0

If $S=a_{1}+a_{2}+a_{3}+\ldots+a_{n}+\ldots$, is it true that $$S=a_{1}+0+a_{2}+0+a_{3}+0+\ldots+a_{n}+0+\ldots\ ?$$

I think the second series $S'_{n} \not =S_{n}$, so it is false. But I'm not sure about that because when $n$ becomes larger and larger, it seems to be the same.

Did
  • 279,727
Leilan.Shaw
  • 81

1 Answers1

3

If $S_n=\sum\limits_{t=1}^{n}{a_n}$ converges to $S$.

Let $b_n=a_1+0+a_2+0+\cdots$ upto $n$ terms. Then $$b_n=\left\{\begin{array}{cc}a_{\frac n2}& n \text{ is even} \\ a_{\frac{n+1}{2}} & n \text{ is odd} \end{array}\right.$$

Now observe that $b_{2n}\rightarrow S$ and $b_{2n+1}\rightarrow S$.

Lemma: Let $t_n$ be a sequence, if $t_{2n}\rightarrow t$ and $t_{2n+1}\rightarrow t$, then $t_n \rightarrow t$.

Use the lemma to observe that $b_n\rightarrow S$.

But the question on whether the two sequences are equal can have different answers when we change perspective of equality. The terms are not equal, both converges to same limit etc.

If $S_n$ is not convergent, we can obtain bizarre results like $1+1+1+\cdots =1+(1+-1)+1+(1+-1)+\cdots=2+2+2+\cdots$.

hrkrshnn
  • 6,287
  • I agree that it is completely obvious. But the point seems to be that the new series has twice as many terms, ie $a'{2n-1}=a_n,a'{2n}=0$. So you have to establish that you are free to group terms. If the original series was not absolutely convergent, nailing the proof down for a pedant requires a little work. – almagest Sep 04 '14 at 11:40
  • 1
    But $1+1+1+⋯=2+2+2+⋯$ is not "bizarre", it is obviously true, no? :-) – Did Sep 04 '14 at 12:11
  • @Did actually the two series are divergent – Leilan.Shaw Sep 04 '14 at 14:11
  • @Leilan.Shaw No kidding. :-) – Did Sep 04 '14 at 15:37