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A. $|a_n-a_{n+1}|->0$ as $n->∞$
B. $\sum_{n=1}^∞ |a_n-a_{n+1}| $ is convergent
C. $\sum_{n=1}^∞ n a_n$ is convergent
D. The sequences $(a_{2n+1}),(a_{3n})$ are convergent

1st can be proven wrong by taking $a_n={\sqrt n}$.
For 2nd,I think that convergence of an infinite series $\sum_{n=1}^∞ a_n$ implies that $lim (a_n)=0$. Does this argument ensure that option B and C are correct? For part c, since series is convergent, implies that $lim (na_n)=0$. But this does not imply the convergence of Sequence $(a_n)$.
D is correct if the sub-sequences converges to same limit.

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    For D think about a sequence for which the subsequence $a_{6n+2}$ diverges but converges otherwise then both $a_{2n+1},a_{3n}$ are convergent.Also as you said for D if $\lim a_{2n+1}\not=\lim a_{3n}$. – 2017-01-24
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    Your example for 1) is correct $\sum a_n$ does imply $a_n \rightarrow 0$ so B) implies condition A) which as you pointed out does not imply convergence. At least not with that argument alone. But there is another argument that will show B) is sufficient.. D) is incorrect. The subsequences can converge to same limit but they could also not. And even if they do the subsequence $a_{6n+4}$ need not. – 2017-01-24
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    What about part C? – 2017-01-24
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    What about part C? – 2017-01-24
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    If $n a_n \to 0$, it is immediate that $a_n \to 0$ too. – 2017-01-24
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    Observe that in part B. the sequence $(a_n)$ converges to $$L = a_1 + \sum_{n=1}^\infty (a_{n+1} - a_n).$$ – 2017-01-24
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    but if i take Abel's series, it diverges even when $n(a_n)->0$ – 2017-01-24
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    but $na_n \not \rightarrow 0$. – 2017-01-24
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    what if $na_n = -(n-1)a_{n-1}$. Then $\sum na_n $ converges. Does it follow if $na_n = -(n-1)a_{n-1}$ then $a_n$ converges. Hint: the answer is no. – 2017-01-24
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    Why would $\sum n a_n$ converge? – 2017-01-24
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    If the subsequences $a_{2n+1}$ and $a_{3n}$ both converge, then $a_{6n+3}$ converges to the same value as each of them, so the two have to converge to the same value. – 2017-01-24
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    ... whic is irrelevent as the subsequence $a_{6n+4}$ can do whatever the heck it likes. – 2017-01-24
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    He (?) is addressing the comment that "the subsequences can converge to the same limit but they could also not". – 2017-01-24
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    Already, that was an oversight. But the rest is still valid. The even non-3 multiples can behave any way at all. – 2017-01-24
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    @UmbertoP. "Why would ∑nan converge?" was that addressed to me? Because if $na_n = - (n-1)a_{n-1}$ then $(n-1)a_{n-1} + n a_n = 0$ so the sum is 0. – 2017-01-24
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    @fleablood Are you claiming that the series $\sum (-1)^n$ is convergent? – 2017-01-24
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    No, I was claiming a different series converged for the same reason. Then I thought about and realized I was wrong. – 2017-01-24

1 Answers 1

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A) Your example shows this is not enough.

B) Your argument shows that $|a_n - a_{n+1}| \rightarrow 0$ and that is not enough. But instead $\sum |a_n - a_{n+1}|$ converges means for any $\epsilon$ there is $\forall \epsilon > 0;\exists M| \epsilon >\sum_{n = M}^{\infty} |a_n - a_{n+ 1}| \ge |a_k - a_j|;\forall j,k > M$ so this is a Cauchy sequence and converges.

C) for $\sum b_n$ to converge $b_n \rightarrow 0$. Which means $|b_n| \rightarrow 0$. So $|n*a_n| \rightarrow 0$. For $n>1$ $|n*a_n| > |a_n| \ge 0$ so so $0 = \lim |n*a_n| > \lim a_n > 0$. So $a_n \rightarrow 0$.

D) Although the sequences $a_{2n+1}$ and $a_{3n}$ converge the sequence of even terms that are not a multiple of $3$ (i.e. $a_{6n + 2}$ and $a_{6n+4}$) need not converge at all. They have absolutely no constraints.

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    In fact for D) a simple example is $a_n = 0$ if $n$ is odd or a multiple of $3$, and $1$ otherwise. – 2017-01-24
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    Exactly. That was the purpose of my $\{a_{m| m\ne 2n+1; m \ne 3n}\}$ was attempting to convey-- the subsequence of all terms whose indices are *not* 2n+1 or 3n. They can do... anything. – 2017-01-24
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    For D), because the subsequences overlap at $a_{6n+3}$, if they converge, they have to converge to the same value. – 2017-01-24
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    @Arthur its true I ovelooked that but that is irrelevent as all the terms that are neither a mulitple of 3 nor odd can do whatever they like. – 2017-01-24
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    Perhaps you could correct C). – 2017-01-24
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    Yeah, I don't like to run away and hide from stupid mistakes. Even ones as stupid as this. But yeah I'll fix it. – 2017-01-24