How to find Sn from Un? (sum of series)

[terryds]

How to derive Sn (sum formula of series) from Un (nth-term of series) ?

In my textbook (in the Mathematical Induction chapter), it's shown that
$\frac{1}{1\times 3}+\frac{1}{3\times5}+\frac{1}{5\times7}+...+\frac{1}{(2n-1)(2n+1)}=\frac{n}{2n+1}$

I know how to proof it by mathematical induction,

But, I want to know how to derive the Sn (sum formula of series) from Un..
I mean, how to get Sn = $\frac{n}{2n+1}$
From Un = $\frac{1}{(2n-1)(2n+1)}$ ??

 

[Svein]

Just a small hint: \frac{1}{(2n-1)(2n+1)}=\frac{1}{2n-1}-\frac{1}{2n+1}. Now insert n=1 and n=2 and add...

 

[nasu]

There is a factor of 2 missing, isn't it?

 

[Svein]

There is a factor of 2 missing, isn't it?

Yes, of course. I could pretend that it really was to test you, but I plain forgot. The long and short of it is that the parts cancel out and you are left with the first and last element.

 

[terryds]

Just a small hint: \frac{1}{(2n-1)(2n+1)}=\frac{1}{2n-1}-\frac{1}{2n+1}. Now insert n=1 and n=2 and add...

But, $\frac{1}{2n-1}-\frac{1}{2n+1}=\frac{2}{(2n-1)(2n+1)}$ not $\frac{1}{(2n-1)(2n+1)}$

However, I can write it $\frac{1}{(2n-1)(2n+1)} = \frac{1}{2}\times(\frac{1}{2n-1}-\frac{1}{2n+1})$

Then, what should I do to determine the Sn formula??

For n = 1, then U₁= 1/3
For n = 2, then U₂= 1/15
For n = 3, then U₃ = 1/35

However, the difference between term 1 and term 2 is not the same as the difference between term 2 and term 3
Also, the ratio is different
How to determine Sn??

 

[Svein]

As I said, write it out for n=1, 2, 3. This gives: S_{3}=\frac{1}{2}(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}). Do you see a pattern?

 

[terryds]

As I said, write it out for n=1, 2, 3. This gives: S_{3}=\frac{1}{2}(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}). Do you see a pattern?

Yupp
It seems like S_n = $\frac{1}{2}(1-\frac{1}{2n+1})$
Simplified it S_n = $\frac{n}{2n+1}$

Thanks for your help!