Thread: 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.

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

Now consider that and consequently

So your sum is equal to



So

Thank you so much, that answers it perfectly.