For n ∈ℕ, let an:=(1)/(2)+(1)/(4)+⋯+(1)/(2 n) and bn:=(1)/(1)+(1)/(3)+⋯+(1)/(2 n-1) Show t

For n ∈ℕ, let an:=(1)/(2)+(1)/(4)+⋯+(1)/(2 n) and ...

For $n \in \mathbb{N}$, let
$$
a_{n}:=\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2 n} \quad \text { and } \quad b_{n}:=\frac{1}{1}+\frac{1}{3}+\cdots+\frac{1}{2 n-1}
$$
Show that $a_{n} \rightarrow \infty$ and $b_{n} \rightarrow \infty$. (Hint: Example 2.10 (iii).)

Key Concepts

Harmonic Series

The harmonic series, given by the sum of 1/n over all natural numbers, is a classic example in analysis of an infinite series whose terms tend to zero yet the overall sum diverges. Its divergence is often demonstrated by comparison and grouping techniques, and it serves as a foundational example in understanding the behavior of series in real analysis.

Comparison Test

The comparison test is a method for establishing the convergence or divergence of an infinite series by comparing it with another series whose behavior is already known. If the terms of the series under consideration are bounded below by the corresponding terms of a known divergent series, then the series in question must also diverge. This tool is frequently used to analyze modified or subseries of well-known divergent series.

Subseries Behavior

Subseries behavior refers to the analysis of series formed by selecting a subset of terms from an original series. In the context of divergent series, if a subseries retains a structure that can be compared effectively to the original divergent series (or another divergent series), one can conclude that the subseries will also diverge. This concept is crucial when the series is partitioned, such as into even and odd indexed terms.

Transcript

00:01 For this problem, we are asked to show that 1 over 1 times 2 plus 1 over 2 times 3, plus 1 over 3 times 4 plus dot, dot, dot, up to 1 over n times n plus 1, where we're given the hint that 1 over n times n plus 1 equals 1 over n minus 1 over n plus 1.

00:19 So we can express that summation above as the sum from k equals 1 up to n of 1 over k times k plus 1, which is a sum.

00:33 In turn using the hint would be equal to the sum from k equals 1 up to n of 1 over k minus 1 over k plus 1.

00:45 So if we look at expanding this out as a telescoping series, what we can see is that we'd have, and i'll be, no, i won't be tricky about this actually, we'll have out front just 1 over 1, then we'd have, i'll put it into brackets, i'll put plus minus 1 over 2 plus 1 plus 1 over over 2, then we'd have plus minus 1 over 3 plus 1 over 3, where we are essentially splitting up terms here.

01:16 The 1 plus, or 1 plus negative 1 half would be from k equals 1, and then this set of pairs would be from k equals 2 and so on.

01:24 So we'd see that up to n, we'd have plus dot, dot, dot, actually i'll go up to n minus 1.

01:32 We'd have 1 over n minus 1.

01:34 Or actually, we'd have 1 over n minus 1.

01:37 Or actually, we we would have minus 1 over n minus 1.

01:45 That would be from the n minus 2, or k equals n minus 2 term.

01:50 Then we would have plus 1 over n minus 1.

01:54 And then lastly, we would have a minus or a plus.

02:00 Actually, yeah, we can just write it simply as a minus 1 over...