frac{a^n-b^n}{a-b}=\sum_{k=0}^na^kb^{n-k}

In this case we have a = 1 2 a=\frac12 and b = 1 3 b=\frac13 :

S = 1 a − b ∑ n = 0 ∞ a n − b n = 6 ∑ ( 1 / 2 ) n − ( 1 / 3 ) n = 6 ( 2 − 1.5 ) = 3 S=\displaystyle\frac1{a-b}\sum_{n=0}^\infty a^n-b^n=6\sum(1/2)^n-(1/3)^n=6(2-1.5)=3

3 Helpful 0 Interesting 0 Brilliant 0 Confused
Alan Yan
Oct 12, 2015

Let's first find the closed form of ∑ k = 0 n 1 2 k 3 n − k \sum_{k=0}^{n}{\frac{1}{2^k3^{n-k}}}

Note that this value is just the x n x^{n} coefficient of ( 1 + 1 2 x + 1 4 x 2 + . . . ) ( 1 + 1 3 x + 1 9 x 2 + . . . ) = 1 1 − x 2 ⋅ 1 1 − x 3 = 3 1 − x 2 − 2 1 − x 3 = 3 ∑ k = 0 ∞ ( x 2 ) k − 2 ∑ k = 0 ∞ ( x 3 ) k \begin{aligned} \left(1 + \frac{1}{2}x + \frac{1}{4}x^2 + ...\right)\left(1 + \frac{1}{3}x + \frac{1}{9}x^2 + ...\right) & = \frac{1}{1 - \frac{x}{2}} \cdot \frac{1}{1 - \frac{x}{3}} \\ & = \frac{3}{1-\frac{x}{2}} - \frac{2}{1-\frac{x}{3}}\\ & = 3\sum_{k = 0}^{\infty}{\left(\frac{x}{2}\right)^k} - 2\sum_{k = 0}^{\infty}{\left(\frac{x}{3}\right)^k} \end{aligned}

The x n x^n coefficient is 3 2 n − 2 3 n . \frac{3}{2^n} - \frac{2}{3^n}. The desired expression is now just 3 ∑ n = 0 ∞ 1 2 n − 2 ∑ n = 0 ∞ 1 3 n = 6 − 3 = 3 3\sum_{n = 0}^{\infty}{\frac{1}{2^n}} - 2\sum_{n = 0}^{\infty}{\frac{1}{3^n}} = 6 - 3 = \boxed{3}

and we are done.

2 Helpful 0 Interesting 0 Brilliant 0 Confused

You could greatly shorten/simplify your solution. You have the right ideas there, and just need to exploit it.

Hint: Let x = 1 x = 1 .

Calvin Lin Staff - 5 years, 8 months ago

Log in to reply

Since n n ranges across all natural numbers, the answer is just ( 1 + 1 2 + 1 4 + . . . ) ( 1 + 1 3 + 1 9 + . . . ) = 2 ⋅ 3 2 = 3 (1 + \frac{1}{2} + \frac{1}{4} + ...)(1 + \frac{1}{3} + \frac{1}{9} + ...) = 2 \cdot \frac{3}{2} = \boxed{3}

Alan Yan - 5 years, 8 months ago
Chew-Seong Cheong
Oct 13, 2015

∑ n = 0 ∞ ∑ k = 0 n 1 2 k 3 n − k = ∑ n = 0 ∞ [ 1 3 n ∑ k = 0 n ( 3 2 ) k ] G P : ∑ k = 0 n r k = r n + 1 − 1 r − 1 = ∑ n = 0 ∞ [ ( 3 2 ) n + 1 − 1 3 n ( 3 2 − 1 ) ] = 2 ∑ n = 0 ∞ ( 3 2 n + 1 − 1 3 n ) = 3 ∑ n = 0 ∞ ( 1 2 ) n − 2 ∑ n = 0 ∞ ( 1 3 ) n = 3 ( 1 1 − 1 2 ) − 2 1 − 1 3 = 6 − 3 = 3 \begin{aligned} \sum_{n=0}^\infty \sum_{k=0}^n \frac{1}{2^k3^{n-k}} & = \sum_{n=0}^\infty \left[ \frac{1}{3^n}\color{#3D99F6}{\sum_{k=0}^n \left(\frac{3}{2}\right)^k} \right] \quad \quad \small \color{#3D99F6}{GP: \quad \sum_{k=0}^n r^k = \frac{r^{n+1}-1}{r-1}} \\ & = \sum_{n=0}^\infty \left[\frac{\color{#3D99F6}{\left(\frac{3}{2}\right)^{n+1}-1}}{3^n \color{#3D99F6}{\left(\frac{3}{2}-1\right)}} \right] \\ & = 2 \sum_{n=0}^\infty \left(\frac{3}{2^{n+1}} - \frac{1}{3^{n}} \right) \\ & = 3 \sum_{n=0}^\infty \left(\frac{1}{2}\right)^n - 2\sum_{n=0}^\infty \left(\frac{1}{3}\right)^n \\ & = 3 \left(\frac{1}{1-\frac{1}{2}}\right) - \frac{2}{1-\frac{1}{3}} \\ & = 6 - 3 = \boxed{3} \end{aligned}

The painless or less-pain approach \color{#3D99F6}{\text{The painless or less-pain approach}}

∑ n = 0 ∞ ∑ k = 0 n 1 2 k 3 n − k = ∑ n = 0 ∞ ( 1 2 ) n ∑ n = 0 ∞ ( 1 3 ) n For n > > k = 1 1 − 1 2 × 1 1 − 1 3 = 2 × 3 2 = 3 \begin{aligned} \sum_{n=0}^\infty \sum_{k=0}^n \frac{1}{2^k3^{n-k}} & = \sum_{n=0}^\infty \left( \frac{1}{2} \right)^n \color{#3D99F6}{\sum_{n=0}^\infty \left( \frac{1}{3} \right)^n} \quad \quad \small \color{#3D99F6}{\text{For }n >> k} \\ & = \frac{1}{1-\frac{1}{2}} \times \frac{1}{1-\frac{1}{3}} \\ & = 2 \times \frac{3}{2} \\ & = \boxed{3} \end{aligned}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Yup, that's the painful way. Now try and find the painless / less painful way.

Calvin Lin Staff - 5 years, 8 months ago

Log in to reply

Finally got it! Is it right?

Chew-Seong Cheong - 5 years, 7 months ago

Log in to reply

Converting the order of summation (with justification of why it's valid), makes this problem much easier to approach.

That's close, but your first term is wrong. It should be ∑ ( 1 2 ) n \sum \left( \frac{1}{2} \right)^n . Note that in the next line, your GP is also wrong (with 3/2), it should be 1 1 − r \frac{1}{1 - r } , but you have 1 r − 1 \frac{1}{ r - 1 } .

Calvin Lin Staff - 5 years, 7 months ago

Log in to reply

@Calvin Lin – Thanks. Yes, obviously ∑ n = 0 ∞ ( 3 2 ) n \sum_{n=0}^\infty \left(\frac{3}{2}\right)^n does not converge. I have changed the solution.

Chew-Seong Cheong - 5 years, 7 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...