if the sequence=$a_1, a_2, a_3$ has no limit, does that mean that the sequence $a_1 +\frac{a_2}{2} + \frac{a_3}{3}$ also doesn't have a limit?

Question

I concluded that it also doesnt have a limit. I used IZREK 1 (o primerjanju). $I∞k=1 ak = a1 +a2 +... and ∞k=1 bk = b1 +b2 +...,$ where this is true for every $k —> 0≤ak ≤bk$ If the bigger function converges that means that the smaller function also diverges?

bk is the first sequence, and ak the second sequence.

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score 3 · Answer 1 · answered Sep 13 '19 at 11:08

The conclusion does not hold. The sequence $a_n = (-1)^n$ has no limit, but $b_n = \sum_{k=1}^n\frac{a_k}{k}$ has a limit because the series $$ \sum_{k=1}^\infty \frac{a_k}{k} = \sum_{k=1}^\infty \frac{(-1)^k}{k} $$ converges due to the alternating series test. (And the value of this series happens to be $-\ln 2$.)

score 1 · Answer 2 · answered Sep 13 '19 at 11:02

1

The statement is false. We may take $$\{a_1\}=\{1,2,3,4\dots\}\\ \{a_2\}=\{2,4,6,8\dots\}\\ \{a_3\}=\{-6,-12,-18,-24\dots\}$$ None of these sequences has a (finite) limit, but $\{a_1+a_2/2+a_3/3\}$ is zero at every term, and thus has a limit of $0$.

answered Sep 13 '19 at 11:02

Parcly Taxel

103,344

but what if we take a_1=1 a_2=2 .... – sasaako Sep 13 '19 at 11:12
@sasaako Each of $a_1$, $a_2$, $a_3$ is a separate sequence. Your question - and the question body - make no sense. – Parcly Taxel Sep 13 '19 at 11:12
doesnt that make the function without a limit? because a_1=1 a_2=1 ... – sasaako Sep 13 '19 at 11:13
@sasaako See my previous comment. – Parcly Taxel Sep 13 '19 at 11:13

if the sequence=$a_1, a_2, a_3$ has no limit, does that mean that the sequence $a_1 +\frac{a_2}{2} + \frac{a_3}{3}$ also doesn't have a limit?

2 Answers2