Can the Partial Sum of a Difficult Series be Solved with Induction?

[MarkFL]

Hello MHB! (Wave)

A young man in high school I know has been essentially tasked with finding the following partial sum:

\(\displaystyle S=\sum_{k=0}^{n}\left(\frac{2^k}{3^{2^k}+1}\right)\)

I honestly have no idea how to proceed, and I am hoping someone here can provide some insight. (Star)

 

[I like Serena]

Hello MHB! (Wave)

A young man in high school I know has been essentially tasked with finding the following partial sum:

\(\displaystyle S=\sum_{k=0}^{n}\left(\frac{2^k}{3^{2^k}+1}\right)\)

I honestly have no idea how to proceed, and I am hoping someone here can provide some insight. (Star)

Hey MarkFL,

For high school I would suggest to simply write it out:
$$S_0=\frac{2^0}{3^{2^0}+1}=\frac 14 = 0.25 \\
S_1=\frac 14 + \frac{2^1}{3^{2^1}+1}=\frac 14 + \frac 2{10} = \frac 9{20} = 0.45 \\
S_2=\frac 9{20} + \frac{2^2}{3^{2^2}+1}=\frac 9{20} + \frac{4}{82} = \frac{409}{820}\approx 0.49878 \\
S_3=\frac{409}{820} + \frac{2^3}{3^{2^3}+1} = \frac{409}{820} + \frac{8}{6561} \approx 0.4999996
$$

 

[MarkFL]

Hey MarkFL,

For high school I would suggest to simply write it out:
$$S_0=\frac{2^0}{3^{2^0}+1}=\frac 14 = 0.25 \\
S_1=\frac 14 + \frac{2^1}{3^{2^1}+1}=\frac 14 + \frac 2{10} = \frac 9{20} = 0.45 \\
S_2=\frac 9{20} + \frac{2^2}{3^{2^2}+1}=\frac 9{20} + \frac{4}{82} = \frac{409}{820}\approx 0.49878 \\
S_3=\frac{409}{820} + \frac{2^3}{3^{2^3}+1} = \frac{409}{820} + \frac{8}{6561} \approx 0.4999996
$$

The actual sum he's given is:

\(\displaystyle S=\sum_{k=0}^{2006}\left(\frac{2^k}{3^{2^k}+1}\right)\)

And he's given several choices for the result. I was hoping to find the general partial sum in closed form, and then answer the question that way.

W|A tells me the corresponding infinite sum is 1/2, but I don't even know how to show that. (Sweating)

 

[I like Serena]

The actual sum he's given is:

\(\displaystyle S=\sum_{k=0}^{2006}\left(\frac{2^k}{3^{2^k}+1}\right)\)

And he's given several choices for the result. I was hoping to find the general partial sum in closed form, and then answer the question that way.

W|A tells me the corresponding infinite sum is 1/2, but I don't even know how to show that. (Sweating)

I take it that 1/2 is one of the choices?
We can show it by simply expanding the first couple of partial sums, which show that the partial sum converges pretty quickly to 1/2.
Indeed, it's not a proof, but I don't expect that to be required on high-school level. Just to be able to get the result.

Is this about a math olympiad?
I learned a long time ago that many of the problems posed can typically be resolved by just enumerating.

 

[MarkFL]

I take it that 1/2 is one of the choices?
We can show it by simply expanding the first couple of partial sums, which show that the partial sum converges pretty quickly to 1/2.
Indeed, it's not a proof, but I don't expect that to be required on high-school level. Just to be able to get the result.

Is this about a math olympiad?
I learned a long time ago that many of the problems posed can typically be resolved by just enumerating.

Here are the choices:

A.) \(\displaystyle \frac{1}{2}\)

B.) \(\displaystyle \frac{1}{2}-\frac{2^{2006}}{3^{2^{2006}}-1}\)

C.) \(\displaystyle \frac{1}{2}-\frac{2^{2005}}{3^{2^{2005}}-1}\)

D.) \(\displaystyle \frac{1}{2}-\frac{2^{2007}}{3^{2^{2007}}-1}\)

E.) None of the above.

I'm not sure where the problem came from, but it's not for a competition, just a problem given to a high school student in a gifted program.

 

[Opalg]

For high school I would suggest to simply write it out:
$$S_0=\frac{2^0}{3^{2^0}+1}=\frac 14 = 0.25 \\
S_1=\frac 14 + \frac{2^1}{3^{2^1}+1}=\frac 14 + \frac 2{10} = \frac 9{20} = 0.45 \\
S_2=\frac 9{20} + \frac{2^2}{3^{2^2}+1}=\frac 9{20} + \frac{4}{82} = \frac{409}{820}\approx 0.49878 \\
S_3=\frac{409}{820} + \frac{2^3}{3^{2^3}+1} = \frac{409}{820} + \frac{8}{656{\color{red}2}} \approx 0.4999996
$$

Writing those as $$S_0= \frac 14 = \frac12 - \frac14 = \frac12 - \frac28 = \frac12 - \frac{2^1}{3^{2^1} - 1}, \\
S_1= \frac 9{20} = \frac12 -\frac1{20} = \frac12 - \frac4{80} = \frac12 - \frac{2^2}{3^{2^2} - 1}, \\
S_2= \frac{409}{820} = \frac12 - \frac1{820} = \frac12 - \frac8{6560} = \frac12 - \frac{2^3}{3^{2^3} - 1},$$ it begins to look as though \(\displaystyle S_n = \frac12 - \frac{2^{n+1}}{3^{2^{n+1}} - 1}.\) That suggests option D.) as the correct answer for $S_{2006}.$

If you want something more convincing, you can use induction to prove that suggested formula for $S_n$.

 

[MarkFL]

Writing those as $$S_0= \frac 14 = \frac12 - \frac14 = \frac12 - \frac28 = \frac12 - \frac{2^1}{3^{2^1} - 1}, \\
S_1= \frac 9{20} = \frac12 -\frac1{20} = \frac12 - \frac4{80} = \frac12 - \frac{2^2}{3^{2^2} - 1}, \\
S_2= \frac{409}{820} = \frac12 - \frac1{820} = \frac12 - \frac8{6560} = \frac12 - \frac{2^3}{3^{2^3} - 1},$$ it begins to look as though \(\displaystyle S_n = \frac12 - \frac{2^{n+1}}{3^{2^{n+1}} - 1}.\) That suggests option D.) as the correct answer for $S_{2006}.$

If you want something more convincing, you can use induction to prove that suggested formula for $S_n$.

Awesome work, as always Chris! (Yes)

Using your post, where the base case and induction statement have been given, then the induction step would be:

\(\displaystyle S_{n+1}=\frac{1}{2}-\frac{2^{n+1}}{3^{2^{n+1}}-1}+\frac{2^{n+1}}{3^{2^{n+1}}+1}=\frac{1}{2}-\frac{2^{n+2}}{3^{2^{n+2}}-1}\)

And this completes the proof by induction. :)