State whether the following series converges or diverges $\sum\limits_{n=0}^\infty{7^n - 2^n\over(2n)!}$

Question

$\sum\limits_{n=0}^\infty{7^n - 2^n\over(2n)!}$

Thanks in advance for any help you are able to provide.

Answer 1

by using the ratio test for $\frac{a^n}{(2n)!}$ $$\lim_{n\rightarrow \infty }\frac{a^{n+1}}{(2n+2)!}\frac{(2n)!}{a^{n}}=\lim_{n\rightarrow \infty }\frac{a(2n)!}{(2n+2)!}=\lim_{n\rightarrow \infty }\frac{a}{(2n+2)(2n+1)}=0<1$$ so the series converges

Answer 2

$$\dfrac{e^x+e^{-x}}2=\sum_{r=0}^\infty\dfrac{x^{2r}}{(2r)!}$$

Now $\dfrac{7^n}{(2n)!}=\dfrac{(\sqrt7)^{2n}}{(2n)!}$

Answer 3

This converges. It is a useful fact that $$ e^x = \sum_{k = 0}^\infty \frac{x^k}{k!} $$ for every $x \in \mathbb{C}$, and we can see immediately that your series can be split into two converging parts: $$ \sum_{n = 0}^\infty \frac{7^n}{(2n)!} - \sum_{k = 0}^\infty \frac{2^n}{(2n)!} $$

where the terms converge since $$ \sum_{k = 0}^\infty \frac{x^k}{(2k)!} < \sum_{k = 0}^\infty \frac{x^k}{k!} $$

Also, one can write an analytic expression for the sum by using $$ \sum_{k = 0}^\infty \frac{x^{2k}}{(2k)!} = \frac{e^x - e^{-x}}{2}$$ Which can be checked by inspecting the Taylor series for $e^x$. Now substitute $x = \sqrt{y}$ to get $$ \sum_{k = 0}^\infty \frac{y^k}{(2k)!} = \frac{e^{\sqrt{y}} - e^{-\sqrt{y}}}{2}$$

so

$$ \sum_{n = 0}^\infty \frac{7^n - 2^n}{(2n)!} = \frac{e^{\sqrt{7}} - e^{-\sqrt{7}} + e^{\sqrt{2}} - e^{-\sqrt{2}}}{2}$$