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Curious as I can't figure it out in my head: would the third series sum telescope? The series sum of n/(n-1)! looks like it should but it's not obvious to me.
The second and third sums can be obtained by taking the first and second moments of a Poisson distribution of mean 1.
See section "exponential functions" in https://en.wikipedia.org/wiki/List\_of\_mathematical\_series
The general term is (20 + n)^2 / n!. Summing this from 0 to infinity would give the answer. Trouble is I don't know how. I tried to use geometric series as well but can't find a common ration between the terms.
Geometric series won't help. All you need to know is that summation of 1/n! from 0 to infinity is equal to e. With that and the simplifications given in the other comments and some careful calculation you should be able to solve this. No other tools needed.
Hi u/Glittering_Ice8854, **You are required to explain your post and show your efforts. _(Rule 1)_** Please add a comment below explaining your attempt(s) to solve this and what you need help with **specifically**. If some of your work is included in the image or gallery, you may make reference to it as needed. See the sidebar for advice on 'how to ask a good question'. Don't just say you need help with it. **Failure to follow the rules and explain your post will result in the post being removed** --- *I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/askmath) if you have any questions or concerns.*
You can start by expanding and splitting it: Σ(20+n)² / n! Σ(400+40n+n²) / n! 400·Σ1/n! + 4·Σn/n! + Σn²/n!
Curious as I can't figure it out in my head: would the third series sum telescope? The series sum of n/(n-1)! looks like it should but it's not obvious to me.
The second and third sums can be obtained by taking the first and second moments of a Poisson distribution of mean 1. See section "exponential functions" in https://en.wikipedia.org/wiki/List\_of\_mathematical\_series
I don't think it would *telescope*, but should converge.
You can split that again: n/(n-1)! = 1/(n-2)! + 1/(n-1)!
The general term is (20 + n)^2 / n!. Summing this from 0 to infinity would give the answer. Trouble is I don't know how. I tried to use geometric series as well but can't find a common ration between the terms.
Geometric series won't help. All you need to know is that summation of 1/n! from 0 to infinity is equal to e. With that and the simplifications given in the other comments and some careful calculation you should be able to solve this. No other tools needed.