Series are awesome!

Calculus Level 3

$1 + \dfrac 2{3!} + \dfrac3{5!} + \dfrac4{7!} + \cdots = \dfrac en$

Find the value of $n$ .

Notation: $e \approx 2.718$ denotes the Euler's number .

The answer is 2.

1 solution

Brian Moehring
Mar 6, 2017

Note that $\sinh(x) = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \frac{x^7}{7!} + \cdots$ so that $\sqrt{x}\sinh(\sqrt{x}) = x + \frac{x^2}{3!} + \frac{x^3}{5!} + \frac{x^4}{7!} + \cdots$

Taking the derivative of both sides, $\frac{\sinh(\sqrt{x}) + \sqrt{x}\cosh(\sqrt{x})}{2\sqrt{x}} = 1 + \frac{2x}{3!} + \frac{3x^2}{5!} + \frac{4x^3}{7!} + \cdots$ and setting $x=1$ , $1 + \frac{2}{3!} + \frac{3}{5!} + \frac{4}{7!} + \cdots = \frac{\sinh(1) + \cosh(1)}{2} = \frac{e}{2}.$

Therefore, the answer is $n=\boxed{2}$ .

Isn't there a solution involving ordinary algebra without the use of hyperbolic trig.functions? And thanks for posting solution sir.

A Former Brilliant Member - 4 years, 3 months ago

Yes there is intact a simple way each term can be written as t(n)=1/2 $1/(2n-1)!+1/(2n-2)!$ .So summing up gives $S=1/2(1/1!+1/2!+1/3!+.....)=e/2$

Spandan Senapati - 4 years, 3 months ago

Yeah.....that did not just cross my mind.BTW thanks!!!

A Former Brilliant Member - 4 years, 3 months ago

Series are awesome!

The answer is 2.

1 solution

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