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

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