Infinite sum proof by induction

In summary, the conversation is about trying to solve a problem involving the expression \sum_{k=1}^{\infty}\frac{6^{k}}{(3^{k}-2^{k})(3^{k+1}-2^{k+1})}. The speaker has simplified the problem to the expression \sum_{k=1}^{n}\frac{6^{k}}{(3^{k}-2^{k})(3^{k+1}-2^{k+1})}=\frac{6(3^{n}-2^{n})}{(3^{n+1}-2^{n+1})} and is trying to prove the equality 3^{n+1}-2^{n+1
  • #1
Doom of Doom
86
0
So I am trying to solve a problem.

Evaluate [tex]\sum_{k=1}^{\infty}\frac{6^{k}}{(3^{k}-2^{k})(3^{k+1}-2^{k+1})}[/tex].

Essentially, I've boiled it down to this, but I can't quite prove it:
[tex]\sum_{k=1}^{n}\frac{6^{k}}{(3^{k}-2^{k})(3^{k+1}-2^{k+1})}=\frac{6(3^{n}-2^{n})}{(3^{n+1}-2^{n+1})}[/tex]
and the limit of this as n approaches infinity is 2.

I need to be able to prove that [tex]3^{n+1}-2^{n+1}+6^{n-1}=(3^{n}-2^{n})^{2}[/tex] in order for my induction hypothesis to work, and I'm having trouble for some reason. Help?
 
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  • #2
Doom of Doom said:
I need to be able to prove that [tex]3^{n+1}-2^{n+1}+6^{n-1}=(3^{n}-2^{n})^{2}[/tex] in order for my induction hypothesis to work, and I'm having trouble for some reason. Help?

Perhaps beacuse it isn't true, consider n=3 then

3^(3+1)-2^(3+1)+6^(3-1)=81-16+36=101
whereas
(3^3-2^3)^2=(27-8)^2=(19)^2=361
 
  • #3
Wow, I'm sorry. Perhaps that's why I was unable to do it. My notes are all over the place and I guess I kind of lost track of a lot of numbers somewhere.

I meant to have [tex](3^{n-1}-2^{n-1})(3^{n+1}-2^{n+1})+6^{n-1}=(3^{n}-2^{n})^{2}[/tex]

Now I'm able to prove it! Thanks anyway.
 

What is an infinite sum proof by induction?

An infinite sum proof by induction is a mathematical technique used to prove that an infinite sum of numbers converges to a specific value. It involves using mathematical induction to show that the sum of the first n terms of the infinite series converges to a finite value as n approaches infinity.

How does an infinite sum proof by induction work?

The proof starts by showing that the statement holds for the first term of the series. Then, it is assumed that the statement holds for the first n terms of the series and the statement is proven for the (n+1)th term. This process is repeated indefinitely until it can be shown that the statement holds for all terms in the series, thus proving that the infinite sum converges.

What are the steps involved in an infinite sum proof by induction?

The steps involved in an infinite sum proof by induction are as follows:

  1. Prove that the statement holds for the first term of the series.
  2. Assume that the statement holds for the first n terms of the series.
  3. Prove that the statement holds for the (n+1)th term of the series.
  4. Repeat steps 2 and 3 indefinitely until it can be shown that the statement holds for all terms in the series.
  5. Conclude that the infinite sum converges to a specific value.

What types of series can be proven using the infinite sum proof by induction?

The infinite sum proof by induction can be used to prove the convergence of any series that follows a specific pattern, such as geometric series or telescoping series. It can also be used to prove the convergence of series with alternating signs, such as the alternating harmonic series.

Why is the infinite sum proof by induction an important concept in mathematics?

The infinite sum proof by induction is an important concept in mathematics because it allows for the proof of the convergence of infinite series, which are commonly used in many mathematical applications. It also helps to establish the validity of mathematical theorems and can be applied to various mathematical problems involving infinite sums.

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