# Calculate (i.e. express without using infinite sum): $\frac{1}{1!\cdot1}+\frac{2}{2!\cdot2}+\frac{4}{4!\cdot4}+\frac{8}{6!\cdot8}+..$

Calculate (i.e. express without using infinite sum):

$$\frac{1}{1!\cdot1}+\frac{2}{2!\cdot2}+\frac{4}{4!\cdot4}+\frac{8}{6!\cdot8}+..$$

In sum it would be:

$$\sum_{n=0}^{\infty}\frac{2^{n}}{(2n)!\cdot2^{n}} = \sum_{n=0}^{\infty}\frac{1}{(2n)!}$$

But now I somehow need to get rid off the sum symbol because the task clearly asks for an expression without it :P

But how can I remove the sum symbol? Two things come to my mind: Derivation and taylor-formula. But how to use derivation here if we got factorial... How use taylor if there isn't a point given (that one where you analyze at), and also it requires derivative as well... ^^

Please if you answer me please explain it simple I'm about frustrating because I don't understand this for several days and I'm absolutely sure it will be asked in the exam too.

This is $\cosh 1 = \frac{e^1+e^{-1}}{2} = \frac{e^2+1}{2e}$.
• You need to know $e^x$ at a minimum. You can then spot this is the even-power terms at $x=1$. To delete the odd-power terms of any series $f\left( x\right)$, just use $\frac{f\left( x\right)+f\left( -x\right)}{2}$.(Similarly, $\frac{f\left( x\right)+f\left( -x\right)}{2}$ deletes the even-power terms.)
Hint: Your series looks a lot like $$\cosh(x)=1+\dfrac{x^2}{2!}+\dfrac{x^4}{4!}+\cdots$$
• I will ask the same question; Is there no way to get to that solution by doing calculation on paper? For me it sounds like I need to keep some / several well-known series in my mind in order to solve this task. Let's say I don't know that the thing in the task is similar to $\cosh$, what can I do? Sep 19, 2016 at 10:27
• I don't see a way to solve this if you don't know that $e^x=\sum_{k\geq 0}\dfrac{x^k}{k!}$ for instance. And yes you should remember the main Taylor series: exp, sin, cos, $(1+x)^{\alpha}$ and $\ln(1+x)$. Sep 19, 2016 at 10:30