Induction Proof: Sum of Series $ \frac{1}{(2n-1)(2n+1)} = \frac{1}{2}$

[ognik]

Q. Show by induction that $ \sum_{1}^{\infty} \frac{1}{(2n-1)(2n+1)} = \frac{1}{2} $

So, start with base case n=1, $ S_1 = \frac{1}{(2-1)(2+1)} = \frac{1}{3}$? Maybe it's bedtime ...

 

[MarkFL]

I think (if forced to use induction rather than partial fractions and telescoping series) what I would do is begin with the partial sum formula:

\(\displaystyle \sum_{k=1}^n\frac{1}{(2k+1)(2k-1)}=\frac{n}{2n+1}\)

Now prove that by induction, and then take the limit as $n\to\infty$ to get the desired result. :)

 

[ognik]

Sorry, suspect I know how to get the partial sum but can't bring it to mind, the terms would be $ \frac{1}{3}, \frac{1}{15},\frac{1}{35}, \frac{1}{63},... $ I can see your sum is right, but not how to get it ...

 

[MarkFL]

Sorry, suspect I know how to get the partial sum but can't bring it to mind, the terms would be $ \frac{1}{3}, \frac{1}{15},\frac{1}{35}, \frac{1}{63},... $ I can see your sum is right, but not how to get it ...

Okay, given that you have verified the base case, then we state $P_n$:

\(\displaystyle \sum_{k=1}^n\frac{1}{(2k+1)(2k-1)}=\frac{n}{2n+1}\)

Now, for our induction step, I would look at the difference:

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

Can you proceed?

 

[ognik]

Hi - what I'm missing is how you got to $ \frac{n}{2n+1} $ - as far as I can see its neither arithmetic nor geometric series, and I couldn't see how to get $ \frac{n}{2n+1} $ by inspection either? Once I have that 1st step, I know the rest :-)

 

[MarkFL]

Hi - what I'm missing is how you got to $ \frac{n}{2n+1} $ - as far as I can see its neither arithmetic nor geometric series, and I couldn't see how to get $ \frac{n}{2n+1} $ by inspection either? Once I have that 1st step, I know the rest :-)

It was a hypothesis (which I confirmed before posting it) based on the first several sums. :)

 

[ognik]

Suppose that comes with experience... Next question is that the book said $ = \frac{1}{2}$ but the partial sum (and testing the 1st partial sum in the original formula) seems to get \(\displaystyle \frac{1}{3}\)?

 

[MarkFL]

Suppose that comes with experience... Next question is that the book said $ = \frac{1}{2}$ but the partial sum (and testing the 1st partial sum in the original formula) seems to get \(\displaystyle \frac{1}{3}\)?

Once you have proved:

\(\displaystyle S_n=\sum_{k=1}^n\frac{1}{(2k+1)(2k-1)}=\frac{n}{2n+1}\)

Then, use:

\(\displaystyle S_{\infty}=\lim_{n\to\infty}\left(\frac{n}{2n+1}\right)\)

What do you get?

 

[ognik]

Hi - it all looked crystal clear in the morning :-) Thanks for your patience

 

[ognik]

For the benefit of anyone who has the predicament of trying to find a partial sum formula for other than geometric & arithmetic series, without using partial fractions, I figured out the following approach.

Start with a row of the first few n values.
Under each n, calculate the corresponding term of the series
Under each n again, find the nth sum by adding the terms from the previous row. A pattern should appear for which you can find a partial sum formula.

Using this post's problem ( $\sum_{n=1}^{\infty} \frac{1}{(2n-1)(2n+1)}$) as an example, and a matrix structure just for convenience, the above approach gives:

$ \begin{bmatrix}n:&1&2&3&4\\ \frac{1}{(2n-1)(2n+1)}:&\frac{1}{3}&\frac{1}{15}&\frac{1}{35}&\frac{1}{63}\\S_n:&\frac{1}{3}&\frac{2}{5}&\frac{3}{7}&\frac{4}{9}\end{bmatrix}$

The pattern for the partial sums is now clear and can be written as $ \frac{n}{2n+1}$