# Proving $(a_1-a_2)+(a_2-a_3) +...$ converges iff ${a}_n$ converges

Hi there I am having trouble with a problem for whatever reason I get stuck. The question goes as following

Prove that the series $(a_1-a_2)+(a_2-a_3) +...$ converges iff the sequence ${a}_n$ converges.

For the forward direction I noted the the series can be rewritten as $∑_{n=1}^{∞}a_n-a_{n+1}$. Since this series converges then I can say $lim _{n->∞}a_n-a_{n+1}=0$. From here, can I conclude the the $lim _{n->∞}a_n$ has to converge to some value L? Or am I missing something? For the backward direction I have zero idea how to start, but i am thinking either using Cauchy sequences. But am not sure. But if anyone can help i would be very grateful.

• Hint: the partial sums of your series are $s_n=a_1-a_{n+1}$.
– lulu
Oct 15 '16 at 19:27

Denote $\;b_n:=a_n-a_{n+1}\;$ , then

$$\sum_{n=1}^\infty b_n\;\;\text{converges}\implies\;\text{the sequence of partial sums of the series converges finitely}\implies$$

$$B_N:=\sum_{n=1}^Nb_n=b_1+b_2+\ldots+b_N=(a_1-a_2)+\ldots+(a_N-a_{N+1})\xrightarrow[N\to\infty]{}B<\infty$$

But, as with any finite sum, we have associativity, thus

$$B_N=a_1-(a_2+a_2)-(a_3+a_3)-\ldots-(a_N-a_N)+a_{N+1}=a_1-a_{N+1}\implies$$

$$B=\lim_{N\to\infty}B_n=\lim_{N\to\infty}(a_1-a_{N+1})\implies\lim_{N\to\infty}a_{N+1}=a_1-B$$

and this proves the sequence has a limit.

But observe we have up there an expression for the sequence of partial sums of the series, so if the limit of $\;a_n\;$ , or $\;a_{n+1}\;$ , which is the same, exists, so that seq. of partials sums also converges...and we're done