Sum of an infinite series

Calculus Level 4

1 3 × 0 ! + 1 4 × 1 ! + 1 5 × 2 ! + 1 6 × 3 ! + 1 7 × 4 ! + ⋯ = ? \frac 1{3\times{0!}}+\frac 1{4\times{1!}}+\frac 1{5\times{2!}}+\frac 1{6\times{3!}}+\frac 1{7\times{4!}}+\cdots=?

Report your answer to 5 places after decimal


The answer is 0.71828.

Mrigank Shekhar Pathak by Mrigank Shekhar Pathak , 22, India

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1 solution

Consider the following Maclaurin series :

∑ k = 0 ∞ x k + 2 k ! = x 2 e x Integrate both sides w.r.t x ∑ k = 0 ∞ x k + 3 ( k + 3 ) k ! = x 2 e x − 2 x e x + 2 e x + C where C is the constant of integration. 0 = 2 + C Putting x = 0 ⟹ C = − 2 ∑ k = 0 ∞ 1 ( k + 3 ) k ! = e − 2 e + 2 e − 2 Putting x = 1 = e − 2 ≈ 0.71828 \begin{aligned} \sum_{k=0}^\infty \frac {x^{k+2}}{k!} & = x^2e^x & \small \color{#3D99F6} \text{Integrate both sides w.r.t }x \\ \sum_{k=0}^\infty \frac {x^{k+3}}{(k+3)k!} & = x^2e^x - 2xe^x + 2e^x + \color{#3D99F6} C & \small \color{#3D99F6} \text{where }C \text{ is the constant of integration.} \\ 0 & = 2 + C & \small \color{#3D99F6} \text{Putting }x = 0 \implies C = -2 \\ \sum_{k=0}^\infty \frac 1{(k+3)k!} & = e - 2e + 2e - 2 & \small \color{#3D99F6} \text{Putting }x = 1 \\ & = e - 2 \approx \boxed{0.71828} \end{aligned}

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