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Suppose I want to prove that a sequence $(s_n)$ converges, and I don't know much more about $(s_n)$ than a few properties. (That is, I don't know a closed-form formula.)

I have seen a proof written by my professor that began with "we're concerned with only the convergence of $(s_n)$, so suppose $s_0 = 0$."

I believe he was implicitly saying, "without loss of generality, suppose $s_0 = 0$." Certainly there are plenty of sequences with the given properties that don't start at $0$, so it seemed to me that this could not be done. Upon considering it further, though, I believe I may have a theory on the overall argument.

If $(a_n)$ and $(b_n)$ are equivalent, then for any $\epsilon > 0$, there exists some $N$ so that for all $n \geq N$, $|a_n - b_n| \leq \epsilon$. If $(a_n)$ is Cauchy, then $(b_n)$ is Cauchy and vice-versa. Hence, if $(a_n)$ converges, then $(b_n)$ converges, and vice-versa. If the first term is $0$, that will neither impact convergence nor equivalence because we can throw out a finite number of terms anyway. Hence, if I prove a result for an equivalent sequence $(a_n)$ that starts at $a_0 = 0$, the result likewise holds for $(b_n)$.

Because of these, we can in effect say, "without loss of generality, suppose $s_0 = 0$" because proving that an equivalent sequence $(s_n)$ that does start at $0$ convergences also implies that $(s_n)$ converges.

Is this correct? More generally, what are the limits to assuming something ``without loss of generality?''

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You've got the right idea, and almost so for the proof. It's more so that, after finitely many terms, if all subsequent terms are the same, then the sequences have a similar behavior. In other words, $(a_n)$ and $(b_n)$ behave similarly if there exists $N_0$ such that for $n>N_0$, we have $a_n=b_n$. A proof might go like this:

Let $\epsilon>0$. If $(a_n)$ is Cauchy, then there exists $N$ such that $m,n>N$ implies $|a_n-a_m|<\epsilon$. Hence for $m,n>\max\{N,N_0\}, a_m=b_m$ and $a_n=b_n$ so $|a_n-a_m|=|b_n-b_m|<\epsilon$. So $(b_n)$ is Cauchy.

Similar proofs exist for convergence and divergence of both sequences.

jl00
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  • This makes sense. Thank you. If I could ask one more rather silly question: am I assuming $s_0 = 0$ without loss of generality or am I proving, directly, that an equivalent sequence converges and hence that the sequence I'm interested in converges? (Or are these equivalent formulations?) –  Jan 15 '20 at 05:09
  • @Matt.P The latter. The former statement is just a lazy formulation of the latter (which is fine, most of the time people will know what you mean). – jl00 Jan 15 '20 at 05:11