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Prove with induction the identity

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

How can I solve this problem?

Should i set k= (p+1) and n = (p+1), then try to get the left side equal to the right side?

Daniel Andersson
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3 Answers3

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If induction is not mandatory,

let $p(m)=\dfrac{am+b}{(2m-1)(2m+1)}$

$$p(k)-p(k+1)=\dfrac{ak+b}{(2k-1)(2k+1)}-\dfrac{a(k+1)+b}{(2k+1)(2k+3)}$$

$$=\dfrac{ak+4b+a}{(2k-1)(2k+1)(2k+3)}$$

We need $a=0,4b+a=1\iff b=?$

So in essence, we have a Telescoping series

lab bhattacharjee
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  • Very elegant solution! How did you know the fraction could be broken like that?Also, I believe $\frac{ak+b}{(2k-1)(2k+1)}-\frac{a(k+1)+b}{(2k+1)(2k+3)}=\frac{2ak+4b+a}{(2k-1)(2k+1)(2k+3)}$. Therefore, $b=\frac 1{4}$. – s0ulr3aper07 Apr 14 '19 at 08:07
  • @s0ulr3aper07, Series containing terms in arithmetic progression or of the form $$\sum_{r=1}^n(a_0+a_1r+a_2r^2+\cdots)$$ generally yields to this method. Loga Series or exponential Series don't. – lab bhattacharjee Apr 14 '19 at 10:21
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When $n=1$,

$$\begin {align} \sum_{k=1}^n\frac 1{(2k-1)(2k+1)(2k+3)}&=\frac 1{(2.1-1)(2.1+1)(2.1+3)}\\ &=\frac 1{15}\\ &=\frac{1(1+2)}{3(2.1+1)(2.1+3)}\\ &=\frac{n(n+2)}{3(2n+1)(2n+3)} \end {align}$$

Assume the result to be true for $n=m$

We show it is also true for $n=m+1$, $$\begin {align} \sum_{k=1}^{m+1}\frac 1{(2k-1)(2k+1)(2k+3)}&=\sum_{k=1}^{m}\frac 1{(2k-1)(2k+1)(2k+3)}\\ &+\frac 1{(2(m+1)-1)(2(m+1)+1)(2(m+1)+3)}\\ &=\frac {m(m+2)}{3(2m+1)(2m+3)}+\frac{1}{(2m+1)(2m+3)(2m+5)}\\ &=\frac {1}{(2m+1)(2m+3)}\times\bigg(\frac{m(m+2)}{3}+\frac 1{2m+5}\bigg)\\ &=\frac {1}{(2m+1)(2m+3)}\times\bigg(\frac{(m^2+2m)(2m+5)+3}{3(2m+5)}\bigg)\\ &=\frac {1}{(2m+1)(2m+3)}\times\bigg(\frac{2m^3+9m^2+10m+3}{3(2m+5)}\bigg)\\ &=\frac {1}{(2m+1)(2m+3)}\times\frac{(2m+1)(m+1)(m+3)}{3(2m+5)}\\ &=\frac{(m+1)(m+3)}{3(2m+5)(2m+3)}\\ &=\frac{\big(m+1\big)\big((m+1)+2\big)}{3\big(2(m+1)+1\big)\big(2(m+1)+3\big)}\\ \end {align}$$

Hence,

by Principle of Mathematical Induction, the result holds for all $n\geq1$.

s0ulr3aper07
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  • Cool but why do you add the right side when n = m, with the left side when n = m+1 ? – Daniel Andersson Apr 13 '19 at 19:21
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    I guess that gives us the right side when n = m + 1? – Daniel Andersson Apr 13 '19 at 19:44
  • Yes, you're absolutely right. I have simply broken the left side sum of $m+1$ terms into two parts. The first part consists of the first $m$ terms and the second part consists of only the last $m+1$th term. Note that this is commonly done in inductive proofs because we have to isolate the first $m$ terms in order to use our assumption that the result is true for $n=m$. – s0ulr3aper07 Apr 13 '19 at 20:35
  • Okay great :) How did you know that 2m^3 + 9m^2 + 10m + 3 = (2m+1)(m+1)(m+3) ? – Daniel Andersson Apr 14 '19 at 18:14
  • @DanielAndersson I did a bit of back-solving there. I was already aware that the final result should not have $(2m+1)$ in the denominator. Therefore, the only way it could be eliminated would be by cancelling out with another $(2m+1)$ in the numerator. However, the numerator was of the form $2m^3+9m^2+10m+3$, so that means $(2m+1)$ must be a factor of $2m^3+9m^2+10m+3$. Now polynomial division of $2m^3+9m^2+10m+3$ by $(2m+1)$ yields the other two factors. – s0ulr3aper07 Apr 15 '19 at 08:24
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Since $k$ is a dummy variable, you should induct on $n$. In other words, show the claim is right when $n=1$, then show $$\sum_{k=1}^p\tfrac{1}{(2k-1)(2k+1)(2k+3)}=\tfrac{p(p+2)}{3(2p+1)(2p+3)}\implies\sum_{k=1}^{p+1}\tfrac{1}{(2k-1)(2k+1)(2k+3)}=\tfrac{(p+1)(p+3)}{3(2p+3)(2p+5)}.$$

J.G.
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