# Math Help - Converging series with factorials

1. ## Converging series with factorials

It's been a while since calculus and this problem actually is part of probability theory question. I've got the answer down into a series.

∑(1.5^n)/[n*(n-2!)] = ∑[(1.5^n)*(n-1)]/(n!)
both summations from n=2 to infinity. (I proved that both sides are equal, just didn't know which would be easier to solve)

I can see that the denominator will increase much faster than the numerator, leading me to believe that this series will converge. I also verified it by the ratio test and I also computed it in Excel. Without having Excel or a calculator, how can you solve this series by hand?

From Excel, the series converges to 3.24084453516903.

2. Originally Posted by DannyG
It's been a while since calculus and this problem actually is part of probability theory question. I've got the answer down into a series.

∑(1.5^n)/[n*(n-2!)] = ∑[(1.5^n)*(n-1)]/(n!)
both summations from n=2 to infinity. (I proved that both sides are equal, just didn't know which would be easier to solve)

I can see that the denominator will increase much faster than the numerator, leading me to believe that this series will converge. I also verified it by the ratio test and I also computed it in Excel. Without having Excel or a calculator, how can you solve this series by hand?

From Excel, the series converges to 3.24084453516903.
Let us generalize $\sum_{n=2}^{\infty}\frac{x^n}{n(n-2)!}=\sum_{n=2}^{\infty}\frac{(n-1)x^n}{n!}=\sum_{n=2}^{\infty}\frac{nx^n}{n!}-\sum_{n=2}^{\infty}\frac{x^n}{n!}=\sum_{n=0}^{\inf ty}\frac{nx^n}{n!}-x-\left\{\sum_{n=0}^{\infty}\frac{x^n}{n!}-1-x\right\}$

Now consider that $e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}$ and consequently $x\cdot e^x=\sum_{n=0}^{\infty}\frac{nx^n}{n!}$

So your sum is equal to

$\sum_{n=2}^{\infty}\frac{x^n}{n(n-2)!}=xe^x-x-\left(e^x-1-x\right)=xe^x-e^x+1$

So $\sum_{n=2}^{\infty}\frac{\left(\frac{3}{2}\right)^ n}{n(n-2)!}=\frac{3}{2}e^{\frac{3}{2}}-e^{\frac{3}{2}}+1\approx 3.24408$

3. Thank you so much, that answers it perfectly.