Series (Seniors)

Real Analysis
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Tolaso J Kos
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Series (Seniors)

#1

Post by Tolaso J Kos »

Use \( f(x)=\left ( x^2+x \right )e^x\) to evaluate the sum \( \displaystyle \sum_{n=1}^{\infty}\frac{n^2}{n!} \).
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Papapetros Vaggelis
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Re: Series (Seniors)

#2

Post by Papapetros Vaggelis »

The domain of the function \(\displaystyle{f}\) is the set \(\displaystyle{D\,(f)=\mathbb{R}}\) .

The function \(\displaystyle{f}\) is countless times continuously differentiable at every point of it's domain with :

\(\displaystyle{f^\prime(x)=\left(2\,x+1\right)\,e^{x}+\left(x^2+x\right)\,e^{x}=\left(x^2+3\,x+1\right)\,e^{x}\,,x\in\mathbb{R}}\)

\(\displaystyle{f^{\prime \prime}(x)=\left(2\,x+3\right)\,e^{x}+\left(x^2+3\,x+1\right)\,e^{x}=\left(x^2+5\,x+4\right)\,e^{x}\,,x\in\mathbb{R}}\)

\(\displaystyle{f^{(3)}(x)=\left(2\,x+5\right)\,e^{x}+\left(x^2+5\,x+4\right)\,e^{x}=\left(x^2+7\,x+9\right)\,e^{x}\,,x\in\mathbb{R}}\) .

Now, we guess that

\(\displaystyle{f^{(n)}(x)=\left[x^2+\left(2\,n+1\right)\,x+n^2\right]\,e^{x}\,,n\in\mathbb{N}\,,x\in\mathbb{R}}\) .

We give a proof for the above relationship inductively.

For \(\displaystyle{n=1\,,2\,,3}\) it's ok.

Let \(\displaystyle{n\in\mathbb{N}}\) such that \(\displaystyle{f^{(n)}(x)=\left[x^2+\left(2\,n+1\right)\,x+n^2\right]\,e^{x}\,,x\in\mathbb{R}}\) .

For each \(\displaystyle{x\in\mathbb{R}}\) holds :

\(\displaystyle{f^{(n+1)}(x)=\left(2\,x+2\,n+1\right)\,e^{x}+\left[x^2+(2\,n+1)\,x+n^2\right]\,e^{x}=\left[x^2+\left(2\,(n+1)+1\right)+\left(n+1\right)^2\right]\,e^{x}}\)

which is the desired result.

So, \(\displaystyle{f^{(n)}(x)=\left[x^2+\left(2\,n+1\right)\,x+1\right]\,e^{x}\,,n\in\mathbb{N}\,,x\in\mathbb{R}}\) .

We have that \(\displaystyle{f(x)=f(0)+\sum_{n=1}^{\infty}\dfrac{x^{n}\,f^{(n)}(0)}{n!}=\sum_{n=1}^{\infty}\dfrac{x^{n}\,n^2}{n!}}\) and if

\(\displaystyle{a_{x}(n)=\dfrac{x^{n}\,n^2}{n!}}\) , then \(\displaystyle{\dfrac{a_{x}(n+1)}{a_{x}(n)}=\dfrac{x\,n^2}{(n+1)\,n^2}\longrightarrow 0<1}\) .

Therefore,

\(\displaystyle{\sum_{n=1}^{\infty}\dfrac{n^2}{n!}=\sum_{n=1}^{\infty}\dfrac{1^{n}\,n^2}{n!}=f(1)=2\,e}\) .
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