Hyberbolics

Calculus Level 5

n = 1 2 ( n + 1 ) ( 2 ( n 1 ) ) ! \large{\sum _{ n=1 }^{ \infty }{ \frac { 2( n+1 ) }{ \left( 2( n-1 ) \right) ! } } }

If above sum can be expressed as sinh ( A ) + B cosh ( C ) \sinh { \left( A \right) } +B\cosh { \left( C \right) } where A , B , C A,B,C are integers, find the value of A + B + C A+B+C .


The answer is 6.

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1 solution

Aditya Malusare
Dec 6, 2015

The series expansions for sinh ( x ) \sinh(x) and cosh ( x ) \cosh(x) are sinh ( x ) = k = 1 x 2 k 1 ( 2 k 1 ) ! \sinh(x) = \sum_{k=1}^{\infty}\dfrac{x^{2k-1}}{(2k-1)!} cosh ( x ) = k = 0 x 2 k ( 2 k ) ! \cosh(x) = \sum_{k=0}^{\infty}\dfrac{x^{2k}}{(2k)!}

The given sum can be written as n = 1 2 ( n + 1 ) ( 2 ( n 1 ) ) ! = n = 1 2 ( n 1 ) + 4 ( 2 ( n 1 ) ) ! = n = 2 1 ( 2 ( n 1 ) 1 ) ! + n = 1 4 ( 2 ( n 1 ) ) ! \sum_{n=1}^{\infty}\dfrac{2(n + 1)}{(2(n-1))!} = \sum_{n=1}^{\infty}\dfrac{2(n - 1) + 4}{(2(n-1))!} = \sum_{n=2}^{\infty}\dfrac{1}{(2(n-1) - 1)!} + \sum_{n=1}^{\infty} \dfrac{4}{(2(n-1))!}

Adjusting the limits of both sums, they become n = 1 1 ( 2 n 1 ) ! + 4 n = 0 1 ( 2 n ) ! \sum_{n=1}^{\infty}\dfrac{1}{(2n - 1)!} + 4\sum_{n=0}^{\infty} \dfrac{1}{(2n)!} which on comparison with the expansions, is equivalent to the expression sinh ( 1 ) + 4 cosh ( 1 ) \sinh(1) + 4\cosh(1) giving A + B + C = 6 A + B + C = \boxed{6}

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