سوال ۳۳ کنکور ارشد علوم کامپیوتر ۱۳۹۹

We use the Maclaurin series of the hyperbolic functions evaluated at $x=1$:

$\sinh 1 = \sum_{n=0}^{\infty} \frac{1}{(2n+1)!} = \frac{e - e^{-1}}{2}, \qquad \cosh 1 = \sum_{n=0}^{\infty} \frac{1}{(2n)!} = \frac{e + e^{-1}}{2}.$

Split the numerator. Write $n = \tfrac{1}{2}\big[(2n+1) - 1\big]$ so the term becomes

$\frac{n}{(2n+1)!} = \frac{1}{2}\cdot\frac{2n+1}{(2n+1)!} - \frac{1}{2}\cdot\frac{1}{(2n+1)!} = \frac{1}{2}\cdot\frac{1}{(2n)!} - \frac{1}{2}\cdot\frac{1}{(2n+1)!}.$

Sum each piece.

$\sum_{n=0}^{\infty} \frac{n}{(2n+1)!} = \frac{1}{2}\sum_{n=0}^{\infty}\frac{1}{(2n)!} - \frac{1}{2}\sum_{n=0}^{\infty}\frac{1}{(2n+1)!} = \frac{1}{2}\cosh 1 - \frac{1}{2}\sinh 1.$

Evaluate.

$\frac{1}{2}\big(\cosh 1 - \sinh 1\big) = \frac{1}{2}\cdot e^{-1} = \frac{1}{2e}.$

So the series equals $\dfrac{1}{2e}$, which is option 2.

Exam trick: When a factorial series has a polynomial in the numerator, rewrite that polynomial in terms of the factorial's argument (here $2n+1$) so each piece collapses to a known series like $\sum \tfrac{1}{(2n)!}=\cosh 1$ or $\sum \tfrac{1}{(2n+1)!}=\sinh 1$. Note $\cosh 1 - \sinh 1 = e^{-1}$ directly.