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I have the following problem:

Let $a_n$, $n\in\mathbb{N}$ be sequence and let $b_n=a_n+a_{n+1}+a_{n+2}$

Prove that if $\sum_{n=1}^{\infty}a_n$ converges then $\sum_{n=1}^{\infty}b_n$ converges

My attempt:

Let $\sum_{n=1}^{\infty}a_n=a$

$\sum_{n=1}^{\infty}b_n=\sum_{n=1}^{\infty}a_n+a_{n+1}+a_{n+2}$=

$\sum_{n=1}^{\infty}a_n+\sum_{n=1}^{\infty}a_{n+1}+\sum_{n=1}^{\infty}a_{n+2}$=

$\sum_{n=1}^{\infty}a_n$+$\sum_{n=1}^{\infty}a_n-a_1$+$\sum_{n=1}^{\infty}a_n-a_1-a_2$=

$\sum_{n=1}^{\infty}3a_n-2a_1-a_2$

From the linearity of series we know that $\sum_{n=1}^{\infty}3a_n=3a$

And the series convergence isn't affected by a change in finite number of elements of the sum

So $\sum_{n=1}^{\infty}b_n$ converges

Is this any good? Is it sufficient?

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  • 2
    $\begingroup$ Your proof is correct. Here's a more general proof: if $\sum a_n, \sum b_n$ converge, then $\sum a_n + b_n$ converges and equals $\sum a_n + \sum b_n$. Use this and the fact that $\sum\limits_{n=1}^\infty a_{n+k}$ converges for every k if $\sum a_n$ exists. You basically used these statements within your proof. $\endgroup$ – James Yang Jan 22 at 20:32
  • $\begingroup$ Yes your proof is correct $\endgroup$ – Mike Jan 22 at 20:35
  • $\begingroup$ Please put in parentheses. $\endgroup$ – zhw. Jan 22 at 21:04

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