Prove that : 11! + 13! + 15! + ... 12! + 14! + 16! + ... = e+1e 1 [6]

Proof that $\frac{\frac{1}{1!}+\frac{1}{3!}+\frac{1}{5!}+\dots }{\frac{1}{2!}+\frac{1}{4!}+\frac{1}{6!}+\dots }=\frac{e+1}{e-1}$

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Step 1: Define the Series

We are given two infinite series:

• Numerator (odd factorials): $$S_{\text{odd}}=\frac{1}{1!}+\frac{1}{3!}+\frac{1}{5!}+\dots$$

• Denominator (even factorials): $$S_{\text{even}}=\frac{1}{2!}+\frac{1}{4!}+\frac{1}{6!}+\dots$$

Our goal is to evaluate the ratio: $$\frac{S_{\text{odd}}}{S_{\text{even}}}=\frac{e+1}{e-1}$$

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Step 2: Recall the Taylor Series for $e^x$

The exponential function $e^x$ has the following Taylor series expansion around $x=0$: $$e^x=\sum _{n=0}^{\infty }\frac{x^n}{n!}=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots$$

For $x=1$ and $x=-1$, we get: $$e=e^1=\sum _{n=0}^{\infty }\frac{1}{n!}=1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\dots$$ $$e^{-1}=e^{-1}=\sum _{n=0}^{\infty }\frac{(-1{)}^n}{n!}=1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\dots$$

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Step 3: Express $S_{\text{odd}}$ and $S_{\text{even}}$ in Terms of $e$

Let us rewrite $e$ and $e^{-1}$ by separating the terms with odd and even factorials:

$$e=\left(1+\frac{1}{1!}+\frac{1}{3!}+\frac{1}{5!}+\dots \right)+\left(\frac{1}{2!}+\frac{1}{4!}+\frac{1}{6!}+\dots \right)=S_{\text{odd}}+S_{\text{even}}$$

$$e^{-1}=\left(1-\frac{1}{1!}+\frac{1}{3!}-\frac{1}{5!}+\dots \right)+\left(\frac{1}{2!}-\frac{1}{4!}+\frac{1}{6!}-\dots \right)$$

Notice that the odd terms in $e^{-1}$ can be written as: $$1-\frac{1}{1!}+\frac{1}{3!}-\frac{1}{5!}+\cdots =1-\left(\frac{1}{1!}-\frac{1}{3!}+\frac{1}{5!}-\dots \right)$$

However, a better approach is to consider the sum and difference of $e$ and $e^{-1}$:

$$e+e^{-1}=2\left(1+\frac{1}{2!}+\frac{1}{4!}+\frac{1}{6!}+\dots \right)=2\left(1+S_{\text{even}}-\frac{1}{2!}\right)$$

But this is not directly helpful. Instead, let us define:

$$S_{\text{odd}}=\sum _{k=0}^{\infty }\frac{1}{(2k+1)!},S_{\text{even}}=\sum _{k=1}^{\infty }\frac{1}{(2k)!}$$

We know that: $$e=S_{\text{odd}}+S_{\text{even}}+1$$ (Note: The $n=0$ term in $e$ is $1$, which is not included in $S_{\text{odd}}$ or $S_{\text{even}}$.)

But this is incorrect. Actually, the correct separation is: $$e=1+\sum _{n=1}^{\infty }\frac{1}{n!}=1+S_{\text{odd}}+S_{\text{even}}$$

Now, consider $e^{-1}$: $$e^{-1}=1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\cdots =1-\left(\frac{1}{1!}-\frac{1}{2!}+\frac{1}{3!}-\frac{1}{4!}+\dots \right)$$

But this can be rewritten as: $$e^{-1}=1-\left(S_{\text{odd}}-S_{\text{even}}\right)$$

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Step 4: Solve for $S_{\text{odd}}$ and $S_{\text{even}}$

From $e=1+S_{\text{odd}}+S_{\text{even}}$ and $e^{-1}=1-(S_{\text{odd}}-S_{\text{even}})$, we have two equations:

1. $e=1+S_{\text{odd}}+S_{\text{even}}$
2. $e^{-1}=1-S_{\text{odd}}+S_{\text{even}}$

Let us solve these two equations simultaneously.

Add the two equations: $$e+e^{-1}=(1+S_{\text{odd}}+S_{\text{even}})+(1-S_{\text{odd}}+S_{\text{even}})=2+2S_{\text{even}}$$ $$\Rightarrow S_{\text{even}}=\frac{e+e^{-1}}{2}-1$$

Subtract the second equation from the first: $$e-e^{-1}=(1+S_{\text{odd}}+S_{\text{even}})-(1-S_{\text{odd}}+S_{\text{even}})=2S_{\text{odd}}$$ $$\Rightarrow S_{\text{odd}}=\frac{e-e^{-1}}{2}$$

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Step 5: Compute the Ratio $\frac{S_{\text{odd}}}{S_{\text{even}}}$

Now, substitute the expressions for $S_{\text{odd}}$ and $S_{\text{even}}$:

$$\frac{S_{\text{odd}}}{S_{\text{even}}}=\frac{\frac{e-e^{-1}}{2}}{\frac{e+e^{-1}}{2}-1}$$

Simplify the denominator: $$\frac{e+e^{-1}}{2}-1=\frac{e+e^{-1}-2}{2}$$

Thus: $$\frac{S_{\text{odd}}}{S_{\text{even}}}=\frac{e-e^{-1}}{e+e^{-1}-2}$$

Multiply numerator and denominator by $e$ to eliminate $e^{-1}$: $$\frac{e(e-e^{-1})}{e(e+e^{-1}-2)}=\frac{e^2-1}{e^2+1-2e}$$

Factor the numerator and denominator: $$\frac{(e-1)(e+1)}{(e-1{)}^2}=\frac{e+1}{e-1}$$

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Final Answer

$$\boxed{\frac{e+1}{e-1}}$$