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Question Number 214860 by hardmath last updated on 21/Dec/24
Find: 1+ (2^3 /(3!)) + (3^3 /(3!)) + ... + (n^3 /(n!)) = ?
$$\mathrm{Find}: \\ $$$$\mathrm{1}+\:\:\frac{\mathrm{2}^{\mathrm{3}} }{\mathrm{3}!}\:\:+\:\:\frac{\mathrm{3}^{\mathrm{3}} }{\mathrm{3}!}\:\:+\:\:…\:\:+\:\:\frac{\mathrm{n}^{\mathrm{3}} }{\mathrm{n}!}\:\:=\:\:? \\ $$
Commented by mr W last updated on 22/Dec/24
do you want 1+ (2^3 /(2!)) + (3^3 /(3!)) + ... + (n^3 /(n!)) +... = ? or really 1+ (2^3 /(2!)) + (3^3 /(3!)) + ... + (n^3 /(n!)) = ? if you meant the first, then 1+ (2^3 /(2!)) + (3^3 /(3!)) + ... + (n^3 /(n!)) +... = 5e
$${do}\:{you}\:{want} \\ $$$$\mathrm{1}+\:\:\frac{\mathrm{2}^{\mathrm{3}} }{\mathrm{2}!}\:\:+\:\:\frac{\mathrm{3}^{\mathrm{3}} }{\mathrm{3}!}\:\:+\:\:…\:\:+\:\:\frac{\mathrm{n}^{\mathrm{3}} }{\mathrm{n}!}\:+…\:=\:\:? \\ $$$${or}\:{really} \\ $$$$\mathrm{1}+\:\:\frac{\mathrm{2}^{\mathrm{3}} }{\mathrm{2}!}\:\:+\:\:\frac{\mathrm{3}^{\mathrm{3}} }{\mathrm{3}!}\:\:+\:\:…\:\:+\:\:\frac{\mathrm{n}^{\mathrm{3}} }{\mathrm{n}!}\:=\:\:? \\ $$$${if}\:{you}\:{meant}\:{the}\:{first},\:{then} \\ $$$$\mathrm{1}+\:\:\frac{\mathrm{2}^{\mathrm{3}} }{\mathrm{2}!}\:\:+\:\:\frac{\mathrm{3}^{\mathrm{3}} }{\mathrm{3}!}\:\:+\:\:…\:\:+\:\:\frac{\mathrm{n}^{\mathrm{3}} }{\mathrm{n}!}\:+…\:=\:\mathrm{5}{e} \\ $$
Answered by MrGaster last updated on 22/Dec/24
Σ_(k=1) ^n (k^3 /(k!)) =Σ_(k=1) ^n ((k∙k^2 )/(k∙(k−1)!)) =Σ_(k−1) ^n (k^2 /((k−1))) =Σ_(k=1) ^n (((k−1+1)/((k−1)!))∙k) =Σ_(k=1) ^n ((((k−1)∙k)/((k−1)!))+(k/((k−1)!))) =Σ_(k=1) ^n ((k/((k−2)!))+(k/((k−1)!))) =Σ_(k=1) ^n (((k−2+2)/((k−2)!))+((k−1+1)/((k−1)!))) =Σ_(k=1) ^n (((k−2)/((k−2)))+(2/((k−2)))+((k−1)/((k−1)!))+(1/((k−1)!))) =Σ_(k=1) ^n ((1/((k−3)))+(3/((k−2)!))+(1/((k−1)!))) Use the index carefully now because thea fctorial in the denominator cannot bea negtive. Therefore for items whosei factoral is undefined the sum should bed ajusted: =(1/((−2)!))+(3/((−1)!))+(1/0)_(undefined terms removed) +Σ_(k−4) ^n ((1/((k−3)!))+(3/((k−2)!))+(1/((k−1)!)))+((1^3 /(1!))+(2^3 /(2!))+(3^3 /(3!)))_(fist therms handled separately) The undefined terms are not included int he sum. The first three terms aret calculaed directly: =(1^3 /(1!))+(2^3 /(2!))+(3^3 /(3!))+Σ_(k=4) ^n ((1/((k−3)!))+(3/((k−2)!))+(1/((k−1)!))) =1+(8/2)+((27)/6)+Σ_(k=4) ^n ((1/((k−3)!))+(3/((k−2)!))+(1/((k−1)!))) =((15)/2)+Σ_(k=4) ^n ((1/((k−3)!))+(3/((k−2)!))+(1/((k−1)!))) Now you can simplify the sum by movingthe index: =((15)/2)+Σ_(j=1) ^(n−3) ((1/(j!))+(3/((j+1)!))+(1/((j+2)!))) siceΣ_(j=0) ^∞ (1/(j!))=e,with the incredase of n Until the sum of the seies of(n−3) will approach e,Therefore missing items and coefficientsm ust be considered: =((15)/2)+(e−(1+(1/(1!))+(1/(2!))))+3(e−(1+(1/(1!))))+e−1 =((15)/2)+5e−(1+1+(1/2))+3e −(3+3)+e−1 =((15)/2)+5e−((5/2)+7) =((15)/2)+5e−((19)/2) =5e−2
$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{k}^{\mathrm{3}} }{{k}!} \\ $$$$=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{k}\centerdot{k}^{\mathrm{2}} }{{k}\centerdot\left({k}−\mathrm{1}\right)!} \\ $$$$=\underset{{k}−\mathrm{1}} {\overset{{n}} {\sum}}\frac{{k}^{\mathrm{2}} }{\left({k}−\mathrm{1}\right)} \\ $$$$=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\frac{{k}−\mathrm{1}+\mathrm{1}}{\left({k}−\mathrm{1}\right)!}\centerdot{k}\right) \\ $$$$=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\frac{\left({k}−\mathrm{1}\right)\centerdot{k}}{\left({k}−\mathrm{1}\right)!}+\frac{{k}}{\left({k}−\mathrm{1}\right)!}\right) \\ $$$$=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\frac{{k}}{\left({k}−\mathrm{2}\right)!}+\frac{{k}}{\left({k}−\mathrm{1}\right)!}\right) \\ $$$$=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\frac{{k}−\mathrm{2}+\mathrm{2}}{\left({k}−\mathrm{2}\right)!}+\frac{{k}−\mathrm{1}+\mathrm{1}}{\left({k}−\mathrm{1}\right)!}\right) \\ $$$$=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\frac{{k}−\mathrm{2}}{\left({k}−\mathrm{2}\right)}+\frac{\mathrm{2}}{\left({k}−\mathrm{2}\right)}+\frac{{k}−\mathrm{1}}{\left({k}−\mathrm{1}\right)!}+\frac{\mathrm{1}}{\left({k}−\mathrm{1}\right)!}\right) \\ $$$$=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\frac{\mathrm{1}}{\left({k}−\mathrm{3}\right)}+\frac{\mathrm{3}}{\left({k}−\mathrm{2}\right)!}+\frac{\mathrm{1}}{\left({k}−\mathrm{1}\right)!}\right) \\ $$$$\mathrm{Use}\:\mathrm{the}\:\mathrm{index}\:\mathrm{carefully}\:\mathrm{now}\:\mathrm{because}\:\mathrm{thea} \\ $$$$\mathrm{fctorial}\:\mathrm{in}\:\mathrm{the}\:\mathrm{denominator}\:\mathrm{cannot}\:\mathrm{bea} \\ $$$$\mathrm{negtive}.\:\mathrm{Therefore}\:\mathrm{for}\:\mathrm{items}\:\mathrm{whosei} \\ $$$$\mathrm{factoral}\:\mathrm{is}\:\mathrm{undefined}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{should}\:\mathrm{bed} \\ $$$$\mathrm{ajusted}: \\ $$$$=\underset{\mathrm{undefined}\:\mathrm{terms}\:\mathrm{removed}} {\underbrace{\frac{\mathrm{1}}{\left(−\mathrm{2}\right)!}+\frac{\mathrm{3}}{\left(−\mathrm{1}\right)!}+\frac{\mathrm{1}}{\mathrm{0}}}}\:+\underset{{k}−\mathrm{4}} {\overset{{n}} {\sum}}\left(\frac{\mathrm{1}}{\left({k}−\mathrm{3}\right)!}+\frac{\mathrm{3}}{\left({k}−\mathrm{2}\right)!}+\frac{\mathrm{1}}{\left({k}−\mathrm{1}\right)!}\right)+\underset{\mathrm{fist}\:\mathrm{therms}\:\mathrm{handled}\:\mathrm{separately}} {\underbrace{\left(\frac{\mathrm{1}^{\mathrm{3}} }{\mathrm{1}!}+\frac{\mathrm{2}^{\mathrm{3}} }{\mathrm{2}!}+\frac{\mathrm{3}^{\mathrm{3}} }{\mathrm{3}!}\right)}} \\ $$$$\mathrm{The}\:\mathrm{undefined}\:\mathrm{terms}\:\mathrm{are}\:\mathrm{not}\:\mathrm{included}\:\mathrm{int} \\ $$$$\mathrm{he}\:\mathrm{sum}.\:\mathrm{The}\:\mathrm{first}\:\mathrm{three}\:\mathrm{terms}\:\mathrm{aret} \\ $$$$\mathrm{calculaed}\:\mathrm{directly}: \\ $$$$=\frac{\mathrm{1}^{\mathrm{3}} }{\mathrm{1}!}+\frac{\mathrm{2}^{\mathrm{3}} }{\mathrm{2}!}+\frac{\mathrm{3}^{\mathrm{3}} }{\mathrm{3}!}+\underset{{k}=\mathrm{4}} {\overset{{n}} {\sum}}\left(\frac{\mathrm{1}}{\left({k}−\mathrm{3}\right)!}+\frac{\mathrm{3}}{\left({k}−\mathrm{2}\right)!}+\frac{\mathrm{1}}{\left({k}−\mathrm{1}\right)!}\right) \\ $$$$=\mathrm{1}+\frac{\mathrm{8}}{\mathrm{2}}+\frac{\mathrm{27}}{\mathrm{6}}+\underset{{k}=\mathrm{4}} {\overset{{n}} {\sum}}\left(\frac{\mathrm{1}}{\left({k}−\mathrm{3}\right)!}+\frac{\mathrm{3}}{\left({k}−\mathrm{2}\right)!}+\frac{\mathrm{1}}{\left({k}−\mathrm{1}\right)!}\right) \\ $$$$=\frac{\mathrm{15}}{\mathrm{2}}+\underset{{k}=\mathrm{4}} {\overset{{n}} {\sum}}\left(\frac{\mathrm{1}}{\left({k}−\mathrm{3}\right)!}+\frac{\mathrm{3}}{\left({k}−\mathrm{2}\right)!}+\frac{\mathrm{1}}{\left({k}−\mathrm{1}\right)!}\right) \\ $$$$\mathrm{Now}\:\mathrm{you}\:\mathrm{can}\:\mathrm{simplify}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{by}\: \\ $$$$\mathrm{movingthe}\:\mathrm{index}: \\ $$$$=\frac{\mathrm{15}}{\mathrm{2}}+\underset{{j}=\mathrm{1}} {\overset{{n}−\mathrm{3}} {\sum}}\left(\frac{\mathrm{1}}{{j}!}+\frac{\mathrm{3}}{\left({j}+\mathrm{1}\right)!}+\frac{\mathrm{1}}{\left({j}+\mathrm{2}\right)!}\right) \\ $$$$\mathrm{sice}\underset{{j}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{j}!}={e},\mathrm{with}\:\mathrm{the}\:\mathrm{incredase}\:\mathrm{of}\:{n}\:\mathrm{Until}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{seies}\:\mathrm{of}\left({n}−\mathrm{3}\right) \\ $$$$\mathrm{will}\:\mathrm{approach}\:{e},\mathrm{Therefore}\:\mathrm{missing}\:\mathrm{items}\:\mathrm{and}\:\mathrm{coefficientsm} \\ $$$$\mathrm{ust}\:\mathrm{be}\:\mathrm{considered}: \\ $$$$=\frac{\mathrm{15}}{\mathrm{2}}+\left({e}−\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}!}+\frac{\mathrm{1}}{\mathrm{2}!}\right)\right)+\mathrm{3}\left({e}−\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}!}\right)\right)+{e}−\mathrm{1} \\ $$$$=\frac{\mathrm{15}}{\mathrm{2}}+\mathrm{5}{e}−\left(\mathrm{1}+\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\right)+\mathrm{3e}\:−\left(\mathrm{3}+\mathrm{3}\right)+\mathrm{e}−\mathrm{1} \\ $$$$=\frac{\mathrm{15}}{\mathrm{2}}+\mathrm{5e}−\left(\frac{\mathrm{5}}{\mathrm{2}}+\mathrm{7}\right) \\ $$$$=\frac{\mathrm{15}}{\mathrm{2}}+\mathrm{5}{e}−\frac{\mathrm{19}}{\mathrm{2}} \\ $$$$=\mathrm{5}{e}−\mathrm{2} \\ $$
Answered by mr W last updated on 23/Dec/24
e^x =Σ_(n=0) ^∞ (x^n /(n!)) (e^x )′=e^x =Σ_(n=0) ^∞ ((nx^(n−1) )/(n!)) xe^x =Σ_(n=0) ^∞ ((nx^n )/(n!)) (xe^x )′=(1+x)e^x =Σ_(n=0) ^∞ ((n^2 x^(n−1) )/(n!)) (x+x^2 )e^x =Σ_(n=0) ^∞ ((n^2 x^n )/(n!)) (1+3x+x^2 )e^x =Σ_(n=0) ^∞ ((n^3 x^(n−1) )/(n!)) x(1+3x+x^2 )e^x =Σ_(n=0) ^∞ ((n^3 x^n )/(n!)) ⇒Σ_(n=1) ^∞ ((n^3 x^n )/(n!))=x(1+3x+x^2 )e^x with x=1 we get: Σ_(n=1) ^∞ (n^3 /(n!))=(1+3+1)e=5e other examples: with x=2 we get Σ_(n=1) ^∞ ((2^n n^3 )/(n!))=2(1+3×2+2^2 )e^2 =22e^2 with x=(1/3) we get Σ_(n=1) ^∞ (n^3 /(3^n n!))=(1/3)(1+3×(1/3)+(1/3^2 ))e^(1/3) =((19e^(1/3) )/(27))
$${e}^{{x}} =\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} }{{n}!} \\ $$$$\left({e}^{{x}} \right)'={e}^{{x}} =\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{nx}^{{n}−\mathrm{1}} }{{n}!} \\ $$$${xe}^{{x}} =\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{nx}^{{n}} }{{n}!} \\ $$$$\left({xe}^{{x}} \right)'=\left(\mathrm{1}+{x}\right){e}^{{x}} =\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{n}^{\mathrm{2}} {x}^{{n}−\mathrm{1}} }{{n}!} \\ $$$$\left({x}+{x}^{\mathrm{2}} \right){e}^{{x}} =\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{n}^{\mathrm{2}} {x}^{{n}} }{{n}!} \\ $$$$\left(\mathrm{1}+\mathrm{3}{x}+{x}^{\mathrm{2}} \right){e}^{{x}} =\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{n}^{\mathrm{3}} {x}^{{n}−\mathrm{1}} }{{n}!} \\ $$$${x}\left(\mathrm{1}+\mathrm{3}{x}+{x}^{\mathrm{2}} \right){e}^{{x}} =\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{n}^{\mathrm{3}} {x}^{{n}} }{{n}!} \\ $$$$\Rightarrow\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}^{\mathrm{3}} {x}^{{n}} }{{n}!}={x}\left(\mathrm{1}+\mathrm{3}{x}+{x}^{\mathrm{2}} \right){e}^{{x}} \\ $$$${with}\:{x}=\mathrm{1}\:{we}\:{get}: \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}^{\mathrm{3}} }{{n}!}=\left(\mathrm{1}+\mathrm{3}+\mathrm{1}\right){e}=\mathrm{5}{e} \\ $$$${other}\:{examples}: \\ $$$${with}\:{x}=\mathrm{2}\:{we}\:{get} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{2}^{{n}} {n}^{\mathrm{3}} }{{n}!}=\mathrm{2}\left(\mathrm{1}+\mathrm{3}×\mathrm{2}+\mathrm{2}^{\mathrm{2}} \right){e}^{\mathrm{2}} =\mathrm{22}{e}^{\mathrm{2}} \\ $$$${with}\:{x}=\frac{\mathrm{1}}{\mathrm{3}}\:{we}\:{get} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}^{\mathrm{3}} }{\mathrm{3}^{{n}} {n}!}=\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{1}+\mathrm{3}×\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }\right){e}^{\frac{\mathrm{1}}{\mathrm{3}}} =\frac{\mathrm{19}{e}^{\frac{\mathrm{1}}{\mathrm{3}}} }{\mathrm{27}} \\ $$
Answered by MathematicalUser2357 last updated on 25/Dec/24
You had a typo in (2^3 /(3!)). Fixed into (2^3 /(2!)). 1+(2^3 /(2!))+(3^3 /(3!))+...+(n^3 /(n!))=? ?=Σ_(k=1) ^n ((k^3 /(k!))) I didn′t understand the summation rules.
$$\mathrm{You}\:\mathrm{had}\:\mathrm{a}\:\mathrm{typo}\:\mathrm{in}\:\frac{\mathrm{2}^{\mathrm{3}} }{\mathrm{3}!}.\:\mathrm{Fixed}\:\mathrm{into}\:\frac{\mathrm{2}^{\mathrm{3}} }{\mathrm{2}!}. \\ $$$$\mathrm{1}+\frac{\mathrm{2}^{\mathrm{3}} }{\mathrm{2}!}+\frac{\mathrm{3}^{\mathrm{3}} }{\mathrm{3}!}+…+\frac{{n}^{\mathrm{3}} }{{n}!}=? \\ $$$$?=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\frac{{k}^{\mathrm{3}} }{{k}!}\right) \\ $$$$\mathrm{I}\:\mathrm{didn}'\mathrm{t}\:\mathrm{understand}\:\mathrm{the}\:\mathrm{summation}\:\mathrm{rules}. \\ $$

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