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

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Discussion Overview

The discussion revolves around deriving the sum formula \( S_n \) for a series from its nth-term \( U_n \). Participants explore the relationship between the terms of the series and the sum, specifically focusing on the series defined by the terms \( \frac{1}{(2n-1)(2n+1)} \).

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant asks how to derive \( S_n = \frac{n}{2n+1} \) from \( U_n = \frac{1}{(2n-1)(2n+1)} \).
  • Another participant suggests a hint involving partial fractions, stating \( \frac{1}{(2n-1)(2n+1)} = \frac{1}{2n-1} - \frac{1}{2n+1} \) and recommends inserting values for \( n \) to find the sum.
  • A participant questions whether a factor of 2 is missing in the derivation, leading to a discussion about the cancellation of terms in the series.
  • Further clarification is provided that \( \frac{1}{(2n-1)(2n+1)} \) can be expressed as \( \frac{1}{2} \left( \frac{1}{2n-1} - \frac{1}{2n+1} \right) \), prompting a request for guidance on determining \( S_n \).
  • One participant calculates specific terms \( U_1, U_2, U_3 \) and notes inconsistencies in the differences and ratios between them, expressing confusion about deriving \( S_n \).
  • Another participant suggests writing out the sums for specific values of \( n \) to identify a pattern, leading to a proposed formula \( S_n = \frac{1}{2}(1 - \frac{1}{2n+1}) \) which simplifies to \( S_n = \frac{n}{2n+1} \).

Areas of Agreement / Disagreement

Participants express differing views on the derivation process and the presence of factors in the formula. While one participant arrives at a simplified formula for \( S_n \), there remains uncertainty and no consensus on the steps leading to that conclusion.

Contextual Notes

Participants note potential missing factors and the need for careful handling of terms in the series. The discussion reflects various approaches to summing the series without resolving all mathematical steps or assumptions.

Who May Find This Useful

This discussion may be useful for students or individuals interested in series summation techniques, particularly those studying mathematical induction and partial fraction decomposition.

terryds
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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)}## ??
 
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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...
 
There is a factor of 2 missing, isn't it?
 
nasu said:
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.
 
Svein said:
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??
 
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?
 
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Svein said:
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 Sn = ## \frac{1}{2}(1-\frac{1}{2n+1})##
Simplified it Sn = ##\frac{n}{2n+1}##

Thanks for your help!
 

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