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Proof of convergence of a sequence?

I have just attempted this question:

Prove that if a sequence an\displaystyle a_n converges then
an+1an0\displaystyle a_{n+1}-a_n \rightarrow 0

My attempt was to say: we know that if anl\displaystyle a_n \rightarrow l then all sub-sequences of an\displaystyle a_n also converge to l\displaystyle l so by the algebra of limits for sums we have an+1an11=0\displaystyle a_{n+1}-a_n \rightarrow 1-1=0.

I don't know if this is even the right approach let alone rigorous enough, (I haven't proved that if a sequence converges then all sub-sequences converge to the same number, I don't even know how to prove that anyway), any help?
Original post by poorform
I have just attempted this question:

Prove that if a sequence an\displaystyle a_n converges then
an+1an0\displaystyle a_{n+1}-a_n \rightarrow 0

My attempt was to say: we know that if anl\displaystyle a_n \rightarrow l then all sub-sequences of an\displaystyle a_n also converge to l\displaystyle l so by the algebra of limits for sums we have an+1an11=0\displaystyle a_{n+1}-a_n \rightarrow 1-1=0.

I don't know if this is even the right approach let alone rigorous enough, (I haven't proved that if a sequence converges then all sub-sequences converge to the same number, I don't even know how to prove that anyway), any help?


If you're using a result you haven't proved, and don't know how to, then it's unlikely to be what they're looking for.

I'd think they'd be looking for a proof from basics.

I'd start from the definition of l being the limit.

ϵ>0,N0 s.t. n>N0,anl<ϵ\forall \epsilon >0, \exists N_0\text{ s.t. }\forall n>N_0, | a_n-l|<\epsilon

Note that if n > N_0, not only is anl<ϵ|a_n-l|< \epsilon but so is an+1l<ϵ|a_{n+1}-l|< \epsilon.

Then, look at what it means for 0 to be the limit of an+1ana_{n+1}-a_n and see how you can link the two.
Original post by poorform
I have just attempted this question:

Prove that if a sequence an\displaystyle a_n converges then
an+1an0\displaystyle a_{n+1}-a_n \rightarrow 0

My attempt was to say: we know that if anl\displaystyle a_n \rightarrow l then all sub-sequences of an\displaystyle a_n also converge to l\displaystyle l so by the algebra of limits for sums we have an+1an11=0\displaystyle a_{n+1}-a_n \rightarrow 1-1=0.

I don't know if this is even the right approach let alone rigorous enough, (I haven't proved that if a sequence converges then all sub-sequences converge to the same number, I don't even know how to prove that anyway), any help?

Subsequences converge to the same limit: it just falls straight out of the definition of the main sequence tending to a limit. Let ϵ>0\epsilon > 0; then there is N such that for all nNn \geq N, anl<ϵ|a_n - l| < \epsilon. Hence for all njN,anjl<ϵn_j \geq N, |a_{n_j}-l| < \epsilon. ϵ\epsilon was arbitrary, so anjla_{n_j} \to l.

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