3
$\begingroup$

I'm having an issue solving this problem using induction. If possible, could someone add in a very brief explanation of how they did it so it's easier for me to understand?

$$\frac{1}{1\cdot3} + \frac{1}{2\cdot4} + \cdots + \frac{1}{n\cdot(n+2)} = \frac{3}{4} - \frac{(2n+3)}{2(n+1)(n+2)}$$

How do I prove the above equation for all integers where $n\geq1$?

$\endgroup$
  • 3
    $\begingroup$ For information on how to format and structure an induction proof (as well as the basic steps that are common to all induction proofs) I recommend you check out the answers here. As for your particular question, have you considered approaching via partial fraction decomposition? I believe the series will telescope. $\endgroup$ – JMoravitz Aug 9 '15 at 20:29
6
$\begingroup$

First, check the formula for $n=1$. So: $$\frac1{1\cdot 3}=\frac34-\frac{2\cdot1+3}{2(1+1)(2+1)}$$ Since this is true, we have shown the so called base case.

Now substitute in the formula $n$ by $n+1$ to get the statement that you have to show:

$$\frac1{1\cdot 3}+\frac1{2\cdot 4}+\cdots+\frac1{(n+1)(n+3)}=\frac34-\frac{2n+5}{2(n+2)(n+3)}\qquad (*)$$

The good news is that you can (and should) assume that the formula is valid for the $n$ first positive integers, so $$\begin{align}&\left(\frac1{1\cdot 3}+\frac1{2\cdot 4}+\cdots+\frac1{n(n+2)}\right)+\frac1{(n+1)(n+3)}\\ &=\frac34-\frac{2n+3}{2(n+1)(n+2)}+\frac1{(n+1)(n+3)} \end{align}$$ and you should obtain $(*)$ with straightforward computings.

Perhaps my explanation is not very brief but I hope it to be useful.

$\endgroup$
3
$\begingroup$

No induction is necessary: it's a matter of a telescoping sum, if you write: $$\frac1{k(k+2)}=\frac12\Bigl(\frac 1k-\frac1{k+2}\Bigr)$$ Apply this decomposition to the above sum: \begin{align*} \sum_{k=1}^n\frac1{k(k+2)}&=\frac12\sum_{k=1}^n\Bigl(\frac 1k-\frac1{k+2}\Bigr)\\ &=\frac12\Bigl(1\color{red}{-\frac13}+\frac12\color{red}{-\frac14+\frac13-\frac15+\dots+\frac1{n-1}}-\frac1{n+1}\color{red}{+\frac 1n}-\frac1{n+2}\Bigr)\\ &=\frac12\Bigl(1+\frac12-\frac1{n+1}-\frac1{n+2}\Bigr)=\frac34-\frac{n+2+n+1}{2(n+1)(n+2)}. \end{align*}

$\endgroup$
  • $\begingroup$ This proof is nice, but I think that what OP asked for is an explanation about induction. +1, though. $\endgroup$ – ajotatxe Aug 9 '15 at 20:57
  • $\begingroup$ Certainly but I find ridiculous to use induction for proofs that don't require it, unless it leads to a simpler proof. In addition the problem with most induction proofs, is they don't explain where things come from. $\endgroup$ – Bernard Aug 9 '15 at 21:02
2
$\begingroup$

Hint: $$\frac { 1 }{ 2 } \left( \frac { 1 }{ n } -\frac { 1 }{ n+2 } \right) =\frac { 1 }{ n\left( n+2 \right) } $$

$\endgroup$
2
$\begingroup$

Let $$p(n):\displaystyle\frac{1}{1\cdot3} + \frac{1}{2\cdot4} + \cdots + \frac{1}{n\cdot(n+2)} = \frac{3}{4} - \frac{(2n+3)}{2(n+1)(n+2)}$$

Put $n=1\;,$ We get $$\displaystyle \frac{1}{1\cdot 3} = \frac{3}{4}-\frac{5}{2\cdot 2\cdot 3} = \frac{4}{12}$$

So it is true for $n=1$

Now Put $n=k\;,$ We get $$\displaystyle \frac{1}{1\cdot3} + \frac{1}{2\cdot4} + \cdots + \frac{1}{k\cdot(k+2)}=\frac{3}{4} - \frac{(2k+3)}{2(k+1)(k+2)}$$

Now Using $p(k)\;,$ We will prove for $p(k+1)$

So $$\displaystyle p(k+1):\frac{1}{1\cdot3} + \frac{1}{2\cdot4} + \cdots + \frac{1}{(k+1)\cdot(k+3)}=\frac{1}{1\cdot3} + \frac{1}{2\cdot4} + \cdots + \frac{1}{k\cdot(k+2)}+\frac{1}{(k+1)\cdot (k+3)}$$

So $$\displaystyle p(k+1) = \frac{3}{4} - \frac{(2k+3)}{2(k+1)(k+2)}+\frac{1}{(k+1)(k+3)}= \frac{3}{4}-\frac{1}{(k+1)}\left\{\frac{2k+3}{2(k+2)}-\frac{1}{(k+3)}\right\}$$

$$\displaystyle = \frac{3}{4}-\frac{1}{(k+1)}\left\{\frac{2k^2+9k+9-2k-4}{2(k+2)(k+3)}\right\} = \frac{3}{4}-\frac{1}{2(k+1)(k+2)(k+3)}\cdot (2k+5)\cdot (k+1)=\frac{3}{4}-\frac{(2k+5)}{2(k+2)(k+3)}$$

So $p(k)$ We have prove for $p(k+1).$

$\endgroup$
2
$\begingroup$

First, show that this is true for $n=1$:

$\sum\limits_{k=1}^{1}\frac{1}{k(k+2)}=\frac34-\frac{2+3}{2(1+1)(1+2)}$

Second, assume that this is true for $n$:

$\sum\limits_{k=1}^{n}\frac{1}{k(k+2)}=\frac34-\frac{2n+3}{2(n+1)(n+2)}$

Third, prove that this is true for $n+1$:

$\sum\limits_{k=1}^{n+1}\frac{1}{k(k+2)}=$

$\color\red{\sum\limits_{k=1}^{n}\frac{1}{k(k+2)}}+\frac{1}{(n+1)(n+3)}=$

$\color\red{\frac34-\frac{2n+3}{2(n+1)(n+2)}}+\frac{1}{(n+1)(n+3)}=$

$\frac34-\frac{2(n+1)+3}{2(n+2)(n+3)}$

Please note that the assumption is used only in the part marked red.

$\endgroup$
2
$\begingroup$

When dealing with a sum like you are here (summing $n$ terms for some general expression), I would almost always recommend that you use $\Sigma$-notation, for it tidies up a lot of the algebraic mess you have to deal with in your induction proof. With that in mind, you may write your claim as follows.

Claim: For any $n\geq 1$, the statement $$ S(n) : \sum_{i=1}^n\frac{1}{i(i+2)}=\frac{3}{4}-\frac{2n+3}{2(n+1)(n+2)} $$ is true.

Base step ($n=1$): $S(1)$ says that $\sum_{i=1}^1\frac{1}{i(i+2)}=\frac{1}{3}=\frac{3}{4}-\frac{5}{12}$, and this is true.

Induction step ($S(k)\to S(k+1)$): Fix some $k\geq 1$, and assume that $$ S(k) : \sum_{i=1}^k\frac{1}{i(i+2)}=\frac{3}{4}-\frac{2k+3}{2(k+1)(k+2)} $$ is true. To be proved is that $$ S(k+1) : \sum_{i=1}^{k+1}\frac{1}{i(i+2)}=\frac{3}{4}-\frac{2k+5}{2(k+2)(k+3)} $$ follows. Beginning with the left side of $S(k+1)$, \begin{align} \sum_{i=1}^{k+1}\frac{1}{i(i+2)}&= \sum_{i=1}^k\frac{1}{i(i+2)}+\frac{1}{(k+1)(k+3)}\tag{by defn.}\\[1em] &= \left(\frac{3}{4}-\frac{2k+3}{2(k+1)(k+2)}\right)+\frac{1}{(k+1)(k+3)}\tag{by $S(k)$}\\[1em] &= \frac{3}{4}-\frac{2k^2+7k+5}{2(k+1)(k+2)(k+3)}\tag{common denom.}\\[1em] &= \frac{3}{4}-\frac{(2k+5)(k+1)}{2(k+1)(k+2)(k+3)}\tag{factor}\\[1em] &= \frac{3}{4}-\frac{2k+5}{2(k+2)(k+3)},\tag{simplify} \end{align} one arrives at the right side of $S(k+1)$, thereby showing $S(k+1)$ is also true, completing the inductive step.

By mathematical induction, the claim $S(n)$ is true for all $n\geq 1$. $\blacksquare$

$\endgroup$
-1
$\begingroup$

Proof: By mathematical induction.

  • $P(1)$: $\frac{1}{1\cdot 3} = \frac{1}{3} = \frac{1}{3} = \frac{4}{12} = \frac{9}{12}-\frac{5}{12} = \frac{3}{4}-\frac{5}{12} = \frac{3}{4}-\frac{5}{2\cdot 2 \cdot 3} = \frac{3}{4}-\frac{2 \cdot 1+3}{2(1+1)(1+2)}$.

  • $P(n) \Rightarrow P(n+1)$: Assume as the Inductive Hypothesis: $$\small \frac{1}{1\cdot 3}+\frac{1}{2\cdot 4}+\frac{1}{3\cdot 5}+\cdots+\frac{1}{n(n+2)} = \frac{3}{4}-\frac{2n+3}{2(n+1)(n+2)}.$$

    We need to show $$\small \frac{1}{1\cdot 3}+\frac{1}{2\cdot 4}+\frac{1}{3 \cdot 5}+\cdots+\frac{1}{(n+1)((n+1)+2)} = \frac{3}{4}-\frac{2(n+1)+3}{2((n+1)+1)((n+1)+2)}.$$

    Note that the LHS is $$\small \frac{1}{1\cdot 3}+\frac{1}{2\cdot 4}+\frac{1}{3\cdot 5}+\cdots+\frac{1}{(n+1)((n+1)+2)} = \frac{1}{1\cdot 3}+\frac{1}{2\cdot 4}+\frac{1}{3\cdot 5}+\cdots+\frac{1}{(n+1)(n+3)}.$$ Notice the two factors in the denominators of each term increase by $1$ as I go to the right. This means I am able to display the term before $\frac{1}{(n+1)(n+3)}$ by subtracting the two factors by $1$.

$$\begin{align} \mathrm{LHS} &= \frac{1}{1\cdot 3}+\frac{1}{2\cdot 4}+\frac{1}{3\cdot 5}+\cdots+\frac{1}{(n+1)(n+3)} \\ &= \frac{1}{1\cdot 3}+\frac{1}{2\cdot 4}+\frac{1}{3\cdot 5}+\cdots+\frac{1}{n(n+2)}+\frac{1}{(n+1)(n+3)} \\ &= \left( \frac{1}{1\cdot 3}+\frac{1}{2\cdot 4}+\frac{1}{3\cdot 5}+\cdots+\frac{1}{n(n+2)}\right)+\frac{1}{(n+1)(n+3)} \\ &= \frac{3}{4}-\frac{2n+3}{2(n+1)(n+2)}+\frac{1}{(n+1)(n+3)} \\ &= \frac{3}{4}-\frac{(2n+3)(n+3)}{2(n+1)(n+2)(n+3)}+\frac{2(n+2)}{2(n+1)(n+2)(n+3)} \\ &= \frac{3}{4}+\frac{-(2n+3)(n+3)+2(n+2)}{2(n+1)(n+2)(n+3)} \\ &= \frac{3}{4}+\frac{-(2n^2+6n+3n+9)+2n+4}{2(n+1)(n+2)(n+3)} \\ &= \frac{3}{4}+\frac{-2n^2-9n-9+2n+4}{2(n+1)(n+2)(n+3)} \\ &= \frac{3}{4}+\frac{-2n^2-7n-5}{2(n+1)(n+2)(n+3)} \\ &= \frac{3}{4}-\frac{2n^2+7n+5}{2(n+1)(n+2)(n+3)} \\ &= \frac{3}{4}-\frac{2n^2+2n+5n+5}{2(n+1)(n+2)(n+3)} \\ &= \frac{3}{4}-\frac{2n(n+1)+5(n+1)}{2(n+1)(n+2)(n+3)} \\ &= \frac{3}{4}-\frac{(n+1)(2n+5)}{2(n+1)(n+2)(n+3)} \\ &= \frac{3}{4}-\frac{2n+5}{2(n+2)(n+3)} \\ &= \frac{3}{4}-\frac{2n+2+3}{2(n+1+1)(n+1+2)} \\ &= \frac{3}{4}-\frac{2(n+1)+3}{2((n+1)+1)((n+1)+2)}. \end{align}$$

Therefore, $\frac{1}{1\cdot 3}+\frac{1}{2\cdot 4}+\frac{1}{3\cdot 5}+\cdots+\frac{1}{n(n+2)} = \frac{3}{4}-\frac{2n+3}{2(n+1)(n+2)}$ for all $n \geq 1$.

$\endgroup$
  • $\begingroup$ I have attempted to clean this up by introducing MathJax. Please check it over to ensure that I haven't inadvertantly introduced any errors. For more information about typesetting mathematics here please see our MathJax basic tutorial and quick reference. $\endgroup$ – user642796 Aug 10 '15 at 4:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.