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?

Please & thank you!!


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.

  • $\begingroup$ 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? $\endgroup$
    – Jaime
    Apr 23 '13 at 5:33
  • $\begingroup$ @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. $\endgroup$ Apr 23 '13 at 6:11

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