(The ratio test): Let an ≥0. (i) If lim supsumn →∞ an+1 / an<1, show that ∑n=1^∞ an<∞. (ii) If l

(The ratio test): Let an ≥0. (i) If lim supsumn →∞ an+1 /...

(The ratio test): Let $a_n \geq 0$. (i) If $\limsup \operatorname{sum}_{n \rightarrow \infty} a_{n+1} / a_n<1$, show that $\sum_{n=1}^{\infty} a_n<\infty$. (ii) If lim inf ${ }_{n \rightarrow \infty} a_{n+1} / a_n>1$, show that $\sum_{n=1}^{\infty} a_n$ diverges. (iii) Find examples of both a convergent and a divergent series having $\lim _{n \rightarrow \infty} a_{n+1} / a_n=1$.

(The ratio test): Let $a_n \geq 0$.
(i) If $\limsup \operatorname{sum}_{n \rightarrow \infty} a_{n+1} / a_n<1$, show that $\sum_{n=1}^{\infty} a_n<\infty$.
(ii) If lim inf ${ }_{n \rightarrow \infty} a_{n+1} / a_n>1$, show that $\sum_{n=1}^{\infty} a_n$ diverges.
(iii) Find examples of both a convergent and a divergent series having $\lim _{n \rightarrow \infty} a_{n+1} / a_n=1$.

Key Concepts

Convergence and Divergence of Series

Convergence and divergence of series are fundamental concepts in analysis, concerning whether the sum of an infinite sequence of numbers approaches a finite limit (convergence) or not (divergence). Tests like the Ratio Test provide criteria to decide on the behavior of the series based on the comparative growth of successive terms, making them central tools in determining the overall behavior of series.

Borderline Cases in the Ratio Test

Borderline cases occur when the limit of the ratio of consecutive terms equals 1, a scenario where the Ratio Test fails to provide a conclusive answer about a series' convergence or divergence. In such instances, additional methods or tests must be employed to analyze the series, as some series with this limit may converge while others diverge. These examples highlight the limitations of the Ratio Test and the need for a broader set of analytical tools in series convergence analysis.

Ratio Test

The Ratio Test is a method used to determine the convergence or divergence of an infinite series by examining the limit of the ratio of consecutive terms. Specifically, for a series with nonnegative terms, if the limit of a(n+1)/a(n) as n approaches infinity is less than 1, the series converges; if it is greater than 1, the series diverges. This test is especially useful for series where the terms involve factorials, exponentials, or other functions that exhibit multiplicative growth rates.

Limit Superior and Limit Inferior

Limit superior (limsup) and limit inferior (liminf) are concepts that provide bounds on the behavior of sequences that do not necessarily converge. In the context of the Ratio Test, using limsup and liminf allows one to address cases where the limit of a(n+1)/a(n) does not exist in the classical sense. They help in formulating criteria where if limsup of the ratio is less than 1, the series converges, and if liminf is greater than 1, the series diverges.

Transcript

00:02 All right, so to solve this problem, i'm going to start by setting r sub n equal to a series going from k equals n plus 1 to infinity for a .k.

00:21 This is just me rewriting the series.

00:24 And then based on this definition, i can rewrite a subk to be 8 to the n plus 1 times a, n plus 2 over.

00:39 A .n.

00:41 Plus 1 times a .n.

00:45 Plus 3 over a .n.

00:48 Plus 2.

00:51 And so on, becoming ak over ak minus 1.

01:00 And what this allows us to do is rewrite the series in terms of the r series that was given.

01:20 Rn plus 1, rn plus 2, so on, rk minus 1.

01:30 Now we know that rn is a decreasing series, so we can rewrite this whole equation with comparison so that ak is less than and equal to a, n plus 1 times r to the n minus, n plus 1, to the k minus n plus 1 to the k minus n minus n.

01:53 To 1.

01:54 And so that accounts for all the terms.

01:58 So now that we know this, we can rewrite them both as the series in terms of giving r so n.

02:05 That gives us from infinity, starting from k equals n plus 1 to a .k, which is less than or equal to from k equals n plus 1 to infinity for a and plus 1.

02:28 Times r n plus 1 to the k minus n minus 1.

02:37 And then if you actually plug in the values, what you'll see is that this term is the same thing as if we go from some other arbitrary term, we call it j, go to a n plus 1 times rn plus 1 to the j...

Please give Ace some feedback