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Question Number 194183 by tri26112004 last updated on 29/Jun/23

Σ_(n=1) ^∞ (1/(n(n+15)(n+30)))

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{n}\left({n}+\mathrm{15}\right)\left({n}+\mathrm{30}\right)} \\ $$

Answered by ARUNG_Brandon_MBU last updated on 29/Jun/23

(1/(n(n+15)(n+30)))=(1/(450n))−(1/(225(n+15)))+(1/(450(n+30))) S=(1/(450))Σ_(n=0) ^∞ (1/((n+1)))−(1/(225))Σ_(n=0) ^∞ (1/((n+16)))+(1/(450))Σ_(n=0) ^∞ (1/((n+31))) =−(1/(450))ψ(1)+(1/(225))ψ(16)−(1/(450))ψ(31) =−(1/(450))ψ(1)+(1/(225))ψ(16)−(1/(450))((1/(30))+(1/(29))+∙∙∙+(1/(17))+(1/(16))+ψ(16)) =−(1/(450))ψ(1)+(1/(450))ψ(16)−(1/(450))Σ_(n=16) ^(30) (1/n) =−(1/(450))ψ(1)+(1/(450))((1/(15))+(1/(14))+∙∙∙+(1/2)+1+ψ(1))−(1/(450))Σ_(n=16) ^(30) (1/n) =(1/(450))Σ_(n=1) ^(15) (1/n)−(1/(450))Σ_(n=16) ^(30) (1/n)

$$\frac{\mathrm{1}}{{n}\left({n}+\mathrm{15}\right)\left({n}+\mathrm{30}\right)}=\frac{\mathrm{1}}{\mathrm{450}{n}}−\frac{\mathrm{1}}{\mathrm{225}\left({n}+\mathrm{15}\right)}+\frac{\mathrm{1}}{\mathrm{450}\left({n}+\mathrm{30}\right)} \\ $$$${S}=\frac{\mathrm{1}}{\mathrm{450}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)}−\frac{\mathrm{1}}{\mathrm{225}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}+\mathrm{16}\right)}+\frac{\mathrm{1}}{\mathrm{450}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}+\mathrm{31}\right)} \\ $$$$\:\:\:=−\frac{\mathrm{1}}{\mathrm{450}}\psi\left(\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{225}}\psi\left(\mathrm{16}\right)−\frac{\mathrm{1}}{\mathrm{450}}\psi\left(\mathrm{31}\right) \\ $$$$\:\:\:=−\frac{\mathrm{1}}{\mathrm{450}}\psi\left(\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{225}}\psi\left(\mathrm{16}\right)−\frac{\mathrm{1}}{\mathrm{450}}\left(\frac{\mathrm{1}}{\mathrm{30}}+\frac{\mathrm{1}}{\mathrm{29}}+\centerdot\centerdot\centerdot+\frac{\mathrm{1}}{\mathrm{17}}+\frac{\mathrm{1}}{\mathrm{16}}+\psi\left(\mathrm{16}\right)\right) \\ $$$$\:\:\:=−\frac{\mathrm{1}}{\mathrm{450}}\psi\left(\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{450}}\psi\left(\mathrm{16}\right)−\frac{\mathrm{1}}{\mathrm{450}}\underset{{n}=\mathrm{16}} {\overset{\mathrm{30}} {\sum}}\frac{\mathrm{1}}{{n}} \\ $$$$\:\:\:=−\frac{\mathrm{1}}{\mathrm{450}}\psi\left(\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{450}}\left(\frac{\mathrm{1}}{\mathrm{15}}+\frac{\mathrm{1}}{\mathrm{14}}+\centerdot\centerdot\centerdot+\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{1}+\psi\left(\mathrm{1}\right)\right)−\frac{\mathrm{1}}{\mathrm{450}}\underset{{n}=\mathrm{16}} {\overset{\mathrm{30}} {\sum}}\frac{\mathrm{1}}{{n}} \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{450}}\underset{{n}=\mathrm{1}} {\overset{\mathrm{15}} {\sum}}\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{\mathrm{450}}\underset{{n}=\mathrm{16}} {\overset{\mathrm{30}} {\sum}}\frac{\mathrm{1}}{{n}} \\ $$

Commented by mr W last updated on 11/Aug/23

can you check Q195672 with a C^(++) program sir? thanks!

$${can}\:{you}\:{check}\:{Q}\mathrm{195672}\:{with}\:{a}\:{C}^{++} \\ $$$${program}\:{sir}?\:{thanks}! \\ $$

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