Is $\sum_{n=1}^\infty \frac{n^2+1}{n!2^n}$ convergent? If yes, evaluate it.
Using the ratio test, I could show that the series is convergent. What is an easy way to evaluate it? I was thinking about estimating it from below and above by series with equal value but coulnd't find any familar series that I could use for this.
On
First of all, you can separate the sum in 2 convergent series :
$$ \sum_{n=1}^\infty \frac{1}{2^n n!} + \sum_{n=1}^\infty \frac{n^2}{2^n n!} $$
The first term is equal to $e^{1/2}-1$ (see exponential as a sum) and we can write the second as :
$$ \sum_{n=1}^\infty \frac{n^2}{2^n n!} = \sum_{n=1}^\infty \frac{n}{2^n (n-1)!} = \frac{1}{2} \sum_{n=1}^\infty \frac{n-1 + 1}{2^{n-1} (n-1)!}$$
As previously, we can separate and apply the same method :
$$ \sum_{n=1}^\infty \frac{n^2}{2^n n!} = \frac{1}{2}(e^{1/2} + \frac{1}{2} e^{1/2})$$
All together, we have :
$$ \sum_{n=1}^\infty \frac{n^2 + 1}{2^n n!} = e^{1/2} - 1 + \frac{1}{2}(e^{1/2} + \frac{1}{2} e^{1/2}) = \frac{7}{4}e^{1/2} - 1$$
Working it out in as elementary a way as possible.
$\begin{array}\\ \sum_{n=1}^\infty \frac{n^2+1}{n!2^n} &=\sum_{n=1}^\infty \frac{n^2+1}{n!2^n}\\ &=\sum_{n=1}^\infty \frac{n^2}{n!2^n} +\sum_{n=1}^\infty \frac{1}{n!2^n}\\ &=\sum_{n=1}^\infty \frac{n}{(n-1)!2^n} +\sum_{n=0}^\infty \frac{(1/2)^n}{n!}-1\\ &=\sum_{n=0}^\infty \frac{n+1}{n!2^{n+1}} +e^{1/2}-1\\ &=\sum_{n=0}^\infty \frac{n}{n!2^{n+1}}+\sum_{n=0}^\infty \frac{1}{n!2^{n+1}} +e^{1/2}-1\\ &=\sum_{n=1}^\infty \frac{n}{n!2^{n+1}}+\frac12\sum_{n=0}^\infty \frac{1}{n!2^{n}} +e^{1/2}-1\\ &=\sum_{n=1}^\infty \frac{1}{(n-1)!2^{n+1}}+\frac12e^{1/2} +e^{1/2}-1\\ &=\sum_{n=0}^\infty \frac{1}{n!2^{n+2}}+\frac32e^{1/2} -1\\ &=\frac14\sum_{n=0}^\infty \frac{1}{n!2^{n}}+\frac32e^{1/2} -1\\ &=\frac14 e^{1/2}+\frac32e^{1/2} -1\\ &=\frac74 e^{1/2} -1\\ \end{array} $