Mathematics Asked on November 29, 2020

I have two divergent series when summed up to infinite $n$: $frac{1}{n}$ and $frac{2^n – 2}{2^n}$. However, in the limit of increasingly large $n$, the $n$-th term of $frac{1}{n}$ is $0$ and the $n$-th term of the second series goes to $1$ in the limit.

Basically, my question is although both are divergent as infinite series, what is the speed with which the $n$-th term approaches $0$ in the first case and what is the speed with which the $n$-th term approaches $1$ in the second case? My guess is that in the first case the speed of decay is slower than exponential whereas the speed of the growth of the second case is slower than logarithmic, but what is the formal analysis way to express this? Can it be shown that one decays slower than the other grows?

Note that the second one can be expressed as $1-frac{1}{2^{n-1}}$. This means that the second approaches $1$ faster than the first approaches $0$, because $2^{n-1}$ grows faster than $n$.

Correct answer by Joshua Wang on November 29, 2020

Get help from others!

Recent Answers

- Peter Machado on Why fry rice before boiling?
- haakon.io on Why fry rice before boiling?
- Lex on Does Google Analytics track 404 page responses as valid page views?
- Jon Church on Why fry rice before boiling?
- Joshua Engel on Why fry rice before boiling?

Recent Questions

- How can I transform graph image into a tikzpicture LaTeX code?
- How Do I Get The Ifruit App Off Of Gta 5 / Grand Theft Auto 5
- Iv’e designed a space elevator using a series of lasers. do you know anybody i could submit the designs too that could manufacture the concept and put it to use
- Need help finding a book. Female OP protagonist, magic
- Why is the WWF pending games (“Your turn”) area replaced w/ a column of “Bonus & Reward”gift boxes?

© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP