6

I am very confused with complete induction. Because in every task there is something different to do, and I never know what to insert (thats my biggest problem). Here's the example: Proof with complete induction. Please please help me, because I have exams coming up (I am just becoming a primary school teacher..)

For $n\in\mathbb{N}$:

$$\sum^n_{i=1}\frac{1}{(2i-1)(2i+1)}=\frac{n}{2n+1}$$

Community
  • 1
Sophia
  • 335

3 Answers3

9

For a full solution, proceed like this:

$n=1$: $$\sum_{i=1}^1 \frac{1}{(2i-1)(2i+1)} = \frac{1}{(2-1)(2+1)} = \frac{1}{3} = \frac{1}{2 \cdot 1 +1},$$ so it holds for $n=1$.

Assume next that it holds for some generic $n$. You need to show that then it also holds for $n+1$. As it holds for $n$, you can assume that $$\sum_{i=1}^n \frac{1}{(2i-1)(2i+1)} = \frac{n}{2n+1}. \quad (1),$$ and want to show that $$\sum_{i=1}^{n+1} \frac{1}{(2i-1)(2i+1)} = \frac{n+1}{2(n+1)+1}. \quad (2)$$ Then: $$\begin{align} \sum_{i=1}^{n+1} \frac{1}{(2i-1)(2i+1} &= \sum_{i=1}^n \frac{1}{(2i-1)(2i+1)} + \frac{1}{(2(n+1)-1)(2(n+1)+1)} \\ & = \frac{n}{2n+1} + \frac{1}{(2n+1)(2n+3)} \quad \text{using (1)} \\ & = \frac{n(2n+3)}{(2n+1)(2n+3)} + \frac{1}{(2n+1)(2n+3)} \\ & = \frac{2n^2 +3n +1}{(2n+1)(2n+3)} \\ & = \frac{(n+1)(2n+1)}{(2n+1)(2n+3)} = \frac{n+1}{2(n+1)+1},\\ \end{align}$$ which is (2), and was to be shown.

gnometorule
  • 4,640
  • 13
    Looking at the edit records, gnometorule answered the question at 2013-02-20 18:13:32Z and amWhy's edit at 2013-02-20 18:19:07Z shows the same content from "Assume next that it holds for some generic $n$." The edits are more than 5 minutes apart (5:35), so it doesn't appear that gnometorule could have copied from amWhy's edit. – robjohn Feb 20 '13 at 19:13
7

For general inductive proofs it may well be true that "there is something different to do" in each new problem, e.g. it may require genuine ingenuity to devise an appropriate induction hypothesis. However, this is not the case for inductive proofs of sums like the above. As I explained in this answer, many inductive proofs of sums and products are of a very simple inductive type known as telescopy. For inductions of this type one can do the induction uniformly - once and for all - by abstracting it into a theorem that applies to all such problems. For sums this yields

Theorem $\rm \displaystyle\ \ \sum_{i\,=\,1}^n\, f(i)\, =\, g(n)\iff f(1) = g(1)\ {\rm\ and\ }\ f(n) \,=\, g(n)-g(n\!-\!1)\:\ $ for $\rm\,n > 1.$

In your case we have

$$\rm f(n) \,=\, \frac{1}{(2n-1)(2n+1)},\quad g(n) \,=\, \frac{n}{2n+1}$$

Thus, applying the theorem, we check that $\rm\ f(1) = 1/3 = g(1)\ $ and

$$\rm g(n) - g(n\!-\!1) = \frac{n}{2n\!+\!1} - \frac{n\!-\!1}{2n\!-\!1}\, =\, \frac{n(2n\!-\!1)-(2n\!+\!1)(n\!-\!1)}{(2n\!-\!1)(2n\!+\!1)}\, = \frac{1}{(2n\!-\!1)(2n\!+\!1)} = f(n)$$

which completes the proof. Notice that the proof required no ingenuity - only verifying some simple polynomial (or rational function) equalities - a mechanical process.

Community
  • 1
Math Gems
  • 19,574
  • I'm quite puzzled by the downvote. If there is something that I can explain further then please let me know what needs clarification and I will gladly do so. – Math Gems Feb 20 '13 at 20:50
  • 2
    I had thought about presenting a proof using telescoping series. My reason for not posting my proof was that the question was more on the mechanics of induction than on possible simplifications in special cases. Perhaps the downvoter (not I) was thinking this was more confusing to someone just beginning with induction. – robjohn Feb 20 '13 at 21:11
  • 6
    @amWhy I never imagined anyone would consider linking to related posts as "self-advertising". I am here first and foremost to teach mathematics. So, e.g. when I teach telescopic induction I think it is helpful to link to other examples of telescopic induction (which arises in many guises). – Math Gems Feb 20 '13 at 21:18
  • @Rob It did occur to me that the reason for the downvote could be due to differences in pedagogical viewpoints. If so, I encourage such downvoters to leave constructive comments explaining the reason (which will be well-received). Lacking such explanation, readers might be misled into thinking the downvote signifies something else (e.g. a mathmatical error), when that is not the case. – Math Gems Feb 20 '13 at 21:28
  • 2
    @MathGems: I agree. I believe that in most cases, a downvote without a reason is non-constructive criticism. I am of the mind that a constructive comment is better than a downvote anyway. – robjohn Feb 20 '13 at 21:31
  • 3
    @amWhy: Math Gems definitely has a certain style :) – The Chaz 2.0 Feb 22 '13 at 23:55
  • 1
    @The Chaz 2.0 Yes, indeed! :-) – amWhy Feb 22 '13 at 23:59
  • @robjohn In this answer I say more about my view of the role of telescopy in teaching induction. Constructive feedback is always welcome. – Math Gems Feb 24 '13 at 00:55
5

There is one more (easier) way to solve this problem without induction: Expand the summand into partial fractions to obtain (denote $S_n$ the actual sum): $$ S_n = \frac{1}{2} \sum_{k=1}^{n}\bigg(\frac{1}{2k-1}-\frac{1}{2k+1} \bigg)=\frac{1}{2} \bigg(1-\frac{1}{3} +\frac{1}{3} + \cdots - \frac{1}{2n+1} \bigg) = \frac{n}{2n+1} $$

Alex
  • 19,262