I-heard-this-sum-can-be-close-using-the-Digamma-function-Please-help-me-use-it-i-don-t-know-it-sum-of-nth-term-1-1-2-3-1-4-5-6-1-7-8-9-

Question Number 45316 by Tawa1 last updated on 11/Oct/18

$$\mathrm{I}\:\mathrm{heard}\:\mathrm{this}\:\mathrm{sum}\:\mathrm{can}\:\mathrm{be}\:\mathrm{close}\:\mathrm{using}\:\mathrm{the}\:\mathrm{Digamma}\:\mathrm{function}. \\ $$$$\mathrm{Please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{use}\:\mathrm{it}.\:\mathrm{i}\:\mathrm{don}'\mathrm{t}\:\mathrm{know}\:\mathrm{it}.\:\:\: \\ $$$$\:\:\:\mathrm{sum}\:\mathrm{of}\:\mathrm{nth}\:\mathrm{term}:\:\:\:\frac{\mathrm{1}}{\mathrm{1}.\mathrm{2}.\mathrm{3}}\:+\:\frac{\mathrm{1}}{\mathrm{4}.\mathrm{5}.\mathrm{6}}\:+\:\frac{\mathrm{1}}{\mathrm{7}.\mathrm{8}.\mathrm{9}}\:+\:…\: \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18

T_n =(1/({1+(n−1)3}{2+(n−1)3}{3+(n−1)3})) T_n =(1/((3n−2)((3n−1)(3n))) here i am using some formula mentioned Hall and knight Algebra... S_n =C−(1/((3n−1)(3n)×2×1)) n=1 (1/(1×2×3))=C−(1/(2×3×2)) C=(1/6)+(1/(12))=((2+1)/(12))=(1/4) S_n =(1/4)−(1/(2(3n−1)(3n))) recheck... S_1 =(1/4)−(1/(2×2×3))=((3−1)/(12))=(1/6)=T_1 =(1/(1×2×3))=(1/6) conform S_2 =(1/4)−(1/(2×5×6))=((15−1)/(60))=((14)/(60))=(7/(30)) T_2 =S_2 −S_1 =(7/(30))−(1/6)=((7−5)/(30))=(1/(15)) but T_2 =(1/(120)) So my answer is not correct... pls menhion the source of this problem S_n =(1/(1×2×3))+(1/(4×5×6))+(1/(7×8×9))+... =

$${T}_{{n}} =\frac{\mathrm{1}}{\left\{\mathrm{1}+\left({n}−\mathrm{1}\right)\mathrm{3}\right\}\left\{\mathrm{2}+\left({n}−\mathrm{1}\right)\mathrm{3}\right\}\left\{\mathrm{3}+\left({n}−\mathrm{1}\right)\mathrm{3}\right\}} \\ $$$${T}_{{n}} =\frac{\mathrm{1}}{\left(\mathrm{3}{n}−\mathrm{2}\right)\left(\left(\mathrm{3}{n}−\mathrm{1}\right)\left(\mathrm{3}{n}\right)\right.} \\ $$$${here}\:{i}\:{am}\:{using}\:{some}\:{formula}\:{mentioned} \\ $$$${Hall}\:{and}\:{knight}\:{Algebra}… \\ $$$${S}_{{n}} ={C}−\frac{\mathrm{1}}{\left(\mathrm{3}{n}−\mathrm{1}\right)\left(\mathrm{3}{n}\right)×\mathrm{2}×\mathrm{1}} \\ $$$${n}=\mathrm{1} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}×\mathrm{2}×\mathrm{3}}={C}−\frac{\mathrm{1}}{\mathrm{2}×\mathrm{3}×\mathrm{2}} \\ $$$${C}=\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{12}}=\frac{\mathrm{2}+\mathrm{1}}{\mathrm{12}}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${S}_{{n}} =\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{3}{n}−\mathrm{1}\right)\left(\mathrm{3}{n}\right)} \\ $$$${recheck}… \\ $$$${S}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}×\mathrm{2}×\mathrm{3}}=\frac{\mathrm{3}−\mathrm{1}}{\mathrm{12}}=\frac{\mathrm{1}}{\mathrm{6}}={T}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{1}×\mathrm{2}×\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$${conform} \\ $$$${S}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}×\mathrm{5}×\mathrm{6}}=\frac{\mathrm{15}−\mathrm{1}}{\mathrm{60}}=\frac{\mathrm{14}}{\mathrm{60}}=\frac{\mathrm{7}}{\mathrm{30}} \\ $$$${T}_{\mathrm{2}} ={S}_{\mathrm{2}} −{S}_{\mathrm{1}} =\frac{\mathrm{7}}{\mathrm{30}}−\frac{\mathrm{1}}{\mathrm{6}}=\frac{\mathrm{7}−\mathrm{5}}{\mathrm{30}}=\frac{\mathrm{1}}{\mathrm{15}} \\ $$$${but}\:{T}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{120}} \\ $$$${So}\:{my}\:{answer}\:{is}\:{not}\:{correct}… \\ $$$${pls}\:{menhion}\:{the}\:{source}\:{of}\:{this}\:{problem} \\ $$$${S}_{{n}} =\frac{\mathrm{1}}{\mathrm{1}×\mathrm{2}×\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}×\mathrm{5}×\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{7}×\mathrm{8}×\mathrm{9}}+… \\ $$$$= \\ $$

Commented by Tawa1 last updated on 12/Oct/18

The question was asked online. But sir, when can i apply the formular you used to find the sum.

$$\mathrm{The}\:\mathrm{question}\:\mathrm{was}\:\mathrm{asked}\:\mathrm{online}.\:\: \\ $$$$\mathrm{But}\:\mathrm{sir},\:\mathrm{when}\:\mathrm{can}\:\mathrm{i}\:\mathrm{apply}\:\mathrm{the}\:\mathrm{formular}\:\mathrm{you}\:\mathrm{used}\:\mathrm{to}\:\mathrm{find}\:\mathrm{the}\:\mathrm{sum}. \\ $$