Evaluate these series

Question

Please help me to evaluate these series. I am working on my physics project and I have encountered to these series. I know that $\sum\limits_{n = 0}^\infty {\frac{{{{(\alpha )}^{2n}}}}{{(2n)!}}}$ is $\cos$ function, but I have problem with $n$. How do i evaluate these series? $$\sum\limits_{n = 0}^\infty {n\frac{{{{(\alpha )}^{2n}}}}{{(2n)!}}} $$ and $$\sum\limits_{n = 1}^\infty {n\frac{{{{(\alpha )}^{2n - 1}}}}{{(2n - 1)!}}}$$

Answer 1

Hint. Note that for $n\geq 1$, $$ {n\frac{{{{(\alpha )}^{2n}}}}{{(2n)!}}}=\frac{\alpha}{2} \cdot{\frac{{{{(\alpha )}^{2n-1}}}}{{(2n-1)!}}}\quad \text{and}\quad {n\frac{{{{(\alpha )}^{2n - 1}}}}{{(2n - 1)!}}}=\frac{d}{d\alpha}\left(\frac{\alpha}{2} \cdot{\frac{{{{(\alpha )}^{2n-1}}}}{{(2n - 1)!}}}\right).$$ Now recall the expansions of the main hyperbolic functions.

Answer 2

Since $$ \sinh x = \sum\limits_{n = 1}^{ + \infty } {\frac{{x^{2n - 1} }} {{\left( {2n - 1} \right)!}}} $$ you have that $$ x\sinh x = \sum\limits_{n = 1}^{ + \infty } {\frac{{x^{2n} }} {{\left( {2n - 1} \right)!}}} = \sum\limits_{n = 1}^{ + \infty } {2n\frac{{x^{2n} }} {{2n\left( {2n - 1} \right)!}}} = 2\sum\limits_{n = 1}^{ + \infty } {n\frac{{x^{2n} }} {{\left( {2n} \right)!}}} $$ thus $$ \sum\limits_{n = 1}^{ + \infty } {n\frac{{x^{2n} }} {{\left( {2n} \right)!}}} = \frac{1} {2}x\sinh x $$ By derivation you get the other series