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Let $(a_n)_{n=1}^\infty$ be a sequence with $\lim_{n \to \infty} a_n=A$. Show that if $b_n=a_{n+1}-a_n$, then the infinite series $$S=\sum_{n=1}^\infty b_n$$ is convergent. What is $S$ equal to?


Suppose $b_n=a_{n+1}-a_n$. Then $$\lim_{n\to \infty} b_n = \lim_{n\to \infty} (a_{n+1}-a_n)=\lim_{n\to \infty} a_{n+1}-\lim_{n\to \infty} a_n=A-A=0$$ Since $b_n \to 0$, we can consider checking if $S$ is convergent.

Since $b_n=a_{n+1}-a_n$, then \begin{align*} S&=\sum_{n=1}^\infty (a_{n+1}-a_n)\\ &=({a_2}-a_1)+({a_3}-{a_2})+({a_4}-{a_3})+({a_5}-{a_4})+\cdots\\ &=-a_1 \end{align*} So, $S=-a_1$. Thus, $S$ is convergent.

Is this correct?

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    $\begingroup$ Try looking at the sequence of partial sums. If that sequence converges, then $\sum b_n$ converges and equals that limit. $\endgroup$ – Mark Twain Dec 18 '17 at 2:25
  • $\begingroup$ You have to be very careful when pairing terms together in an infinite series the way you did. It isn't always allowed. $\endgroup$ – Mark Twain Dec 18 '17 at 2:26
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It is not equal to $-a_1$. Did you not find weird that $A$ didn't appear in your result? Just note that $$\sum_{n=1}^N b_n = a_{N+1}-a_1\to A-a_1.$$

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  • $\begingroup$ How are you getting this? Even if I take the parenthesis off in the summation, I'm still getting that everything except $-a_1$ cancels out. $\endgroup$ – Al Jebr Dec 18 '17 at 2:29
  • $\begingroup$ $$\sum_{n=1}^N( a_{n+1} - a_n) = \sum_{n=1}^N a_{n+1} - \sum_{n=2}^N a_n - a_1 = a_{N+1}-a_1. $$ $\endgroup$ – Ivo Terek Dec 18 '17 at 2:31
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Let $(s_n)$ be the sequence of partial sums for $\sum b_k$. Then for every $n \geq 1$, \begin{align*}s_n = \sum_{k=1}^{n} b_k &= \sum_{k=1}^{n} (a_{k+1}-a_k) \\&= (a_2-a_1)+(a_3-a_2)+\cdots + (a_{n+1}-a_n)\\[1ex]&= a_{n+1}-a_1.\end{align*} Since $\lim a_{n+1}=\lim a_n = A$, $$\sum_{k=1}^{\infty} b_k = \lim s_n = \lim {}(a_{n+1}-a_1) = \lim a_{n+1}-\lim a_1 = A - a_1.$$ So $\sum b_k$ indeed converges; its sum is $A - a_1$.

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For $1 \leq k < \infty$, let $$S_k = \sum_{n=1}^{k}b_n\text{.}$$ Regardless of the value of $k$, we have $$S_k = b_1+b_2 + \cdots + b_k = a_2 - a_1+a_3-a_2+\cdots+a_{k+1}-a_k=a_{k+1}-a_1\text{.}$$ The sum you desire, $S$, is equal to $S_{\infty} = \lim\limits_{k \to \infty}S_k$. Observe that $a_1$ is constant with respect to $k$. Given that $\lim\limits_{k \to \infty}a_{k+1} = \lim\limits_{k \to \infty}a_{k}=A$, we obtain $$S = S_{\infty} = \lim\limits_{k \to \infty}(a_{k+1}-a_1)=\left(\lim\limits_{k \to \infty}a_{k+1}\right)-a_1=A-a_1\text{.}$$

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A series is convergent if and only if its sequence of partial sums is convergent.

Let $(B_N)$ be the sequence of partial sums and consider, for sake of an example, the partial sum $B_4$: \begin{align*} B_4 &=\sum_{n=1}^4 b_n\\ &=\sum_{n=1}^4 a_{n+1}-a_n\\ &=a_2-a_1+a_3-a_2+a_4-a_3+a_5-a_4\\ &=-a_1+a_5\\ &=a_5-a_1. \end{align*}

Now, if we consider any partial sum $B_N$, we can see that $B_N=a_{N+1}-a_1$.

So, $$\lim_{N\to \infty} B_N=\lim_{N\to \infty} a_{N+1}-a_1=A-a_1.$$

So, the sequence of partial sums $(B_N)$ converges to $A-a_1$ and thus $$S=\sum_{n=1}^\infty b_n=\lim_{N\to \infty} \sum_{n=1}^N b_n=\lim_{N\to \infty}B_N=A-a_1.$$

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