# Does an alternating sequence diverge when its magnitude does not converge to 0?

Given $$a_n = b_n * (-1)^n$$, if $$b_n$$ converges to $$L$$ (which is not 0) then $$a_n$$ does not converge(diverges).

I was learning about alternating series by myself, and I came up with this statement which would be very helpful if is true.

Intuitively, it seems to be true, but I don't know how to prove it clearly.

I came up with an idea "if n is large enough, $$b_n$$ would be very close to $$L$$ and $$a_n$$ would be alternating between (some value) and (some value + $$L$$) so $$a_n$$ does not converge", but this explanation seems to be too vague and not logical.

## 2 Answers

You are correct. To make it rigorous:

If $$b_n \to L \ne 0$$, take $$\varepsilon$$ such that $$|L| > \varepsilon > 0$$. Then for sufficiently large even $$n$$ we have $$|a_n - L| < \varepsilon$$ and for sufficiently large odd $$n$$, $$|a_n + L| < \varepsilon$$.
If $$\lim_{n \to \infty} a_n = R$$ existed, we'd need both $$|R - L| \le \varepsilon$$ and $$|R+L| \le \varepsilon$$, and this is impossible because the intervals $$[L-\varepsilon, L+\varepsilon]$$ and $$[-L-\varepsilon, -L+\varepsilon]$$ are disjoint.

• Thank you for your answer! I just figured out if the sign of b_n is constant, then since "If lim a_n = R then lim abs(a_n) = R" so we get the statement "if lim b_n =/= 0 then lim a_n =/= 0". I wonder if this works too when the sign of b_n is not constant. – jkuk5046 Feb 5 at 17:09

Without loss of generality, I'll assume $$L>0$$.

Just note that, given any $$\varepsilon>0$$, there exists $$n_0(\varepsilon)$$ such that for any $$n>n_0$$ we have:

$$|b_n-L|<\varepsilon$$

$$b_n\in [L-\varepsilon,L+\varepsilon]$$

Now for odd $$n$$, $$a_n=-b_n\in [-L-\varepsilon,-L+\varepsilon]$$ While for even $$n$$, $$a_n=b_n\in [L-\varepsilon,L+\varepsilon]$$

Then if I were to assume that there exists $$C$$ such that for any $$n>n_1$$:

$$|a_n-C|<\varepsilon_1$$

Thus $$C\in[-L-\varepsilon,-L+\varepsilon]$$ and $$C\in[L-\varepsilon,L+\varepsilon]$$

If I assume $$\varepsilon = L/2$$, then: Thus $$C\in[-3L/2,-L/2]$$ and $$C\in[L/2,3L/2]$$

Since these intervals are disjoint, we've reached a contradiction, so there can be no such $$C$$.