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Question Number 157351 by physicstutes last updated on 22/Oct/21

Prove by mathematical induction Σ_(r=1) ^n (1/(r(r+1))) = (n/(n+1))

$$\mathrm{Prove}\:\mathrm{by}\:\mathrm{mathematical}\:\mathrm{induction} \\ $$$$\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{r}\left({r}+\mathrm{1}\right)}\:=\:\frac{{n}}{{n}+\mathrm{1}} \\ $$

Commented by hknkrc46 last updated on 22/Oct/21

★ (1/(r(r + 1))) = (((r + 1) − r)/(r(r + 1))) = ((r + 1)/(r(r + 1))) − (r/(r(r + 1))) = (1/r) − (1/(r + 1)) ★ Σ_(r=1) ^n (1/(r(r+1))) = Σ_(r=1) ^n ((1/r) − (1/(r + 1))) = ((1/1) − (1/2))+((1/2) − (1/3))+ ∙∙∙ +((1/n) − (1/(n + 1))) = (1/1) − (1/(n + 1)) = (n/(n + 1))

$$\bigstar\:\frac{\mathrm{1}}{\boldsymbol{{r}}\left(\boldsymbol{{r}}\:+\:\mathrm{1}\right)}\:=\:\frac{\left(\boldsymbol{{r}}\:+\:\mathrm{1}\right)\:−\:\boldsymbol{{r}}}{\boldsymbol{{r}}\left(\boldsymbol{{r}}\:+\:\mathrm{1}\right)} \\ $$$$=\:\frac{\boldsymbol{{r}}\:+\:\mathrm{1}}{\boldsymbol{{r}}\left(\boldsymbol{{r}}\:+\:\mathrm{1}\right)}\:−\:\frac{\boldsymbol{{r}}}{\boldsymbol{{r}}\left(\boldsymbol{{r}}\:+\:\mathrm{1}\right)} \\ $$$$=\:\frac{\mathrm{1}}{\boldsymbol{{r}}}\:−\:\frac{\mathrm{1}}{\boldsymbol{{r}}\:+\:\mathrm{1}}\: \\ $$$$\bigstar\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{r}\left({r}+\mathrm{1}\right)}\:=\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\frac{\mathrm{1}}{\boldsymbol{{r}}}\:−\:\frac{\mathrm{1}}{\boldsymbol{{r}}\:+\:\mathrm{1}}\right) \\ $$$$=\:\left(\frac{\mathrm{1}}{\mathrm{1}}\:−\:\frac{\mathrm{1}}{\mathrm{2}}\right)+\left(\frac{\mathrm{1}}{\mathrm{2}}\:−\:\frac{\mathrm{1}}{\mathrm{3}}\right)+\:\centerdot\centerdot\centerdot\:+\left(\frac{\mathrm{1}}{\boldsymbol{{n}}}\:−\:\frac{\mathrm{1}}{\boldsymbol{{n}}\:+\:\mathrm{1}}\right) \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{1}}\:−\:\frac{\mathrm{1}}{\boldsymbol{{n}}\:+\:\mathrm{1}}\:=\:\frac{\boldsymbol{{n}}}{\boldsymbol{{n}}\:+\:\mathrm{1}} \\ $$

Answered by ajfour last updated on 22/Oct/21

S=Σ_(r=1) ^n (1/T_r )=Σ_(r=1) ^n ((1/r)−(1/(r+1))) =Σ_(r=1) ^n (t_r −t_(r+1) )=1−(1/(n+1)) ⇒ S=(n/(n+1))

$${S}=\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{T}_{{r}} }=\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\frac{\mathrm{1}}{{r}}−\frac{\mathrm{1}}{{r}+\mathrm{1}}\right) \\ $$$$\:\:\:=\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\left({t}_{{r}} −{t}_{{r}+\mathrm{1}} \right)=\mathrm{1}−\frac{\mathrm{1}}{{n}+\mathrm{1}} \\ $$$$\Rightarrow\:\:{S}=\frac{{n}}{{n}+\mathrm{1}} \\ $$$$ \\ $$

Answered by physicstutes last updated on 22/Oct/21

• prove for n = 1. LHS = Σ_(r=1) ^1 (1/(r(r+1))) = (1/(1(1+1))) = (1/2) RHS = (1/(1+1)) =(1/2) ⇒ true for n=1. • Assume it is true for n=k ⇒ Σ_(r=1) ^k (1/(r(r+1))) = (k/(k+1)) • Prove for n= k+1. Σ_(r=1) ^(k+1) (1/(r(r+1))) = Σ_(r=1) ^k (1/(r(r+1))) + (1/((k+1)(k+2))) = (k/(k+1))+(1/((k+1)(k+2))) = (1/(k+1))(k+(1/(k+2))) =(1/(k+1))(((k^2 +2k+1)/(k+2))) = (1/(k+1))((((k+1)^2 )/(k+2))) = ((k+1)/(k+2)) ⇒ true for n= k+1 hence true ∀ n ∈Z

$$\bullet\:\mathrm{prove}\:\mathrm{for}\:{n}\:=\:\mathrm{1}.\: \\ $$$$\mathrm{LHS}\:=\:\underset{{r}=\mathrm{1}} {\overset{\mathrm{1}} {\sum}}\frac{\mathrm{1}}{{r}\left({r}+\mathrm{1}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{1}\left(\mathrm{1}+\mathrm{1}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{RHS}\:=\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{1}}\:=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{true}\:\mathrm{for}\:{n}=\mathrm{1}. \\ $$$$\bullet\:\mathrm{Assume}\:\mathrm{it}\:\mathrm{is}\:\mathrm{true}\:\mathrm{for}\:{n}={k} \\ $$$$\Rightarrow\:\underset{{r}=\mathrm{1}} {\overset{{k}} {\sum}}\frac{\mathrm{1}}{{r}\left({r}+\mathrm{1}\right)}\:=\:\frac{{k}}{{k}+\mathrm{1}} \\ $$$$\bullet\:\mathrm{Prove}\:\mathrm{for}\:{n}=\:{k}+\mathrm{1}. \\ $$$$\underset{{r}=\mathrm{1}} {\overset{{k}+\mathrm{1}} {\sum}}\frac{\mathrm{1}}{{r}\left({r}+\mathrm{1}\right)}\:=\:\underset{{r}=\mathrm{1}} {\overset{{k}} {\sum}}\frac{\mathrm{1}}{{r}\left({r}+\mathrm{1}\right)}\:+\:\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)\left({k}+\mathrm{2}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{{k}}{{k}+\mathrm{1}}+\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)\left({k}+\mathrm{2}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{1}}{{k}+\mathrm{1}}\left({k}+\frac{\mathrm{1}}{{k}+\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{{k}+\mathrm{1}}\left(\frac{{k}^{\mathrm{2}} +\mathrm{2}{k}+\mathrm{1}}{{k}+\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{1}}{{k}+\mathrm{1}}\left(\frac{\left({k}+\mathrm{1}\right)^{\mathrm{2}} }{{k}+\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{{k}+\mathrm{1}}{{k}+\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{true}\:\mathrm{for}\:{n}=\:{k}+\mathrm{1}\: \\ $$$$\mathrm{hence}\:\mathrm{true}\:\forall\:{n}\:\in\mathbb{Z} \\ $$

Answered by som(math1967) last updated on 22/Oct/21

To prove (1/(1×2)) +(1/(2×3)) +(1/(3×4)) +...+(1/(n(n+1)))=(n/(n+1)) p(1) L.H.S=(1/(1×2))=(1/2) R.H.S=(1/(1+1))=(1/2) ∴true for p(1) let true for p(m) ∴ (1/(1×2)) +(1/(2×3)) +...+(1/(m(m+1)))=(m/(m+1)) now p(m+1) =(1/(1×2)) +(1/(2×3)) +...+(1/(m(m+1))) +(1/((m+1)(m+2))) =(m/(m+1)) +(1/((m+1)(m+2))) ★ =((m(m+2)+1)/((m+1)(m+2))) =(((m+1)^2 )/((m+1)(m+2)))=((m+1)/(m+2)) ∴true for p(m+1) ∴Σ^n _(r=1) (1/(r(r+1))) =(n/(n+1)) ★ (1/(1×2))+(1/(2×3)) +...+(1/(m(m+1)))=(m/(m+1))

$${To}\:{prove}\: \\ $$$$\frac{\mathrm{1}}{\mathrm{1}×\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}×\mathrm{3}}\:+\frac{\mathrm{1}}{\mathrm{3}×\mathrm{4}}\:+...+\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)}=\frac{{n}}{{n}+\mathrm{1}} \\ $$$${p}\left(\mathrm{1}\right)\:{L}.{H}.{S}=\frac{\mathrm{1}}{\mathrm{1}×\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${R}.{H}.{S}=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\therefore{true}\:{for}\:{p}\left(\mathrm{1}\right) \\ $$$${let}\:{true}\:{for}\:{p}\left({m}\right) \\ $$$$\therefore\:\frac{\mathrm{1}}{\mathrm{1}×\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}×\mathrm{3}}\:+...+\frac{\mathrm{1}}{{m}\left({m}+\mathrm{1}\right)}=\frac{{m}}{{m}+\mathrm{1}} \\ $$$${now}\:{p}\left({m}+\mathrm{1}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}×\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}×\mathrm{3}}\:+...+\frac{\mathrm{1}}{{m}\left({m}+\mathrm{1}\right)}\:+\frac{\mathrm{1}}{\left({m}+\mathrm{1}\right)\left({m}+\mathrm{2}\right)} \\ $$$$=\frac{{m}}{{m}+\mathrm{1}}\:+\frac{\mathrm{1}}{\left({m}+\mathrm{1}\right)\left({m}+\mathrm{2}\right)}\:\bigstar \\ $$$$=\frac{{m}\left({m}+\mathrm{2}\right)+\mathrm{1}}{\left({m}+\mathrm{1}\right)\left({m}+\mathrm{2}\right)} \\ $$$$=\frac{\left({m}+\mathrm{1}\right)^{\mathrm{2}} }{\left({m}+\mathrm{1}\right)\left({m}+\mathrm{2}\right)}=\frac{{m}+\mathrm{1}}{{m}+\mathrm{2}} \\ $$$$\therefore{true}\:{for}\:{p}\left({m}+\mathrm{1}\right) \\ $$$$\therefore\underset{\boldsymbol{{r}}=\mathrm{1}} {\overset{\boldsymbol{{n}}} {\sum}}\:\frac{\mathrm{1}}{\boldsymbol{{r}}\left(\boldsymbol{{r}}+\mathrm{1}\right)}\:=\frac{\boldsymbol{{n}}}{\boldsymbol{{n}}+\mathrm{1}} \\ $$$$\bigstar\:\frac{\mathrm{1}}{\mathrm{1}×\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}×\mathrm{3}}\:+...+\frac{\mathrm{1}}{{m}\left({m}+\mathrm{1}\right)}=\frac{{m}}{{m}+\mathrm{1}} \\ $$

Commented by peter frank last updated on 22/Oct/21

great

$$\mathrm{great} \\ $$

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