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Find a formula for $\sum_{i=1}^{n} \frac{1}{(2i-1)(2i+1)}$ and prove that it holds for all $n \geq 1$ I don't know how to solve this particular problem, can someone help me please. Thanks

Michael Joyce
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Alex
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    Note that our expression is $\frac{1}{2}\left(\frac{1}{2i-1}-\frac{1}{2i+1}\right)$. So we are looking at a sum of the shape $1/2$ times $\left(1-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\cdots+ \left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)$. Note the cancellations (telescoping). – André Nicolas Apr 12 '14 at 02:44
  • The telescoping approach is nice and does not require induction to prove the formula, but it's worth noting that it is also possible to figure out a formula and prove it without being clever enough to recognize the telescoping nature of the series. – Michael Joyce Apr 12 '14 at 02:47

2 Answers2

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Hint: Expand the summand by partial fractions to see that the partial sums telescope:

$$\frac{1}{(2k-1)(2k+1)}=\frac{1}{2(2k-1)}-\frac{1}{2(2k+1)}\\ \sum_{k=1}^n\frac{1}{(2k-1)(2k+1)}=\sum_{k=1}^n\left(\frac{1}{2(2k-1)}-\frac{1}{2(2k+1)}\right)\\ =\frac{1}{2(2\cdot 1-1)}-\frac{1}{2(2n+1)}$$

David H
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Here are the steps to take to prove a formula for the sum by induction. I'll leave it to you to work out the details.

1) Compute values of the sum for small values of $n$. Doing so for 5 or more values of $n$ will hopefully reveal a pattern that you can conjecture holds for all $n$.

2) Now that you have a conjectural formula for the sum, check to make sure that it holds in the base case when $n=1$. This should be easy after step 1.

3) Now the key step is that you will assume your conjectured formula is correct for an arbitrary value of $n$. Under that assumption, prove that the formula holds for $n+1$. Having done that, you will have proven that if your sum formula holds for a particular value for $n$, then it also holds for all values of $n+1$.

4) You can now conclude that your conjectured formula holds for $n$ by the Principle of Mathematical Induction (PMI). Indeed, the PMI says that if a statement is true for $n=1$ (what you prove in step 2) and if it is also the case that whenever a statement holds for a value of $n$, it also holds for the value $n+1$ (what you prove in step 3), then the statement holds for all natural numbers.

Michael Joyce
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