# How to prove a telescoping series converges ???

Let ${a_k}$ be a real sequence, such that $a_k \rightarrow 3,$as k$$$\rightarrow \infty . Prove that \sum_{k=1}^\infty (a_k- a_{k+3}) = a_1 + a_2 +a_3 - 9. I know this is a telescoping series and I can split the summation into: \sum_{k=1}^\infty a_k - \sum_{k=1}^\infty a_{k+3}, so I think \sum_{k=1}^\infty a_k = a_1 + a_2 + a_3 + a_4 + a_5 + a_6 +...+ a_k and \sum_{k=1}^\infty a_{k+3} = \sum_{k=4}^\infty a_k = a_4 + a_5 + a_6 +...+ a_k So \sum_{k=1}^\infty a_k - \sum_{k=1}^\infty a_{k+3} would clearly cancel all terms \geq a_4 and leave a_1 + a_2 + a_3, but I don't know where the -9 would come from. I know it has something to do with the fact that a_k \rightarrow 3, as k$$\rightarrow \infty$, but I do not know how this comes into play.

Am I approaching this correctly, or should I be thinking about it differently?

Start with $$\sum_{k=1}^n (a_k-a_{k+3})=a_1+a_2+a_3-a_{n+1}-a_{n+2}-a_{n+3}$$ and take the limit as $n$ goes to infinity.

• so my error was where I interpretted $\sum_{k=1}^\infty a_{k+3}$ = $\sum_{k=4}^\infty a_k$ = $a_4 + a_5 + a_6 +...+ a_k$, instead of $\sum_{k=1}^\infty a_{k+3}$ = $\sum_{k=4}^\infty a_k$ = $a_4 + a_5 + a_6 +...+ a_k + a_{k+1} + a_{k+2} + a_{k+3}$ ? I now can see how $$\sum_{k=1}^n (a_k-a_{k+3})=a_1+a_2+a_3-a_{n+1}-a_{n+2}-a_{n+3}$$. Do you think it is sufficient to say that $\lim_{n \rightarrow \infty} a_{n+1}-a_{n+2}-a_{n+3}$ = -3 -3 -3 = -9, or should I show the summations more clearly? – Jaime Apr 23 '13 at 5:33
• @Jaime As by definition $\sum_{k=1}^\infty (a_k-a_{k+3})=\lim_{n\to\infty}\sum_{k=1}^n (a_k-a_{k+3})$ there is little room to be more clearly. – Hagen von Eitzen Apr 23 '13 at 6:11