0
$\begingroup$

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!!

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Hagen von Eitzen Apr 23 '13 at 6:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.