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$$\lim_{n\to \infty} S_n=\lim_{n\to \infty}\sum_{k=1}^n\frac{1}{n!}$$

I need to show that this series is a convergent series. How do I show this series to be convergent? My book says that this is convergent. Please provide sone hint.

The Learner
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  • I know that {$\frac{1}{n}$} when n tends to $\infty$ is 0 by epsilon definition of congruence. – The Learner Nov 23 '19 at 12:42
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    ratio test${}$? – Angina Seng Nov 23 '19 at 12:44
  • I dont know much about this as I am new but I think you are telling me to take the the ratio $\frac{a_n+1}{a_n}$ , that is, ratio of two successive terms I hope I am correct @Lord Shark The Unknown. – The Learner Nov 23 '19 at 12:46
  • @Martin R I want the ratio test , my question is different than the one you are saying – The Learner Nov 23 '19 at 13:25
  • @TheLearner: Your question does not say that you want the ratio test. And if you want the ratio test, why do you accept an answer using the direct comparison test (which identical to the answer given here https://math.stackexchange.com/a/2955458/42969)? – Martin R Nov 23 '19 at 13:28
  • Why do you want to close this question also? Are you having sone personal offense with me @Martin R? My last question also you closed – The Learner Nov 23 '19 at 13:35
  • @TheLearner: If a question has been asked and answered before then it can be closed as a duplicate. That has nothing to do with you personally. See also https://math.stackexchange.com/help/duplicates: “The fundamental goal of closing duplicate questions is to help people find the right answer by getting all of those answers in one place.” – Martin R Nov 23 '19 at 13:38
  • But I will lose the reputation I earned on this site @Martin R. Please consider that. – The Learner Nov 23 '19 at 13:41
  • You don't lose reputation if your question is closed as a duplicate. And please note that I can only vote to close, it takes 5 people to close a question, unless (as in your last question) you confirm that it is a duplicate. – Martin R Nov 23 '19 at 13:45
  • Okay sir no offense – The Learner Nov 23 '19 at 14:04

1 Answers1

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By direct comparison test since eventually, notably for $n\ge 4$, $n! \ge n^2$

$$\sum_{k=4}^n\frac{1}{n!}\le \sum_{k=4}^n\frac{1}{n^2}$$

and therefore the series converges by monotone convergence theorem.

user
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  • Thanks again @user – The Learner Nov 23 '19 at 12:49
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    @TheLearner You are welcome. Direct comparison test for positive series is one of the first method you need to learn. As suggested also ratio and root test are very important you should also learn to use them. – user Nov 23 '19 at 12:50
  • I want to ask u one thing. I am a first year college student. How much time it would take for me to answer the questions posted on this site. The topics on which these questions are posted are not taught to me till date. Please tell me. I am highly interested in Math but I am not able to answer any question posted here. I feel very bad. – The Learner Nov 23 '19 at 12:52
  • I also want to have a decent reputation man. – The Learner Nov 23 '19 at 12:53
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    @TheLearner Don't feel bad for that. You can read the answers and learn a lot from other users. Also note that often also basic question are posed, eg. on algebra and precalculus, then feel free to try to answer and help other users. – user Nov 23 '19 at 12:58