A calculus problem by Parnab Ghosh

Calculus Level 3

1 3 ⋅ 1 ! + 1 4 ⋅ 2 ! + 1 5 ⋅ 3 ! + ⋯ + 1 ( n + 2 ) n ! + ⋯ = ? \dfrac1{3 \cdot 1!} + \dfrac1{4\cdot2!} + \dfrac1{5 \cdot 3!} + \cdots + \dfrac 1{(n+2) n!} + \cdots = \, ?


The answer is 0.5.

Parnab Ghosh by Parnab Ghosh , 20, India

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1 solution

Rishabh Jain
Jan 27, 2017

Relevant wiki: Telescoping Series - Sum

Given sum is:

∑ n = 1 ∞ 1 ( n + 2 ) ⋅ n ! = ∑ n = 1 ∞ n + 1 ( n + 2 ) ! = ∑ n = 1 ∞ ( n + 2 ) − 1 ( n + 2 ) ! = ∑ n = 1 ∞ { ( n + 2 ) ( n + 2 ) ! − 1 ( n + 2 ) ! } = ∑ n = 1 ∞ { 1 ( n + 1 ) ! − 1 ( n + 2 ) ! } ( T e l e s c o p i c S e r i e s ) = 1 2 ! = 0.5 \begin{aligned}\sum_{n=1}^{\infty}\dfrac{1}{(n+2)\cdot n!}=&\sum_{n=1}^{\infty}\dfrac{n+1}{ (n+2)!}\\=&\sum_{n=1}^{\infty}\dfrac{(n+2)-1}{(n+2)!}\\=&\sum_{n=1}^{\infty}\left\{\dfrac{(n+2)}{(n+2)!}-\dfrac{1}{(n+2)!}\right\} \\=&\sum_{n=1}^{\infty}\left\{\dfrac{1}{(n+1)!}-\dfrac{1}{(n+2)!}\right\}\\& (\small{\color{cyan}{\mathcal{Telescopic~Series}}})\\\large =&\dfrac{1}{2!}=\boxed{\color{#D61F06}{0.5}}\end{aligned}

6 Helpful 0 Interesting 0 Brilliant 0 Confused

Great solution! I am just learning series and your solution has given me another way to solve thanks!

Sravanth C. - 4 years, 4 months ago

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Thanks... :-)

Rishabh Jain - 4 years, 4 months ago

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