Hello Sabo Originally Posted by

**Sabo** I have huge problems with induction exercises, so if I could get a pointer or two on the following exercise (not a full solution preferably) I would appreciate it a lot.

What is the sum of $\displaystyle \sum_{k=1}^n \frac{1}{(2k-1)(2k+1)}$? Prove the assumption (assumption the correct term?) by means of induction.

Sorry, but I'm not clear exactly what method you are supposed to use to solve this question. As you have posed it (where the sum is not known), this isn't really an induction question. Until we have an Induction Hypothesis, we must use other methods. I will show you another method, and (once we have a solution) the Induction method.

First, then, we can solve this directly by means of Partial Fractions.

Let$\displaystyle \frac{1}{(2k-1)(2k+1)} = \frac{A}{2k-1}+\frac{B}{2k+1}$

$\displaystyle \Rightarrow 1 \equiv A(2k+1) +B(2k-1)$

$\displaystyle \Rightarrow A = \tfrac12, B = \tfrac12$

$\displaystyle \Rightarrow \frac{1}{(2k-1)(2k+1)} = \frac{1}{2(2k-1)}-\frac{1}{2(2k+1)}$

Therefore$\displaystyle \sum_{k=1}^n \frac{1}{(2k-1)(2k+1)}=\sum_{k=1}^n\Big(\frac{1}{2(2k-1)}-\frac{1}{2(2k+1)}\Big)$$\displaystyle =\frac12\sum_{k=1}^n\Big(\frac{1}{2k-1}-\frac{1}{2k+1}\Big)$

$\displaystyle =\frac12\Big(\sum_{k=1}^n\frac{1}{2k-1}-\sum_{k=1}^n\frac{1}{2k+1}\Big)$

$\displaystyle =\frac12\Big(\sum_{k=1}^n\frac{1}{2k-1}-\sum_{k=2}^{n+1}\frac{1}{2k-1}\Big)$

$\displaystyle =\frac12\Big(1+\sum_{k=2}^n\frac{1}{2k-1}-\sum_{k=2}^{n}\frac{1}{2k-1}-\frac{1}{2n+1}\Big)$

$\displaystyle =\frac12\Big(1-\frac{1}{2n+1}\Big)$

$\displaystyle =\frac{n}{2n+1}$

However, if the question had said:Prove by induction that $\displaystyle \sum_{k=1}^n \frac{1}{(2k-1)(2k+1)}=\frac{n}{2n+1}$

then the method is as follows:

Let $\displaystyle P(n)$ be the proposition $\displaystyle \sum_{k=1}^n \frac{1}{(2k-1)(2k+1)}=\frac{n}{2n+1}$

Then $\displaystyle P(n) \Rightarrow \sum_{k=1}^n \frac{1}{(2k-1)(2k+1)}+\frac{1}{(2n+1)(2n+3)}=\frac{n}{2n+1}+\f rac{1}{(2n+1)(2n+3)}$$\displaystyle \Rightarrow \sum_{k=1}^{n+1} \frac{1}{(2k-1)(2k+1)}=\frac{n(2n+3)+1}{(2n+1)(2n+3)}$$\displaystyle =\frac{2n^2+3n+1}{(2n+1)(2n+3)}$

$\displaystyle =\frac{(2n+1)(n+1)}{(2n+1)(2n+3)}$

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

$\displaystyle \Rightarrow P(n+1)$

Now $\displaystyle P(1)$ is $\displaystyle \sum_{k=1}^1 \frac{1}{(2k-1)(2k+1)}=\frac{1}{3}$, which is true. Therefore by Induction $\displaystyle P(n)$ is true for all $\displaystyle n \ge 1$.

Grandad

P.S. Ah, I see that Soroban has given you a solution as well. So, you've got lots to look at!